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GMAT Quantitative Foundations Prep Course

Welcome to the Foundations Course! This course is designed to provide you with a strong foundation in the core concepts and skills necessary for success in your academic and professional endeavors. The course will cover a wide range of topics, including critical thinking, problem-solving, communication, and teamwork.

Here's why BobPrep's course is your ultimate weapon for taming the GMAT Quant beast:

★ By the end of this course, you will be able to:

Think critically and creatively:

Analyze information from multiple perspectives, evaluate arguments, and generate innovative solutions.

Solve problems effectively:

Identify and define problems, develop and implement The GMAT is a standardized test designed to assess the quantitative and analytical writing skills of business professionals seeking admission to executive MBA programs. The Quantitative Reasoning section of the GMAT tests your ability to solve basic math problems involving algebra, geometry, and data analysis.

If you're feeling rusty on your math skills, don't worry! BobPrep's GMAT GMAT Quantitative Foundations Prep Course is designed to help you brush up on the essential math concepts you need to know for the GMAT.

★ With BobPrep's GMAT Quantitative Foundations Prep Course, you can:

1. Improve your math skills and gain the confidence you need to ace the EA Quantitative Reasoning section.

2. Increase your chances of getting accepted to your top choice executive MBA program.

3. Invest in your future and open up new career opportunities.

★ Why choose BobPrep:

Expert instructors:
Our instructors are GMAT experts who have years of experience teaching and helping students succeed on the GMAT.

Comprehensive curriculum:
Our course covers all the essential math topics tested on the GMAT.

Personalized learning:
Our adaptive learning platform tailors the course content to your individual needs.

Flexible learning:
You can access our course materials anytime, anywhere, on any device.

Guaranteed results:
We are so confident in our course that we offer a money-back guarantee.

★ What you'll learn:

Algebra:
Review basic concepts like fractions, decimals, percentages, exponents, and roots. Learn how to solve linear equations, inequalities, and systems of equations.

Geometry:
Brush up on your knowledge of geometric shapes, angles, and area and volume formulas.

Data analysis:
Learn how to interpret tables, charts, and graphs, and how to calculate basic statistics like mean, median, and mode.

Problem-solving strategies:
Develop effective strategies for tackling quantitative reasoning problems on the GMAT.

Test-taking tips:
Learn how to manage your time effectively and avoid common mistakes on the GMAT.

★ Benefits of taking BobPrep's GMAT Quantitative Foundations Prep Course:

Improved problem-solving skills:

The skills you learn in our course will not only help you on the GMAT, but they will also be valuable in your business career.

Enhanced critical thinking skills:

Our course will help you develop the critical thinking skills you need to analyze complex information and make sound decisions.

Increased confidence:

By taking our course, you will gain the confidence you need to succeed on the GMAT and achieve your goals.

★  Our Insights:

We believe that everyone has the potential to succeed on the GMAT. With the right preparation, you can achieve your score goals and get into the executive MBA program of your dreams.

What is BobPrep?


Unsure how to get begin your GMAT journey? Hesitant about spending hundreds if not thousands of dollars on GMAT prep and private tutoring? Is there a less expensive, but equally effective way?


You’re not alone, and these are the questions you should be asking. Too long has the GMAT Prep industry equated higher-prices with better quality. Fortunately, the days of sky-high prep costs are gone and access to the methods taught by the world’s best GMAT tutors is now available to all.


Foundations:


Foundations is for students who need to review the mechanics of the GMAT quant and verbal sections. This course focuses on how to solve the math behind the quant section versus our more advanced offerings, which focus more on strategies and logical reasoning. This course is ideal for students who need a refresher on GMAT math and verbal sections and should be used as a building block to move on to our more advanced materials. 


Foundations is for students looking to learn the core concepts needed for the GMAT quant and verbal sections. This course is the perfect building block for students who want to get the most out of our advanced materials later on. Used alone,  Foundations  can get you a GMAT score of up to 550. 

Course Outcomes

GMAT Quantitative Foundations Prep Course

Welcome to the Foundations Course! This course is designed to provide you with a strong foundation in the core concepts and skills necessary for success in your academic and professional endeavors. The course will cover a wide range of topics, including critical thinking, problem-solving, communication, and teamwork.

Here's why BobPrep's course is your ultimate weapon for taming the GMAT Quant beast:

★ By the end of this course, you will be able to:

Think critically and creatively:

Analyze information from multiple perspectives, evaluate arguments, and generate innovative solutions.

Solve problems effectively:

Identify and define problems, develop and implement The GMAT is a standardized test designed to assess the quantitative and analytical writing skills of business professionals seeking admission to executive MBA programs. The Quantitative Reasoning section of the GMAT tests your ability to solve basic math problems involving algebra, geometry, and data analysis.

If you're feeling rusty on your math skills, don't worry! BobPrep's GMAT GMAT Quantitative Foundations Prep Course is designed to help you brush up on the essential math concepts you need to know for the GMAT.

★ With BobPrep's GMAT Quantitative Foundations Prep Course, you can:

1. Improve your math skills and gain the confidence you need to ace the EA Quantitative Reasoning section.

2. Increase your chances of getting accepted to your top choice executive MBA program.

3. Invest in your future and open up new career opportunities.

★ Why choose BobPrep:

Expert instructors:
Our instructors are GMAT experts who have years of experience teaching and helping students succeed on the GMAT.

Comprehensive curriculum:
Our course covers all the essential math topics tested on the GMAT.

Personalized learning:
Our adaptive learning platform tailors the course content to your individual needs.

Flexible learning:
You can access our course materials anytime, anywhere, on any device.

Guaranteed results:
We are so confident in our course that we offer a money-back guarantee.

★ What you'll learn:

Algebra:
Review basic concepts like fractions, decimals, percentages, exponents, and roots. Learn how to solve linear equations, inequalities, and systems of equations.

Geometry:
Brush up on your knowledge of geometric shapes, angles, and area and volume formulas.

Data analysis:
Learn how to interpret tables, charts, and graphs, and how to calculate basic statistics like mean, median, and mode.

Problem-solving strategies:
Develop effective strategies for tackling quantitative reasoning problems on the GMAT.

Test-taking tips:
Learn how to manage your time effectively and avoid common mistakes on the GMAT.

★ Benefits of taking BobPrep's GMAT Quantitative Foundations Prep Course:

Improved problem-solving skills:

The skills you learn in our course will not only help you on the GMAT, but they will also be valuable in your business career.

Enhanced critical thinking skills:

Our course will help you develop the critical thinking skills you need to analyze complex information and make sound decisions.

Increased confidence:

By taking our course, you will gain the confidence you need to succeed on the GMAT and achieve your goals.

★  Our Insights:

We believe that everyone has the potential to succeed on the GMAT. With the right preparation, you can achieve your score goals and get into the executive MBA program of your dreams.

Course Topics are followed Below:

1 Approximation
N/A

  • In mathematical expression, which includes division and multiplication of decimal values of large number, it becomes quite complicated to solve these expressions.
  • So, to reduce this complexity we use approximation method. In approximation method, we need not calculate the exact value of an expression, but we calculate the nearest (round off) values.
  • When we use approximation method, then final result obtained is not equal to the exact result, but it is very close to the final result (either a little less or little more).

Basic Rules to Solve the Problems by Approximation:

Rule1:

To solve the complex mathematical expression, take the nearest value of numbers given in the expression.

Rule 2:

To multiply large number, we can take the approximate value (round off) of numbers by increasing one number and decreasing the other accordingly, so that the calculation is eased.

Rule 3:

When we divide large number with decimals, then we can increase or decrease both numbers accordingly.

Rule 4:

To find the percentage of any number, we can use the following shortcut methods

To calculate 10% of any number, we simply put a decimal after a digit from the right end.

To calculate 1% of any number, we simply put a decimal after two digits from right end.

To calculate 25% of any number, we simply divide the number by 4.

Examples:

  1. 89% of (599.88 + 30 x 400) + 50 =?

Solution:

 2.  

Solution: 393 x 197 + 5600 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>390</mn><mo>&#xD7;</mo><mn>200</mn><mo>+</mo><mn>5600</mn><mo>&#xD7;</mo><mfrac><mn>5</mn><mn>4</mn></mfrac><mo>+</mo><mn>8200</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>78000</mn><mo>+</mo><mn>7000</mn><mo>+</mo><mn>8200</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>93200</mn></mtd></mtr></mtable></math>

 3. (9.5)2 =?

Solution: (9.5)2 = 90.25 = 90

 4. 183.5 ÷ 273.5 x 63.5 = 2?

Solution:  183.5 ÷ 273.5 x 63.5 = 2?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mfenced separators="|"><mrow><mfrac><mn>18</mn><mn>27</mn></mfrac><mo>&#xD7;</mo><mn>6</mn></mrow></mfenced><mn>3.5</mn></msup><mo>=</mo><msup><mn>2</mn><mo>?</mo></msup></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mn>4</mn><msup><mo>)</mo><mn>3.5</mn></msup><mo>=</mo><msup><mn>2</mn><mo>?</mo></msup></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mfenced separators="|"><msup><mn>2</mn><mn>2</mn></msup></mfenced><mn>3.5</mn></msup><mo>=</mo><msup><mn>2</mn><mo>?</mo></msup></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mn>2</mn><mn>7.0</mn></msup><mo>=</mo><msup><mn>2</mn><mo>?</mo></msup></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>?</mo><mo>=</mo><mn>7</mn></mtd></mtr></mtable></math>

 5. 125.009 + 69.999 + 104.989 =?

Solution:

125.009 + 69.999 + 104.989 =?

 Each value is approximated to nearest whole number

=> ? =125 + 70 + 105

=> ? = 300


2 Chain Rule
N/A

  1. Direct Proportion:

Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increase (or decreases) to the same extent.

Example1: Cost is directly proportional to the number of articles (More Articles, More Cost)

Example2: Work done is directly proportional to the number of men working on it. (More Men, More work)

  1. Indirect Proportion:

Two quantities are said to be indirectly proportional, if on the increase of the one, the other decreases to the same extent and vice-versa.

Example1: The time taken by a car in covering a certain distance is inversely proportional to the speed of the car.

(More speed, less is the time taken to cover a distance)

Example2: Time taken to finish a work is inversely proportional to the number of persons working at it.

(More persons, less is the time taken to finish a job)

Note: In solving questions by chain rule, we compare every item with the term to be found out.

Examples:

  1. If the price of 6 toys is $ 264.37, what will be the approximate price of 5 toys?

Solution: Let the required price be $ x.

Then, less toys, less cost (Direct Proportion)

Therefore, 6: 5:: 264.37: x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>6</mn><mi>x</mi><mo>=</mo><mo>(</mo><mn>5</mn><mo>&#xD7;</mo><mn>264.37</mn><mo>)</mo></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>5</mn><mo>&#xD7;</mo><mn>264.37</mn></mrow><mn>6</mn></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>220.308</mn></mtd></mtr></mtable></math>

 2. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?

Solution: Let the required number of days be x.

Then, less men, more days (Indirect proportion)

Therefore, 27: 36:: 18: x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>27</mn><mi>x</mi><mo>=</mo><mn>36</mn><mo>&#x2217;</mo><mn>18</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>36</mn><mo>&#x2217;</mo><mn>18</mn></mrow><mn>27</mn></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>24</mn></mtd></mtr></mtable></math>

 3. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

Solution:

  • Let the required number of days b x.
  • Less cows, More days (Indirect Proportion)
  • Less bags, Less days (Direct Proportion)

Therefore, 1 * 40 * x = 40 * 1 * 40

 x = 40

  1. If the cost of x meters of wire is d dollars, then what is the cost of y metres of wire at the same rate?

Solution:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2022;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>cos</mi><mi>t</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>x</mi><mo>&#xA0;</mo><mi>m</mi><mi>e</mi><mi>t</mi><mi>r</mi><mi>e</mi><mi>s</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>$</mo><mo>&#xA0;</mo><mi>d</mi><mspace linebreak="newline"/><mo>&#x2022;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mtext>Cost of&#xA0;</mtext><mn>1</mn><mtext>&#xA0;metre&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mfrac><mi>d</mi><mi>x</mi></mfrac></mfenced><mspace linebreak="newline"/><mo>&#x2022;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mtext>Cost of&#xA0;</mtext><mi mathvariant="normal">y</mi><mtext>&#xA0;metre&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mfrac><mi mathvariant="normal">d</mi><mi mathvariant="normal">x</mi></mfrac><mo>&#x2217;</mo><mi mathvariant="normal">y</mi></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mfrac><mi>yd</mi><mi mathvariant="normal">x</mi></mfrac></mfenced></math>

 5. If  of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?

Solution:

  • Let the required time be ‘x’ seconds.
  • Part filled =  Remaining part 
  • Less part, Less time (Direct proportion)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mi>f</mi><mi>o</mi><mi>r</mi><mi>e</mi><mo>,</mo><mfrac><mn>3</mn><mn>5</mn></mfrac><mo>:</mo><mfrac><mn>2</mn><mn>5</mn></mfrac><mo>:</mo><mo>:</mo><mn>60</mn><mo>:</mo><mi>x</mi><mo>&#x21D4;</mo><mo>(</mo><mfrac><mn>3</mn><mn>5</mn></mfrac><mo>*</mo><mi>x</mi><mo>)</mo><mo>=</mo><mo>(</mo><mfrac><mn>2</mn><mn>5</mn></mfrac><mo>*</mo><mn>60</mn><mo>)</mo></math>


3 Stock and Shares
N/A

To start a big business or an industry, a large amount of money is needed.  It is beyond the capacity of or two persons to arrange such a huge amount. However, some persons associate together to form a company. They, then, draft a proposal, issue a prospectus (in the name of the company), explaining the plan of the project and invite the public to invest money in this project. They, thus, pool up the funds from the public, by assigning them shares of company.

Important Formulae:

  1. Stock-capital: The total amount of money needed to run the company is called the stock-capital.
  2. Shares or Stock: The whole capital is divided in to small units, called shares or stock.

For each investment, the company issues a share-certificate, showing the value of each share and the number of shares held by a person.

The person who subscribes in shares or stock is called a shareholder or stock holder.

  1. Dividend: The annual profit distributed among shareholders is called dividend.

Dividend is paid annually as per share or as a percentage.

  1. Face Value: The value of a share or stock printed on the share-certificate is called its Face Value or Nominal Value or Par value.
  2. Market Value: The stocks of different companies are sold and bought in the open market through brokers at stock-exchanges. A share (or stock) is said to be:
  1. At premium, if its market value is more than its face value
  2. At par, if its market value is the same as its face value
  3. At discount or Below par, if its market value is less than its face value.

Thus, if a $100 stock is quoted at a premium of 18, then market value of the stock = $(100 + 18) = $ 118

Likewise, if a $ 100 stocked at a discount of 5, then the market value of the stock = $ (100 – 5) = $ 95

  1. Brokerage: The broker’s charge is called brokerage
  1. When stock is purchased, brokerage is added to the cost price.
  2. When stock is sold, brokerage is subtracted from the selling price.

NOTE:

  • The face value of a share always remains the same
  • The market value of a share changes from time to time
  • Dividend is always paid on the face value of a share.
  • Number of shares held by a person

Examples:

  1. Find the annual income derived from $ 2500, 8% stock at 106.

Solution: Income from $ 100 stock = $ 8

Income from $ 2500 stock = $ = $ 200

 2. A man buys $ 25 shares in a company which pays 9 % dividend. The money invested is such that it gives 10% on investment. At what price did he buy the shares?

Solution: Suppose he buys each share for $ x.

x = 22.50

Cost of each share = $ 22.50

 3. A man invests in a 16% stock at 128. The interest obtained by him is?

Solution: By investing $ 128, income derived = 16

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>By investing&#xA0;</mtext><mi mathvariant="normal">$</mi><mn>100</mn><mtext>, income derived&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mfrac><mn>16</mn><mn>128</mn></mfrac><mo>&#xD7;</mo><mn>100</mn></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mn>12.5</mn></math>

Therefore, interest obtained = 12.5%

 4. To produce an annual income of $ 1200 from 12% stock at 90, the amount of stock needed is:

Solution: For an income of $ 12, stock needed = $ 100

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;For an income of&#xA0;</mtext><mi mathvariant="normal">$</mi><mn>1200</mn><mo>,</mo><mtext>&#xA0;stock needed&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mfrac><mn>100</mn><mn>12</mn></mfrac><mo>&#xD7;</mo><mn>1200</mn></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mn>10</mn><mo>,</mo><mn>000</mn></math>

 5. A invested some money in 10% stock at 96. If B wants to invest in an equally good 12% stock, he must purchase a stock worth of:

Solution: For an income of $ 10, investment = $ 96

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;For an income of&#xA0;</mtext><mi mathvariant="normal">$</mi><mn>12</mn><mtext>, investment&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mfrac><mn>96</mn><mn>10</mn></mfrac><mo>&#xD7;</mo><mn>12</mn></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mn>115.20</mn></math>


4 Speed, Time and Distance
N/A

Speed

The rate at which a body or an object travels to cover a certain distance is called speed of that body.

Time

The duration in hours, minutes or seconds spent to cover a certain distance is called the time. Distance

The length of the path travelled by any object or a person between two places is known as distance.

Relation between Speed, Time and Distance:

Speed is the distance covered by an object in unit time. It is calculated by dividing the distance by the time taken.

Distance = Speed x time

Formulae:

  1. If speed is kept constant, then the distance covered by an object is proportional to time.

Distance  Time (speed constant) or 

 2. If time is kept constant, then the distance covered by an object is proportional to speed.

Distance  Speed (speed constant) or 

 3. If distance is kept constant, then the speed covered by a body is inversely proportional to time.

Speed  (distance constant) or S1 T1 = S2 T2 = S3 T3 = ….

 4. When two bodies A and B are moving with speed a miles/h and b miles/h respectively, then the relative speed of two bodies is

  1.  (a + b) miles/h (if they are moving in opposite direction)
  2. (a - b) miles/h (if they are moving in same direction)

 5. When a body travels with different speeds for different durations, then average speed of that body for the complete Journey is defined as the total distance covered by the body divided by the total time taken to cover the distance.

Average speed

Note: If a body covers a distance D1 at S1 miles/h, D2 at S2 miles/h, D3 at S3 miles/h and so on up to Dn at Sn, then Average speed =

Examples:

  1. A train covers a distance of 200 miles with a speed of 10 miles/h. What time is taken by the train to cover this distance?

Solution: Given, speed = 10 miles/h and distance = 200 miles

Therefore, required time = 20 h

 2. A person covers a distance of 12 miles, while walking at a speed of 4 miles/h. How much distance he would cover in same time, if he walks at a speed of 6 miles/h?

Solution: Given that, D1 = 12 miles, S1 = 4 miles/h, D2 =? and S2 = 6 miles/h

Since, the time is kept constant.

According to the formula, 

 D2 = 18 miles

Therefore, the person will cover 18 miles.

 3. A person covers a certain distance with a speed of 18 miles/h in 8 min. If he wants to cover the same distance in 6 min, what should be his speed?

Solution: We know that, Speed = 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Speed to cover&#xA0;</mtext><mfrac><mn>12</mn><mn>5</mn></mfrac><mtext>&#xA0;miles in&#xA0;</mtext><mn>6</mn><mi>m</mi><mi>i</mi><mi>n</mi><mo>=</mo><mfrac><mtext>&#xA0;Distance&#xA0;</mtext><mtext>&#xA0;Time&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mfrac><mn>12</mn><mn>5</mn></mfrac><mfrac><mn>1</mn><mn>10</mn></mfrac></mfrac><mo>=</mo><mfrac><mn>12</mn><mn>5</mn></mfrac><mo>&#xD7;</mo><mn>10</mn><mo>=</mo><mn>12</mn><mo>&#xD7;</mo><mn>2</mn><mo>=</mo><mn>24</mn><mi>m</mi><mi>i</mi><mi>l</mi><mi>e</mi><mi>s</mi><mo>/</mo><mi>h</mi></math>

 4. Two trains are running in the same direction. The speeds of two trains are 5 miles/h and 15 miles/h, respectively. What will be the relative speed of second train with respect to first?

Solution: We know that, if two trains are running in same direction, then difference in speeds is the required relative speed.

Required relative speed = 15 - 5 = 10 miles/h

 5. A person covers a distance of 20 miles by bus in 35 min. After deboarding the bus, he took rest for 20 min and covers another 10 miles by a taxi in 20 min. Find his average speed for the whole journey?

Solution: Total distance covered = (20 + 10) miles = 30 miles

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Total time taken&#xA0;</mtext><mo>=</mo><mo>(</mo><mn>35</mn><mo>+</mo><mn>20</mn><mo>+</mo><mn>20</mn><mo>)</mo><mo>min</mo><mo>=</mo><mn>75</mn><mi>m</mi><mi>i</mi><mi>n</mi><mo>=</mo><mfrac><mn>5</mn><mn>4</mn></mfrac><mi>h</mi></math>

According to the formula, Average speed = 

So, the average speed of the person for the whole journey is 24 miles/h


5 Races and Games of Skill
N/A

A race or a game of skill includes the contestants in a contest and their skill in the concerned contest/game.

Important Terms

Race

A race is a contest of speed in running, driving, riding, sailing or rowing.

Race Course

The ground/path on which a contest is organised in a systematic way, is called a race course. Starting Point

The exact point/place from where a race begins, is called starting point.

Start

If two persons A and B are contesting a race and before the start of the race, A is at the starting point and B is ahead of A by 20 m, then it is said that A gives B a start of 20 m.

Finishing Point

The exact point/place where a race ends, is known as finishing point.

Winning Point/Goal

 A person who reaches the finishing point first, is called the winner.

Note

For a winner, finishing point is as same as the winning point/goal

Dead Heat Race

A race is said to be a dead heat race, if all the contestants reach the finishing point exactly at the same time.

Game

A game of 100 means that the contestant who scores 100 points first, is declared the winner.

Some Facts about Race

 For Two Contestants A and B

  1. If A beats B by x m, then

Distance covered by A (winner) = L m Distance covered by B (loser) = (L - x) m

 2. If B starts from x m ahead of A (or A gives B a start of x m), then

A start from M and B starts from Z. Distance covered by B = (L - x) m

 3. If A beats B by T s, then

A and B both start from point M.

Time taken by A (winner) = Time taken by B (loser) - T

It means that A completes the race in T s less time than that of B

 4. If B starts the race T s before the time A starts (or if A gives B a start of T s), then

In such case, we say that A starts T s after the time B starts. 5. If both of the contestants get at the finishing point at the same time, then Difference in time of defeat = 0; Difference in distance of defeat = 0

Examples:

  1. In a game of 100 points, A scores 100 points, while B scores only 75 points. In this game, how many points can A give to B? S

Solution: Given

  • Score of A = 100 points;
  • Score of B = 75 points

... A can give (100 - 75) = 25 points to B

  1. In a 100 m race, Jack runs at the speed of 4 km/h. Jack gives Bob a start of 4 m and still beats him by 15 s. Find the speed of Bob?

Solution:

  • Time taken by Jack to cover 100 m = 
  • Therefore, Bob covers (100 – 4) = 96 m in (90 + 5) = 105 s

So, Bob’s speed = 

 3. A 10 km race is organised at 800 m circular race course. P and Q are the contestants of this race. If the ratio of the speeds of P and Q is 5: 4, how many times will the winner overtake the loser?

Solution: Given Speed of P: Speed of Q = 5: 4

  • Time taken by P to cover 5 rounds = Time taken by Q to cover 4 rounds
  • Distance covered by P in 5 rounds 
  • Distance covered by Q in 4 rounds 
  • In 5 rounds, P will overtake Q every time.
  • It means that after covering 4 km, P will overtake Q one time.

Therefore, after covering 10 km P will overtake 

 4. In a 200 m race, A can beat B by 50 m and B can beat C by 8 m. In the same race, A can beat C by what distance?

Solution:

  • From given data we can write, A: B = 200: 150 and B: C = 200: 192
  • So, A beats C by (200 – 144) m = 56 m

 5. In a game of billiards, A can give 20 points in the game of 120 points and he can give C 30 points in the game of 120 points. How many points can B give C in a game of 90?

Solution:

  • If A scores 120 points, then B scores 100 points and C scores 90 points.
  • When B scores 100 points, then C scores 90 points.
  • When B scores 90 points, then C scores points = 81 points.
  • Therefore, B can give C, 9 points in a game of 90.


6 Problems Based on Trains
N/A

Problems based on trains are same as the problems related to 'Speed, Time and Distance' and some concepts of 'Speed, Time and Distance' are also applicable to these problems.

The only difference is that the length of the moving object (train) is taken into consideration in these types of problems

Rules:

Rule 1:  Speed of train (S) = 

Here, unit of speed is m/s or km/h

  1. a km/h
  2. a m/s

Rule 2:  

The distance covered by train in passing a pole or a standing man or a signal post or any other object (of negligible length) is equal to the length of the train.

Rule 3:  

If a train passes a stationary object (bridge, platform etc;) having some length, then the distance covered by train is equal to the sum of the lengths of train and that particular stationary object which it is passing

Rule 3:  

If two trains are moving in opposite directions, then their relative speed is equal to the sum of the speeds of both the trains.

Rule 4:  

If two trains are moving in the same direction, then the relative speed is the difference of speeds of both trains.

Rule 5:  

If two trains of lengths x and y are moving in opposite directions with speeds of u and v respectively, then time taken by the trains to cross each other is equal to 

Rule 6:  

If two trains of lengths x and y are moving in the same direction with speeds of u and v respectively, then time taken by the faster train to cross the slower train is equal to 

Rule 7:  

If two trains start at the same time from points P and Q towards each other and after crossing each other, they take t1, and t2 time in reaching points Q and P respectively, then (P's speed): (Q's speed) = t1: t2

Examples:

  1. A train takes 9 s to cross a pole. If the speed of the train is 48 km/h, then length of the train is

Solution:

  • Let the length of the train be x m.
  • Now, speed = 48 km/h 
  • Train takes 9 s to cross a pole.
  • Therefore, length of train(x) = speed x time 

 2. The distance between two stations P and Q is 145 km. A train with speed of 25 km/h leaves station at 8:00 am towards station Q. Another train with speed of 35 km/h leaves station Q at 9:00 am towards station P. Then, at what time both trains meet?

Solution:

  • First train leaves at 8:00 am and second at 9:00 am.
  • So, first train i.e., from P to Q has covered 25 km distance in 1 h.
  • So, distance left between the station =145-25 = 120 km
  • Now, trains are travelling in opposite directions
  • So, relative speed = 25 + 35 = 60 km/h
  • Time taken to cover 120 km 

Therefore, the time at which both the trains will meet, is 2 h after second train left i.e., 9:00 am + 2h = 11:00 am

 3. A train A is 180 m long, while another train B is 240 m long. A has a speed of 30 km/h and B's speed is 40 km/h. If the trains move in opposite directions, find when will A pass B completely?

Solution: Let the total distance = x + y = 180 + 240 = 420 m

            Active speed 

Therefore, required time 

 4. Train A crosses a pole in 25 s and train B crosses the pole in 1 min 15 s. Length of train A is half the length of train B. What is the ratio between the speeds of A and B, respectively?

Solution:

  • Let the lengths of trains A and B be l and 2l respectively
  • When a train crosses a pole, it covers the distance equal to its length

Therefore, required ratio of speeds 

 5. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/h and the other one walks at 5.4 km/h. The train needs 8.4 s and 8.5 s respectively, to overtake them. What is the speed of the train, if both the persons are walking in the same direction as the train?

Solution: Speed of 1st person = 4.5 km/h 

Speed of 2nd person = 5.4 km/h 

Let the speed of the train be k m/s

Then, (k – 1.25) x 8.4 = (k – 1.5) x 8.5

8.4k – 10.5 = 8.5k – 12.75

0.1k = 2.25

k = 22.5

Therefore, speed of train 

 


7 Escalator-Time, Speed and Distance
N/A

Escalator problems are similar to the upstream and downstream problems.

Here the escalator moves in both the directions but in a stream, the direction of flow of water is constant.

These escalator questions are bit confusing compared to other topics in time, speed and distance.

Escalator:

  • An escalator is a moving stair, i.e., it moves continuously up or down.
  • The total number of stairs are fixed in an escalator always.
  • The distance is said in terms of the number of steps.
  • The number of steps taken by the individual and the escalator is equal to the total number of steps on the escalator.

Important Things:

  1. If we are moving with escalator, we have to climb less steps as escalator will push us forward on its own.
  2. If we are moving against the escalator, we have to climb more steps as escalator will pull us back on its own.

Moving with escalator:

The total number of steps will be the addition of the two (individual and escalator) in case the individual is moving in the same direction as that of the escalator.

Total number of steps = steps climbed by individual + steps of escalator

Moving against escalator:

The total number of steps will be the subtraction of the two (individual and escalator) in case the escalator and the individual are moving in the opposite direction.

Total number of steps = steps climbed by individual - steps of escalator

Note:

The time taken by the individual to climb up or down the escalator is equal to the time for which escalator is moving.

Common Questions based on escalator:

  1. Speed scenario
  2. Steps scenario
  3. Time scenario

Examples:

  1. A boy is walking down a downward-moving escalator and steps down 15 steps to reach the bottom. Just as he reaches the bottom of the escalator, he remembers that he forgot his wallet in a shop on the floor above. He runs back up the downward moving escalator at a speed 3 times that which he walked down. He covers 45 steps in reaching the top. How many steps are visible on the escalator when it is switched off?

Solution:

Given that boy runs back at a speed 3 times that which he walked down

Let the boy’s speed with the escalator = s

And boy’s speed against the escalator = 3s

And speed of escalator = u

We know that total number of steps in both cases are equal.

15 + (15/s) times u = 45 – (45/3s) times u

(30/s) times u = 45 -15

(1/s) times u = 30/30 = 1

Placing the value in any of the equation = 15 + 15 = 30

Number of steps visible on the escalator when it is switched off = 30

  1. A lady walks up on an escalator which is moving up at a rate of one step per second, thirty steps bring her top. The other day she goes up at three steps per second, reaching the top in 45 steps. Find the number of steps in the escalator?

Solution: The time taken on two days are 30s and 15s which are in ratio 2: 1.

So, the steps thrown out by the escalator are in ratio 2: 1

 30 + 2x = 45 + x

  x = 15

Therefore, total steps = 30 + 2(15) = 60

  1. Richard takes 60 seconds to walk up on an upward moving escalator but he takes 90 seconds to walk up on a downward moving escalator. Calculate the time that Richard will take to walk up if the escalator is not moving?

Solution:

Let us consider the speed of Richard be u

Let us consider the speed of escalator be v

As the distance is constant, the three speeds i.e., u + v, u, u – v will be in arithmetic progression.

Since time is inversely proportional to speed, the time taken in each case will be in harmonic progression.

Calculating the harmonic mean of the given time taken = the time taken by Richard to walk up if the escalator is not moving = (2 * 60 * 90)/ (60 + 90) = 10800/150 = 72

  1. Bob is going up the escalator. It takes Bob 60 seconds to walk up the escalator which is moving upwards and 180 seconds to walk up the escalator which is moving downwards. Calculate the time taken by Bob to climb the escalator which is stationary.

Solution:

  • Let speed of Bob be "s" steps per second.
  • Let speed of escalator be "e" steps per second.
  • No. of steps (N)= 60s + 60e

Since Bob is moving upwards, 90e will be added

  • No. of steps= 180s-180e

Since Bob is moving downwards, 150e will be subtracted

By equating the number of steps,

60s + 60e = 180s -180e

s = 2e

  • N = 60s + 60e = 60s + 60(s/2) = 60s + 30s = 90s
  • Time taken when escalator is stationary = 90s/s = 90 seconds

N=80a+80x= 80a+16a=96a Time taken when the escalator is stationary= 96a/a= 96seconds.

  1. An escalator moves downward from first floor to ground floor at a constant rate. When the escalator is turned off, Harry takes 90 steps to descend from first floor to ground floor. When the escalator is turned on, Harry needs only 40 steps to descend from first to ground floor. If Harry begins at ground floor and walks upward, against the escalator’s downward movement, how many steps will he need to take to reach first floor Assuming that Harry walks at a constant rate in all scenarios.

Solution:

  Let the distance from first to ground floor = d

  Let speed be u with which Harry moves when escalator is still

  Let the speed be v with which escalator moves up or down

When escalator is stand still, he needs 90 steps

 Therefore d/u = 90 => u = d/90 ............. (I)

Also, when both escalator and Harry moves in same direction, he needs 40 steps

  Therefore d/ (u +v) = 40 => u + v = d/40 ........... (II)

  Solving both (i) & (ii)

we get v= d/ (72) ..........(iii)

   Hence u - v = d/ (35x72)

Therefore, Number of steps required when escalator and Harry are moving in opposite directions = d/(u-v) = d/ (d/(35x72)) = 35x72 = 2520


8 Relative Speed – Time, Speed and Distance
N/A

Relative Speed:

It is defined as the speed of a moving object with respect to another.

It mainly includes:

  Two bodies moving in the same direction

  Two bodies moving in the opposite direction

Relative Speed Formulas:

  • When two objects are moving in the same direction, relative speed is given as their difference.

  If two objects are moving with speeds X and Y, their relative speed is X – Y, if they are moving in the same direction

  • When the two objects are moving in opposite directions, relative speed is given by adding the two speeds.

  If two objects are moving with speeds X and Y, their relative speed is X + Y, if they are moving in the opposite direction

Time and Distance Formula

Distance = Speed × Time.

Speed is inversely proportional to the time taken when distance travelled is constant.

So, when speed increases, time decreases and vice versa.

Note:

There will be no change in the relative speed equation, if both the bodies reverse their direction at the same instant

The description of the motion of the two bodies between two consecutive meetings will also be governed by the proportionality between speed and distance – since the time of movement between any two meetings will be constant

Examples:

  1. Two friends Jack and Jimmy 130 miles apart from each other, start travelling towards each other at the same time. If the Jack covers 7 miles per hour to the Jimmy’s 6 miles per hour, how far will Jimmy have travelled when they meet?

Solution: Time taken to meet = (Distance between them)/ (Relative speed)

Here moving in opposite direction

Relative speed = x + y = 7 + 6 = 13 miles per hour

Time taken to meet = 130/ 13 = 10 hours

Therefore, Jimmy travels = 10 x 6 = 60 miles

  1. Two trains for station A leave station B at 8 am and 8.48 am and travel at 100 kmph and 145 kmph respectively. How many kilometres from station B will the two trains be together?

Solution:

Distance gained by first train started at 8 am till 8:48 am = 100*(48/60) = 80 km

Relative speed = 145 – 100 = 45

Time = relative distance / relative speed = 80/45 hours

distance travelled by first train = 80 + (80/ 45) *100 = 257.77 km

  1. The distance between two hills P and Q is 65 km. A car racer starts from P towards Q at 6 am at a speed of 5 km/hr. Another car racer starts from Q towards P at 7 am at a speed of 15 km/hr. At what time will they cross each other?

Solution: In an hour, the earlier racer covers a distance of 5 km.

  • The distance between the two = 65 – 5 = 60 km.
  • Therefore, the two racers approach each other with a relative speed of 5 + 15 = 20 kmph.
  • time taken = 60/ 20 = 3 hrs.
  • So, they meet each other at (7 + 3) = 10 am.
  1. Two cars start at the same time from X and Y and proceed towards each other at the rate of 20 km/hr and 24 km/hr respectively. When they meet, it is found that one car has travelled 80 km more than the other. The distance between the X and Y is

Solution:

Both the cars are coming closer to each other by (24 – 20) = 4 km/ hr.

They would meet after 80/4 = 20 hrs.

Hence, the distance between the X and Y = (24 + 20) x 20 = 44 x 20 = 880 km.

  1. Two trains of length 300m and 500m run on parallel lines. when they run in the opposite direction, they take 20 sec to cross each other and when they run in same direction, they take 80 sec to cross each other. Find the speed of faster train.

Solution:

When they run in same direction relative speed = p – q

When they run in opposite direction relative speed = p + q

We know that (p – q) x 80 = (p + q) x 20

  80p – 80q = 20p + 20q

  60p = 100q

  3p = 5q – (1)

We know that (p – q) x 80 = 800 m

p – q = 10 – (2)

Solving 1 & 2

p – (3/5) p = 10

5p – 3p = 50

2p = 50

p = 25, q = 15

The speed of the faster train = 25 m/sec


9 Number Series
N/A

A number series is a sequence of numbers written from left to right in a certain pattern. To solve the questions on series, we have to detect/find the pattern that is followed in the series between the consecutive terms, so that the wrong/missing term can be find out.

Types of Series

There can be following types of series 

 1. Prime Number Series:

The number which is divisible by 1 and itself, is called a prime number. The series formed by using prime number is called prime number series.

Example: Find out the next term in the series 7,11,13,17, 19....

Sol. Given series is a consecutive prime number series.

 Therefore, the next term will be 23

 2. Addition Series:

The series in which next term is obtained by adding a specific number to the previous term, is known as addition series.

Addition series are increasing order series and difference between consecutive term is equal.

Example: Find out the missing term in the series 2, 6, 10, 14, _, 22.

Solution: Here, every next term is obtained by adding 4 to the previous term.

So, Required term = 14 + 4 = 18

 3. Difference Series:

Difference series are decreasing order series in which next term is obtained by subtracting a fixed/specific number from the previous term.

Example: Find out the missing term in the series 108, 99, 90, 81, _, 63.

Solution: Here, every next number is 9 less than the previous number.

So, required number = 81 - 9 = 72

 4. Multiple Series:

When each term of a series is obtained by multiplying a number with the previous term, then the series is called a multiplication series.

Note: Number which is multiplied to consecutive terms, can be fixed or variable

Example: Find out the missing term in the series 4, 8,16, 32, 64, _, 256.

Solution: Here, every next number is double the previous number.

So, required number = 64 x 2 = 128

 5. Division Series:

 Division series are those in which the next term is obtained by dividing the previous term by a number.

Note Number which divides consecutive terms can be fixed or variable

Example: Find out the missing term in the series 10080,1440, 240, _, 12,4

Series pattern 

Hence, missing term is 48

 6. n2 Series

When a number is multiplied with itself, then it is called as square of a number and the series formed by square of numbers is called n2 series.

Example: Find out the missing term in the series 4,16, 36, 64, _, 144.

Solution: This is a series of squares of consecutive even numbers.

Let us see 22 = 4, 42 = 16, 62 = 36, 82 = 64, 102 = 100, 122 = 144

Hence, missing term is 100

 7. (n2 + 1) Series

If in a series each term is a sum of a square term and 1, then this series is called (n2 +1) series. Example: Find out the missing term in the series 10, 17, 26, 37, _, 65.

Solution: Series pattern 32 + 1 = 10, 42 + 1 = 17, 52 + 1 = 26, 62 + 1 = 37, 72 + 1 = 50, 82 + 1 = 65

 So, required number = 50

 8. (n2 - 1) Series

In a series, if each term is obtained by subtracting 1 from square of a number, then such series    is known as (n2 -1) series.

Example: Find out the missing term in the series 0, 3, 8,15, 24, _, 48.

Solution: Series pattern 12 - 1 = 0, 22 - 1 = 3, 32 - 1 = 8, 42 - 1 = 15, 52 - 1 = 24, 62 - 1 = 35, 72 - 1 = 48

So, correct answer is 35.

 9. (n2 + n) Series

Those series in which each term is a sum of a number with square of that number, is called as (n2 + n) series.

Example: Find out the missing term in the series 12, 20, 30, 42, _, 72.

Solution: Series pattern 32 + 3, 42 + 4, 52 + 5, 62 + 6, 72 + 7, 82 + 8

So, required number =72 + 7 = 56

 10. (n2 - n) Series

Series in which each term is obtained by subtracting a number from square of that number, is known as (n2 - n) series.

Example: Find out the missing term in the series 42, 30,_,12, 6.

Solution: Series pattern 72 - 7, 62 - 6, 52 - 5, 42 - 4, 32 - 3

So, required number = 52 - 5 = 20

 11. n3 Series

A number when multiplied with itself twice, then the resulting number is called the cube of a number and series which consist of cube of different number following a specified sequence, is called as n3 series.

Example: Find out the missing term in the series 1, 8, 27, _, 125, 216.

Solution: Series pattern 13, 23, 33, 43, 53, 63

 So, required number = 43 = 64

 12(n3 + 1) Series

Those series in which each term is a sum of a cube of a number and 1, are known as (n3 +1) series.

Example: Find out the missing term in the series 126, 217, 344, _, 730.

Solution: Series pattern 53 + 1, 63 + 1, 73 + 1, 83 + 1, 93 + 1 So, required number = 83 + 1 = 513

 13(n3 - 1) Series

 Series in which each term is obtained by subtracting 1 from the cube of a number, is known as (n3 -1) series.

Example: Find out the missing term in the series 0, 7, 26, 63, 124, 215, ....

Solution: Series pattern 13 – 1, 23 - 1, 33 - 1, 43 - 1, 53 – 1, 63 - 1, 73 - 1

So, required number = 73 - 1 = 342

 14. (n3 + n) Series

When each term of a series is a sum of a number with its cube, then the series is known as (n3 + n) series.

Example: Find out the missing term in the series 2,10, 30, ..., 130, 222.

Solution: Series pattern 13 + 1, 23 + 2, 33 + 3, 43 + 4, 53 + 5, 63 + 6

 So, required number = 43 + 4 = 68

 15. (n3 - n) Series

When each term of a series is a sum of a number with its cube, then the series is known as (n3 + n) series.

Example: Find out the missing term in the series 2,10, 30, ..., 130, 222.

Solution: Series pattern 13 + 1, 23 + 2, 33 + 3, 43 + 4, 53 + 5, 63 + 6

So, required number = 43 + 4 = 68

 16. Alternating Series

In alternating series, successive terms increase and decrease alternately.

The possibilities of alternating series are if it is a combination of two different series.

Two different operations are performed on successive terms alternately.

Example: Find the next term in the series 15,14,19,11,23,8, ...

Series Pattern

 17.  Arithmetic Progression (AP)

The progression of the form a, a + d, a + 2d, a + 3d, ...is known as an arithmetic progression with first term a and common difference d.

Then, nth term Tn = a + (n -1) d

Example: In series 359, 365, 371..., what will be the 10th term?

Solution:

The given series is in the form of AP.

Since, common difference i. e., d is same.

Here, a = 359 and d = 6

10th term = a + (n - 1) d= 359 + (10 - 1) 6= 359 + (9 x 6) = 413.

 18.  Geometric Progression (GP)

The progression of the form a, ar, ar2, ar3... is known as a GP with first term a and common ratio = r.

Then, nth term of GP Tn = arn-1

Example: In the series 7,14, 28…..., what will be the 10th term?

Solution: The given series is in the form of GP.

 Since, common ration i. e., r is same.

Here, a = 7 and r = 2, 10th term = arn-1 = 7(2)10-1 = 7 x 29 = 3584

Examples:

 1.  4, 8, 24, 60, 124, ?

Solution: 

Series pattern

 4 + 22 =8

 8 + 42 = 24

24 + 62 = 60

60 + 82 = 124

124 + 102 =224

So, number is 224

 2.  8, 12, 24, 46, 72, 108, 52. which one is odd man out?

Solution: 8 + 4 = 12

12 + 12 = 24

24 + 20 = 44

should come in place of 46

44 + 28 = 72

72+ 36 = 108

108 + 44 = 152

So, 44 should come in place of 46

  1. 13, 25, 40, 57, 79, 103, 130. What one is odd man out?

Solution:

Series pattern

 13 + 12 = 25

 25+ 15 = 40

40 + 18 = 58

should come in place of 57

58 + 21 = 79

79 + 24 = 103

103 + 27 = 130

  1. 13, 35, 57, 79, 911, ?

Solution:

Series pattern

The elements of the given series are the numbers formed by joining together consecutive odd numbers in order, i.e., 1 and 3, 3 and 5, 5 and 7, 7 and 9, 9 and 11, ...

Therefore, Missing term = Number formed by joining 11 and 13 = 1113

  1. 7, 9, 16, 25, 41, 68, 107, 173

Solution:

Series pattern

 7+9 = 16;

9 + 16=25;

16 + 25 = 41;

25 + 41 = 66;

Should come in place of 68

41 + 66 = 107;

 66 + 107 = 173

Clearly, 68 is wrong and will be replaced by 66.


10 Last two digits of a number
N/A

Last two digits of a number is generally the tens place and units place digit of that number.

Let the number be 1345, the last two digits of this number are 4 and 5.

Method to calculate last two digits of a product:

Let us take the product of two numbers P and Q (Let us take P is 1448 and Q is 2677).

Let u and v, respectively represent the digits in the ten’s place and one’s place of P and same way x and y respectively represent the digits in the ten’s place and one’s place of Q, then

  1. Unit’s digit of P x Q is given by the unit’s digit in the product of v and y results in more than 1-digit, excess digit will be carried over to the left. i.e., in 1448 x 2677, multiply 8 and 7 giving 56. So, 6 is unit’s digit and 5 goes as carry to step 2
  2. Next, we cross multiply u and y & x and v, then add the resulting products. i.e., in 1448 x 2677 => 4 x 7 + 8 x 7 = 28 + 56 = 84
  3. If a carry is generated in step 1, add that with the result obtained in step 2. i.e., 5 + 84 = 89
  4. The unit’s digit in the result obtained in step 3 forms the tens digit in the product of P and Q i.e., the units digit in 89 which is 9 becomes the tens digit of 1448 x 2677.

So, the last two digits of 1448 x 2677 is 9 and 6

Before finding last two digits, let us see the binomial theorem for calculations

(x + a) n = nC0 an + nC1 an – 1 x + nC2 an – 2 x2 + …... where nCr

Method of finding last two digits:

Let the number be in the form p q

Last two digits of numbers ending in 1

If p ends in 1, then p raised to q, ends in 1 and its digit is obtained by multiplying the tens digit in p with the unit’s digit in q.

Example: find the last two digits of 81236

Solution: Since the base 81 ends in 1, 81236 ends in 1 and the tens place digit is obtained from the unit’s digit in 8 x 6 which is 8. Hence the last two digits of 81236 are 8 and 1.

Last two digits of numbers ending in 3, 7 or 9

Convert the number by repeatedly squaring until we get the unit digit as 1, and then apply the same process of finding the last two digits of number with unit digit 1.

Alternative way:

  1. When p ends in 9 raised to the power q.
  • Raise the base divide the exponent by 2
  • We know that number ending in 9 and raised to the power 2 ends in 1
  • As the base now ends in ‘1’ we can follow the method of finding last two digits of numbers ending in 1.
  1. When p ends in 3 raised to the power q.
  • Raise the base divide the exponent by 4
  • We know that number ending in 3 and raised to the power 4 ends in 1
  • As the base now ends in ‘1’ we can follow the method of finding last two digits of numbers ending in 1.
  1. When p ends in 7 raised to the power q.
  • Raise the base divide the exponent by 4
  • We know that number ending in 7 and raised to the power 4 ends in 1
  • As the base now ends in 1 we can follow the method of finding last two digits of numbers ending in 1.

Last two digits of numbers ending in 2, 4, 6 or 8

When p ends with the even numbers, we can find the last two digits of the number raised to the power q by this method

We know that 210 = 24

  • (24) odd number always ends in 24
  • (24) even number always ends in 76
  • (76) number always ends in 76

Let us see an example:

236 = 230 x 26

      = (210)3 x 26

      = 243 x 26

     = 24 x 64 (since 24oddnumber always ends in 24)

    = 36(last two digits)

Last two digits of numbers ending in 0 or 5

  • When any number with its unit digit as 0 raised to any power gives 00 in last two digits.
  • When p ends with the 5, we can find the last two digits of the number raised to the power q by this method

Example: (65)265

Here ten’s digit of base is even and exponent is odd, last two digits of (65)265 is 2 and 5.

Examples:

  1. Find the last two digits of 5135

Solution: ending with 1

The unit digit is 1 itself

In the last two digits ten’s digit is given by taking product of ten’s digit in base and units digit in component i.e., 5 x 5 = 25

So, take 5 as ten’s digit

So, the last two digits are 51 

  1. Find the last two digits of 17180

Solution: 17180 = (172)90

                           = (89)90 [last two digits of 172]

                        = (21)45 [last two digits of 892]

Here base is ending with 1

Ten’s digit of base is 2 and unit’s digit of power is 5.

 2 x 5 = 10 or 0

Therefore, last two digits of 17180 = 01

  1. Find the last two digits of 145157

Solution: here ten’s digit of base is even and exponent is odd

So, last two digits of 145157 will be 25

  1. Find the last two digits of 8244

Solution: 8244 can be written as 244 x 4144

244 = (210)4 x 24 = 244 x 16 = 76 x 16 = 16

Base is 1: 4144

4 x 4 = 16

So last two digits are 61

Therefore, last two digits of 8244 are 16 x 61 = 76


11 Factors of a number
N/A

Number of factors of a number:

If a number N can be represented as <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>P</mi><mn>1</mn><mi>a</mi></msubsup><mo>&#xD7;</mo><msubsup><mi>P</mi><mn>2</mn><mi>b</mi></msubsup><mo>&#xD7;</mo><msubsup><mi>P</mi><mn>3</mn><mi>c</mi></msubsup></math> where P1, P2, P3 are prime factors of N then number of factors of N can be found as (a + 1) * (b + 1) * (c + 1).

50 = 2 * 52, so number of factors of 50 = (1+1) * (2+1) =6. Which are 1, 2, 5, 10, 25and 50.

These are various combinations of the prime factors 2, 5 and 5.

  • Take 544; By using the divisibility rules we can easily see 2, 4, 8 are factors of 544. Take the highest power of each prime (here 8) for division.

544/8 = 68, divisible by 2, 4 and 17.

Take the highest power of each prime

68/ (4 * 17) = 1.

544 = 8 * 4 * 17 = 2* 17

Number of factors = 6 * 2 = 12

  • Take 1080; Applying divisibility rules, we get 2, 3, 4, 5, 8, 9 divides the number. Take the highest power of each prime (here 8, 5, 9) for division.

1080 / (8 * 5 * 9) = 3 (can stop dividing here as we got a prime, no need to write an obvious step to get unity by dividing with the same prime!)

1080 = 8 * 5 * 9 * 3 = 23* 33* 5

Number of factors = (3 + 1) * (3 + 1) * (1 + 1) = 4 * 4 * 2 = 32

Sum of factors of a number

Let’s take an example:  40 = 23 * 5,

number of factors = (3 + 1) * (1 + 1) = 4 * 2 = 8.

Sum of the factors = 1 + 2 + 4 + 8 + 5 + 10 + 20 + 40 = 90

This is same as (20 + 21 + 22 + 23) (50 + 51) = 15 * 6 = 90.

  • Find the sum of all the powers of each prime numbers, starting from 0 to the highest power contained in the standard form. Product of all such sums will give us the sum of the factors.
  • We have already seen that 1080 = 23 * 3* 5
  • In this case, sum of the factors will be (2+ 21 + 22 + 23) * (3+ 31 + 32 + 33) * (50 + 51) = 15 * 40 * 6 = 3600

This also explains how we find the numbers of factors. It is the product of the number of factors in each bracket (here, 4 * 4 * 2 = 32).

Product of factors of a number

N = <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>P</mi><mn>1</mn><mi>a</mi></msubsup><mo>&#xD7;</mo><msubsup><mi>P</mi><mn>2</mn><mi>b</mi></msubsup><mo>&#xD7;</mo><msubsup><mi>P</mi><mn>3</mn><mi>c</mi></msubsup></math>

Then the product of all the factors of N = N (a + 1) (b + 1) (c + 1)/2

e.g., 50 = 2 * 52;

Product of all factors of 50 = 50 (6) / 2 = 503

Number of odd factors

  • To get the number of odd factors, ignore the powers of 2.
  • Take 1080, 1080 = 23 * 33 * 5
  • Number of odd factors of 1080 = (3 + 1) * (1 + 1) = 4 * 2 = 8, here we considered only the powers of 3 and 5 to get the number of odd factors.

Sum of odd factors of a number

  • Sum of factors of 1080 = (2+ 21+ 22+ 23) * (30+ 31+ 32+ 33) * (50+ 51), where each term in the expansion represents a factor of 1080.
  • We need at least one 2 to get an even factor.  Remove all terms that has a 2 in it and we will remove all even factors.
  •  Even factors in 1080 are 21, 22and 23.
  • The only power of 2 remaining is 2(=1), which can be removed as it will not make any impact. Means, you ignore all powers of 2 to get the sum of all odd factors.
  • Now we need to remove even factors from the expression so that the result will be the sum of all odd factors.
  • Sum of odd factors of 1080 = (30+ 31+ 32+ 33) * (50+ 51) = 240

Number of even factors of a number

Let a number is of the form, N = 2a x   ...

  • Then Number of even factors of N = a * (b + 1) * (c + 1) …

For e.g., 1080 = 2* 33 * 5

Number of even factors of 1080 = 3 * (3 + 1) * (1 + 1) = 24

Sum of even factors of a number

  • To get the sum of the even factors we need to ensure that all the terms in the expression for the sum of factors are even.
  • All members in the 2’s bracket except 20 will give an even number.
  •  Remove 2from the group and that’s all it takes!

For e.g., 1080 = 2* 33 * 5

Sum of even factors of 1080 = (21 + 2+ 23) * (3+ 31 + 3+ 33) * (5+ 51) = 3360

Other conditions

What if we are asked to find the sum and number of factors of 1080 which are divisible by 15?

  • We don’t have to use anything else other than the logic we used for odd and even factors. We know number of terms in the expression is the number of factors.
  • So, when a condition is given, number of factors is the number of terms in the expression satisfying the given condition and considers only those factors to get the sum.
  • 1080 = (2+ 21+ 22+ 23) * (30+ 31+ 32+ 33) * (50+ 51)
  • Now 15 = 3 * 5, so we need to ensure that every term in the expression has a 3 and 5 in it. Now look at the bracket of 3 and find the terms that don’t yield a 3 in the expression.

Also check for bracket of 5 for the terms that don’t give us a 5. 30and 50are the culprits. Remove those entries and we get the expression we need.

Examples:

  1. Find Sum of factors of 1080 which are divisible by 15

 = (2+ 21+ 22+ 23) * (31+ 32+ 33) * (51)

= 15 * 39 * 5 = 2730

  2. Find the sum of factors of 544 which are perfect squares

  • 544 = 8 * 4 * 17 = 2* 17
  • Our expression, (20+ 21+ 22+ 23+2+25) (170+171)
    Ensure all terms in the expression satisfy the condition, perfect squares.
  • So, each term in the expression should be of even power. We have to remove all odd powers in the expression
  • Required sum is (20+ 2+24) (170) = 21.
  • Number of factors = 3 * 1 (number of terms in each bracket) = 3.
  1. For the number 1000 find the below values
  • The number of divisors = (3 + 1) * (3 + 1) = 16 (as 1000 = 23* 53)
  • The product of divisors = 100016/2 = 10008
  • The sum of divisors= (20+ 21+ 22+ 23) (5+ 51+ 52+ 53) = 2340
  • The sum and number of odd divisors

we need to remove all factors that can yield a 2 from the two’s bracket.

Sum of odd factors = (20) * (50+ 51+ 52+ 53) = 156

Number of odd factors = 1 * 4 = 4

  • The sum and number of even divisors

We need to remove all factors that cannot yield a 2 from the two’s bracket (which is 20)

Sum of even factors = (21+ 22+ 23) (50+ 51+ 52+ 53) = 2184

Number of even factors = 3 * 4 = 12

  • The sum and number of divisors which are perfect squares

We need to remove all the factors which are not perfect squares

Sum of factors which are perfect squares = (20+ 22) (50+ 52) = 130

Number of factors which are perfect squares = 2 * 2 = 4

  • The sum and number of divisors which are not perfect squares

Total sum – sum of divisors which are perfect squares = 2340 – 130 = 2210

Total factors – number of divisors which are perfect squares = 16 – 4 = 12

  • The sum and number of divisors which are divisible by 125

We need to remove all the factors that cannot yield 125 (125 = 53)

Sum of factors which are divisible by 125 = (20+ 21+ 22+ 23) (53) = 1875

Number of factors which are divisible by 125 = 4 * 1 = 4

  • The sum and number of divisors which are divisible by 100

100 = 2* 52

We need to remove all factors that cannot yield a 2from two’s bracket and also remove all factors that cannot yield a 5from five’s bracket.

Sum of factors which are divisible by 100 = (22+ 23) (52+ 53) = 1800

Number of factors which are divisible by 125 = 2 * 2 = 4


12 Average
N/A

Average: An average or an arithmetic mean of given data is the sum of the given observations divided by number of observations.

Properties of Average:

1.     Average of a given data is less than the greatest observation and greater than the smallest observation of the given data.

Ex: Average of 3, 7, 9, and 13 = ((3 + 7 + 9 + 13)/4) = (32/4) = 8

Here Clearly, 8 is less than 13 and greater than 3.

2.      If the observations of given data are equal, then the average will also be the same as observations.

Ex: Average of 6, 6, 6, and 6 = ((6 + 6 + 6 + 6)/4) = (24/4) = 6

3.      If 0 (zero) is one of the observations of a given data, then that 0 (zero) will also be included while calculating average.

Ex: Average of 3, 6, and 0 = ((3 + 6 + 0)/3) = (9/3) = 3

NOTE:

·       If all the numbers get increased by a, then their average must be increased by a

·       If all the numbers get decreased by a, then their average must be decreased by a

·       If all the numbers are multiplied by a, then their average must be multiplied by a

·       If all the numbers are divided by a, then their average must be divided by a

Important Formulae Related to Average of Numbers:

  1. Average of first n natural numbers - <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow><mn>2</mn></mfrac></mfenced></math>
  2. Average of first n even numbers = (n + 1)
  3. Average of first n odd numbers = n 
  4. Average of consecutive numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;First number&#xA0;</mtext><mo>+</mo><mtext>&#xA0;Last number&#xA0;</mtext></mrow><mn>2</mn></mfrac></mfenced></math> 
  5. Average of 1 to n odd numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;Last odd number&#xA0;</mtext><mo>+</mo><mn>1</mn></mrow><mn>2</mn></mfrac></mfenced></math> 
  6. Average of 1 to n even numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;Last even number&#xA0;</mtext><mo>+</mo><mn>2</mn></mrow><mn>2</mn></mfrac></mfenced></math> 
  7. Average of squares of first n natural numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mo>(</mo><mi>n</mi><mo>+</mo><mn>1</mn><mo>)</mo><mo>(</mo><mn>2</mn><mi>n</mi><mo>+</mo><mn>1</mn><mo>)</mo></mrow><mn>6</mn></mfrac></mfenced></math>
  8.  Average of the cubes of first n natural numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>n</mi><mo>(</mo><mi>n</mi><mo>+</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mn>4</mn></mfrac></math> 
  9. Average of n multiples of any number = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;Number&#xA0;</mtext><mi>x</mi><mo>(</mo><mi>n</mi><mo>+</mo><mn>1</mn><mo>)</mo></mrow><mn>2</mn></mfrac></mfenced></math>

 Examples:

1.      Find out the average of 4, 7,10,13, ..., 28, 31.

    Solution. Here, the difference between any two numbers written in continuous sequence is 3.

Hence, this is a series of consecutive numbers.

As, we know, average of consecutive numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mtext>&#xA0;First number + Last number&#xA0;</mtext><mn>2</mn></mfrac></mfenced></math>

Here, first number = 4 and last number = 31

Therefore, required average = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mn>4</mn><mo>+</mo><mn>31</mn></mrow><mn>2</mn></mfrac></mfenced><mo>=</mo><mfenced separators="|"><mfrac><mn>35</mn><mn>2</mn></mfrac></mfenced><mo>=</mo><mn>17.5</mn></math>

2.      Find the average of all the odd numbers and average of all the even numbers from 1 to 45.

Solution: According to the formula,

Average of 1 to n odd numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;Last odd number&#xA0;</mtext><mo>+</mo><mn>1</mn></mrow><mn>2</mn></mfrac></mfenced></math>

Here, the last odd number = 45

Therefore, Average of 1 to 45 odd numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mn>45</mn><mo>+</mo><mn>1</mn></mrow><mn>2</mn></mfrac></mfenced><mo>=</mo><mfenced separators="|"><mfrac><mn>46</mn><mn>2</mn></mfrac></mfenced><mo>=</mo><mn>23</mn></math>

Again, according to the formula,

Average of 1 to n even numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mtext>&#xA0;Last even number&#xA0;</mtext><mo>+</mo><mn>2</mn></mrow><mn>2</mn></mfrac></mfenced></math>

Here, the last odd number = 44

Therefore, Average of 1 to 44 odd numbers = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mn>44</mn><mo>+</mo><mn>2</mn></mrow><mn>2</mn></mfrac></mfenced><mo>=</mo><mfenced separators="|"><mfrac><mn>46</mn><mn>2</mn></mfrac></mfenced><mo>=</mo><mn>23</mn></math>

 

3.      The average salary of the entire staff in an office is $200 per day. The average salary of officers is $550 and that of non-officers is $120. If the number of officers is 16, then find the numbers of non-officers in the office.

Solution: Let number of non-officers = x

Then, 120x + 550 x 16 = 200 (16 + x)

=> 12x + 55 x 16 = 20 (16 + x)

=> 3z + 55 x 4 = 5 (16 + x)

=> 3x + 220 = 80 + 5x

=> 5x - 3x = 220 – 80

=> 2x = 140

=> x = 70

4.      The average runs scored by a cricketer in 42 innings, is 30. The difference between his maximum and minimum scores in an innings is 100. If these two innings are not taken into consideration, then the average score of remaining 40 innings is 28. Calculate the maximum runs scored by him in an innings?

Solution: Let the minimum score = x

Maximum score = x + 100

x + (x + 100) = (30 x 42) – (40 x 28)

2x + 100 = 1260 – 1120

2x + 100 = 140

2x = 140 – 100

2x = 40

x = 20

Hence, the maximum score = x + 100 = 20 + 100 = 120

5.      The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students of A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg, respectively. If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg, what is the ratio of the number of students in sections A and D?

Solution:

Let number of students in the sections A, B, C and D be a, b, c and d, respectively.

Then, total weight of students of section A = 45a

Total weight of students of section B = 50b

Total weight of students of section C = 72c

Total weight of students of section D = 80d

According to the question, Average weight of students of sections A and B = 48 kg

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore,&#xA0;</mtext><mfenced separators="|"><mfrac><mrow><mn>45</mn><mi>a</mi><mo>+</mo><mn>50</mn><mi>b</mi></mrow><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow></mfrac></mfenced><mo>=</mo><mn>48</mn><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>45</mn><mi>a</mi><mo>+</mo><mn>50</mn><mi>b</mi><mo>=</mo><mn>48</mn><mi>a</mi><mo>+</mo><mn>48</mn><mi>b</mi></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>3</mn><mi>a</mi><mo>=</mo><mn>2</mn><mi>b</mi></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>15</mn><mi>a</mi><mo>=</mo><mn>10</mn><mi>b</mi></mtd></mtr></mtable></math>

 

And average weight of students of sections B and C=60kg

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mfrac><mrow><mn>50</mn><mi>b</mi><mo>+</mo><mn>72</mn><mi>c</mi></mrow><mrow><mi>b</mi><mo>+</mo><mi>c</mi></mrow></mfrac></mfenced><mo>=</mo><mn>60</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x21D2;</mo></math> 50b + 72c = 60b + 60c

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x21D2;</mo></math> 10b = 12c

Now, average weight of students of A, B, C and D = 60 kg

45a + 50b + 72c + 80d = 60(a + b + c + d)

 => 15a + 10b - 12c - 20d = 0

 => 15a = 20d

 

=> a: d = 4 :3 

 


13 PERCENTAGE
N/A

 

Percentage:

 

The term per cent means 'for every hundred'.

 

It can be defined as follows

 

 "A per cent is a fraction whose denominator is 100 and the numerator of the fraction is called the rate per cent." Per cent is denoted by the sign '%’.

 

Conversion of Per Cent into Fraction:

 

Expressing per cent (x%) into fraction.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Required fraction&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mfrac><mi>x</mi><mn>100</mn></mfrac></mfenced></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example: Express&#xA0;</mtext><mn>25</mn><mi mathvariant="normal">%</mi><mtext>&#xA0;in fraction&#xA0;</mtext><mo>=&gt;</mo><mn>25</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mn>25</mn><mn>100</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>4</mn></mfrac></math>

Conversion of fraction into Percentage:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Expressing a fraction&#xA0;</mtext><mfenced separators="|"><mfrac><mi>x</mi><mi>y</mi></mfrac></mfenced><mtext>&#xA0;in per cent.</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Required percentage&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mfrac><mi>x</mi><mi>y</mi></mfrac><mo>&#x2217;</mo><mn>100</mn></mrow></mfenced><mi mathvariant="normal">%</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example: convert&#xA0;</mtext><mfenced separators="|"><mrow><mfrac><mn>3</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mn>100</mn></mrow></mfenced><mi mathvariant="normal">%</mi><mo>=</mo><mn>37.5</mn><mi mathvariant="normal">%</mi></math>

Expressing One Quantity as a Per Cent with Respect to Other:

 

To express a quantity as a per cent with respect to other quantity following formula is used

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>T</mi><mi>h</mi><mi>e</mi><mo>&#xA0;</mo><mi>q</mi><mi>u</mi><mi>a</mi><mi>n</mi><mi>t</mi><mi>i</mi><mi>t</mi><mi>y</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mo>&#xA0;</mo><mi>b</mi><mi>e</mi><mo>&#xA0;</mo><mi>e</mi><mi>x</mi><mi>p</mi><mi>r</mi><mi>e</mi><mi>s</mi><mi>s</mi><mi>e</mi><mi>d</mi><mo>&#xA0;</mo><mi>i</mi><mi>n</mi><mo>&#xA0;</mo><mi>p</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mi>c</mi><mi>e</mi><mi>n</mi><mi>t</mi></mrow><mrow><mn>2</mn><mi>n</mi><mi>d</mi><mo>&#xA0;</mo><mi>q</mi><mi>u</mi><mi>a</mi><mi>n</mi><mi>t</mi><mi>i</mi><mi>t</mi><mi>y</mi><mo>(</mo><mi>i</mi><mi>n</mi><mo>&#xA0;</mo><mi>r</mi><mi>e</mi><mi>s</mi><mi>p</mi><mi>e</mi><mi>c</mi><mi>t</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>w</mi><mi>h</mi><mi>i</mi><mi>c</mi><mi>h</mi><mo>&#xA0;</mo><mi>t</mi><mi>h</mi><mi>e</mi><mo>&#xA0;</mo><mi>p</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mi>c</mi><mi>e</mi><mi>n</mi><mi>t</mi><mo>&#xA0;</mo><mi>h</mi><mi>a</mi><mi>s</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mo>&#xA0;</mo><mi>b</mi><mi>e</mi><mo>&#xA0;</mo><mi>o</mi><mi>b</mi><mi>t</mi><mi>a</mi><mi>i</mi><mi>n</mi><mi>e</mi><mi>d</mi><mo>)</mo></mrow></mfrac><mo>&#xA0;</mo><mo>&#xD7;</mo><mo>&#xA0;</mo><mn>100</mn><mi>%</mi></math>

Note: To apply this formula, both the quantities must be in same metric unit

Example: 70 kg is what per cent of 280 kg?

 

Solution: According to formula,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>R</mi><mi>e</mi><mi>q</mi><mi>u</mi><mi>i</mi><mi>r</mi><mi>e</mi><mi>d</mi><mi>p</mi><mi>e</mi><mi>r</mi><mi>c</mi><mi>e</mi><mi>n</mi><mi>t</mi><mi>a</mi><mi>g</mi><mi>e</mi><mo>=</mo><mfrac><mrow><mi>T</mi><mi>h</mi><mi>e</mi><mo>&#xA0;</mo><mi>q</mi><mi>u</mi><mi>a</mi><mi>n</mi><mi>t</mi><mi>i</mi><mi>t</mi><mi>y</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mo>&#xA0;</mo><mi>b</mi><mi>e</mi><mo>&#xA0;</mo><mi>e</mi><mi>x</mi><mi>p</mi><mi>r</mi><mi>e</mi><mi>s</mi><mi>s</mi><mi>e</mi><mi>d</mi><mo>&#xA0;</mo><mi>i</mi><mi>n</mi><mo>&#xA0;</mo><mi>p</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mi>c</mi><mi>e</mi><mi>n</mi><mi>t</mi></mrow><mrow><mn>2</mn><mi>n</mi><mi>d</mi><mo>&#xA0;</mo><mi>q</mi><mi>u</mi><mi>a</mi><mi>n</mi><mi>t</mi><mi>i</mi><mi>t</mi><mi>y</mi><mo>(</mo><mi>i</mi><mi>n</mi><mo>&#xA0;</mo><mi>r</mi><mi>e</mi><mi>s</mi><mi>p</mi><mi>e</mi><mi>c</mi><mi>t</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>w</mi><mi>h</mi><mi>i</mi><mi>c</mi><mi>h</mi><mo>&#xA0;</mo><mi>t</mi><mi>h</mi><mi>e</mi><mo>&#xA0;</mo><mi>p</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mi>c</mi><mi>e</mi><mi>n</mi><mi>t</mi><mo>&#xA0;</mo><mi>h</mi><mi>a</mi><mi>s</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mo>&#xA0;</mo><mi>b</mi><mi>e</mi><mo>&#xA0;</mo><mi>o</mi><mi>b</mi><mi>t</mi><mi>a</mi><mi>i</mi><mi>n</mi><mi>e</mi><mi>d</mi><mo>)</mo></mrow></mfrac><mo>&#xA0;</mo><mo>&#xD7;</mo><mo>&#xA0;</mo><mn>100</mn><mi>%</mi><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mn>70</mn><mn>280</mn></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>100</mn><mn>4</mn></mfrac><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mn>25</mn><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

Formula to calculate Per Cent:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>I</mi><mi>f</mi><mo>&#xA0;</mo><mi>w</mi><mi>e</mi><mo>&#xA0;</mo><mi>h</mi><mi>a</mi><mi>v</mi><mi>e</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mo>&#xA0;</mo><mi>f</mi><mi>i</mi><mi>n</mi><mi>d</mi><mo>&#xA0;</mo><mi>y</mi><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>x</mi><mo>,</mo><mi>t</mi><mi>h</mi><mi>e</mi><mi>n</mi><mo>&#xA0;</mo><mi>y</mi><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>x</mi><mo>=</mo><msup><mi>x</mi><mo>&#x2217;</mo></msup><mfrac><mi>y</mi><mn>100</mn></mfrac></math>

Some quick Results:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>5</mn><mi>%</mi><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>20</mn></mfrac><mspace linebreak="newline"/><mn>10</mn><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>10</mn></mfrac><mspace linebreak="newline"/><mn>12</mn><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>6</mn></mfrac><mspace linebreak="newline"/><mn>20</mn><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>5</mn></mfrac><mspace linebreak="newline"/><mn>25</mn><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>4</mn></mfrac><mspace linebreak="newline"/><mn>50</mn><mi>%</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mi>n</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi><mo>=</mo><mfrac><mrow><mi>N</mi><mi>u</mi><mi>m</mi><mi>b</mi><mi>e</mi><mi>r</mi></mrow><mn>2</mn></mfrac></math>

Examples:

1.     The price of a computer is $ 20000. What will be the price of computer after reduction of 25%?

Solution: Here, x = $ 20000 and y = 25%

 

According to the formula,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;New price&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#x2212;</mo><mi>y</mi></mrow><mn>100</mn></mfrac><mo>&#x2217;</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#x2212;</mo><mn>25</mn></mrow><mn>100</mn></mfrac><mo>&#xD7;</mo><mn>20000</mn><mo>=</mo><mfrac><mn>75</mn><mn>100</mn></mfrac><mo>&#xD7;</mo><mn>20000</mn><mo>=</mo><mn>75</mn><mo>&#xD7;</mo><mn>200</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>15000</mn></math>

2.     The salary of a worker is first increased by 5% and then it is decreased by 5%. What is the change in his salary?

Solution: Let the initial salary of the worker be $ 100.

 

Firstly, the salary of worker is increased by 5%.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;So, increased salary&#xA0;</mtext><mo>=</mo><mn>105</mn><mi mathvariant="normal">%</mi><mtext>&#xA0;of&#xA0;</mtext><mn>100</mn><mo>=</mo><mfrac><mrow><mn>105</mn><mo>&#xD7;</mo><mn>100</mn></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>105</mn></math>

 

Now, the salary is reduced by 5% after the increase.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Reduced salary&#xA0;</mtext><mo>=</mo><mn>95</mn><mi mathvariant="normal">%</mi><mtext>&#xA0;of&#xA0;</mtext><mn>105</mn><mo>=</mo><mfrac><mrow><mn>95</mn><mo>&#xD7;</mo><mn>105</mn></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>99.75</mn></math>

Therefore, required change is a decrease i.e., 100 – 99.75 = 0.25

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;So, required percentage decrease in salary&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>0.25</mn><mo>&#xD7;</mo><mn>100</mn></mrow><mn>100</mn></mfrac><mi mathvariant="normal">%</mi><mo>=</mo><mn>0.25</mn><mi mathvariant="normal">%</mi></math>

3.     Population of a city in 2014 was 1000000. If in 2015 there is an increment of 15%, in 2016 there is a decrement of 35% and in 2017 there is an increment of 45%, then find the population of city at the end of year 2017.

Solution:

 

Given that, P = 1000000, R1 = 15%, R2 = 35% (decrease) and R3 = 45%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Population of city at the end of year&#xA0;</mtext><mn>2017</mn><mo>=</mo><mi>P</mi><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>1</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mrow><mi>R</mi><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>3</mn></mrow><mn>100</mn></mfrac></mrow></mfenced></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mn>1000000</mn><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>15</mn><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>35</mn><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>45</mn><mn>100</mn></mfrac></mrow></mfenced></mtd></mtr><mtr><mtd><mo>=</mo><mn>1000000</mn><mo>&#xD7;</mo><mfenced separators="|"><mfrac><mn>115</mn><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mfenced separators="|"><mfrac><mn>65</mn><mn>100</mn></mfrac></mfenced><mfenced separators="|"><mfrac><mn>145</mn><mn>100</mn></mfrac></mfenced></mtd></mtr><mtr><mtd><mo>=</mo><mn>1083875</mn></mtd></mtr></mtable></math>

4.     A student was asked to measure the length and breadth of a rectangle. By mistake, he measured the length 20% less and the breadth 10% more. If its original area is 200 sq cm, then find the area after this measurement?

 

Solution:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Net effect on area&#xA0;</mtext><mo>=</mo><mfenced open="[" close="]" separators="|"><mrow><mo>&#x2212;</mo><mn>20</mn><mo>+</mo><mn>10</mn><mo>+</mo><mfrac><mrow><mo>(</mo><mo>&#x2212;</mo><mn>20</mn><mo>)</mo><mo>(</mo><mn>10</mn><mo>)</mo></mrow><mn>100</mn></mfrac></mrow></mfenced><mi mathvariant="normal">%</mi><mo>=</mo><mo>(</mo><mo>&#x2212;</mo><mn>10</mn><mo>&#x2212;</mo><mn>2</mn><mo>)</mo><mi mathvariant="normal">%</mi><mo>=</mo><mo>&#x2212;</mo><mn>12</mn><mi mathvariant="normal">%</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Now, after this mistake new area&#xA0;</mtext><mo>=</mo><mo>(</mo><mn>100</mn><mo>&#x2212;</mo><mn>12</mn><mo>)</mo><mi mathvariant="normal">%</mi><mtext>&#xA0;of&#xA0;</mtext><mn>200</mn><mo>=</mo><mfrac><mn>88</mn><mn>100</mn></mfrac><mo>&#xD7;</mo><mn>200</mn><mo>=</mo><mn>176</mn><mi>s</mi><mi>q</mi><mi>c</mi><mi>m</mi></math>

5.       Due to an increase of 30% in the price of eggs, 6 eggs less are available for $7.80. The present rate of eggs per dozen is?

 

Solution: Let the original price per egg be x.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Then, increased price&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mfrac><mn>130</mn><mn>100</mn></mfrac><mi>x</mi></mrow></mfenced></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;According to the question,&#xA0;</mtext><mfrac><mn>7.80</mn><mi>x</mi></mfrac><mo>&#x2212;</mo><mfrac><mn>780</mn><mrow><mn>130</mn><mi>x</mi></mrow></mfrac><mo>=</mo><mn>6</mn><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>1014</mn><mo>&#x2212;</mo><mn>780</mn><mo>=</mo><mn>6</mn><mo>&#x2217;</mo><mn>130</mn><mi>x</mi></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>780</mn><mi>x</mi><mo>=</mo><mn>234</mn></mtd></mtr></mtable><mspace linebreak="newline"/><mspace linebreak="newline"/><mtext>&#xA0;Therefore,&#xA0;</mtext><mi>x</mi><mo>=</mo><mfrac><mn>234</mn><mn>780</mn></mfrac><mo>=</mo><mn>0.3</mn><mspace linebreak="newline"/><mspace linebreak="newline"/><mtext>&#xA0;Therefore, present price per dozen&#xA0;</mtext><mo>=</mo><mi mathvariant="normal">$</mi><mfenced separators="|"><mrow><mn>12</mn><mo>&#xD7;</mo><mfrac><mn>130</mn><mn>100</mn></mfrac><mo>&#xD7;</mo><mn>0.3</mn></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mn>4.68</mn><mo>.</mo></math>

 


14 Profit and Loss
N/A

Profit and Loss: Profit and loss are the terms related to monetary transactions in trade and business. Whenever a purchased article is sold, then either profit is earned or loss is incurred.

Cost Price (CP) This is the price at which an article is purchased or manufactured.

Selling Price (SP) This is the price at which an article is sold.

Overhead Charges

Such charges are the extra expenditures on purchased goods apart from actual cost price. Such charges include freight charges, rent, salary of employees, repairing cost on purchased articles etc.

Note:

 If overhead charges are not specified in the question, then they are not considered

 Profit (SP>CP) When an article is sold at a price more than its cost price, then profit is earned,

 Loss (CP>SP) When an article is sold at a price lower than its cost price, then loss is incurred.

Basic Formulae Related to Profit and Loss:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;1.&#xA0;&#xA0;Profit&#xA0;</mtext><mo>=</mo><mi>S</mi><mi>P</mi><mo>&#x2212;</mo><mi>C</mi><mi>P</mi><mspace linebreak="newline"/><mtext>&#xA0;2.&#xA0;</mtext><mi>L</mi><mi>o</mi><mi>s</mi><mi>s</mi><mo>=</mo><mi>C</mi><mi>P</mi><mo>&#x2212;</mo><mi>S</mi><mi>P</mi><mspace linebreak="newline"/><mtext>&#xA0;3. Profit&#xA0;</mtext><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mtext>&#xA0;Profit&#xA0;</mtext><mtext>&#xA0;Cost price&#xA0;</mtext></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mi>P</mi><mrow><mi>C</mi><mi>P</mi></mrow></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mspace linebreak="newline"/><mtext>&#xA0;4.&#xA0;&#xA0;Loss%&#xA0;</mtext><mo>=</mo><mfrac><mtext>&#xA0;Loss&#xA0;</mtext><mtext>&#xA0;Cost price&#xA0;</mtext></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mi mathvariant="normal">L</mi><mi>CP</mi></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mspace linebreak="newline"/><mtext>&#xA0;5.&#xA0;</mtext><mi>SP</mi><mo>=</mo><mfenced separators="|"><mfrac><mrow><mn>100</mn><mo>+</mo><mtext>&#xA0;Gain&#xA0;</mtext><mi mathvariant="normal">%</mi></mrow><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mi>CP</mi><mspace linebreak="newline"/><mtext>&#xA0;6.&#xA0;</mtext><mspace width="1em"/><mi>CP</mi><mo>=</mo><mfenced separators="|"><mfrac><mn>100</mn><mrow><mn>100</mn><mo>+</mo><mtext>&#xA0;Gain&#xA0;</mtext><mi mathvariant="normal">%</mi></mrow></mfrac></mfenced><mo>&#xD7;</mo><mi>SP</mi><mspace linebreak="newline"/><mtext>&#xA0;7.&#xA0;</mtext><mi>SP</mi><mo>=</mo><mfenced separators="|"><mfrac><mrow><mn>100</mn><mo>&#x2212;</mo><mtext>&#xA0;Loss&#xA0;</mtext><mi mathvariant="normal">%</mi></mrow><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mi>CP</mi><mspace linebreak="newline"/><mtext>&#xA0;8.&#xA0;</mtext><mi>CP</mi><mo>=</mo><mfenced separators="|"><mfrac><mn>100</mn><mrow><mn>100</mn><mo>&#x2212;</mo><mi>Loss</mi><mi mathvariant="normal">%</mi></mrow></mfrac></mfenced><mo>&#xD7;</mo><mi>SP</mi></math>

Things to Keep in Mind:

·       Profit and loss are always calculated on cost price unless otherwise stated in the question.

·        If an article is sold at a certain gain (say 45%), then SP = 145% of CP

 

·        If an article is sold at a certain loss (say 25%), then SP =75% of CP 

Examples:

          1.    Find the SP, when CP is $100 and gain is 20%.

Sol. Given, CP = $100 and gain = 20%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>we know that,&#xA0;</mtext><mi>S</mi><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mrow><mn>100</mn><mo>+</mo><mtext>&#xA0;Gain&#xA0;</mtext><mi mathvariant="normal">%</mi></mrow><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mi>C</mi><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mrow><mn>100</mn><mo>+</mo><mn>20</mn></mrow><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><mfenced separators="|"><mfrac><mn>120</mn><mn>100</mn></mfrac></mfenced><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>120</mn></math>

2.     A person sold a table at $2000 and got a loss of 20%. At what price should he sell it to gain 20%?

Sol. Given SP = $2000 and loss = 20%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>We know that,&#xA0;</mtext><mi>C</mi><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mn>100</mn><mrow><mn>100</mn><mo>&#x2212;</mo><mi>Loss</mi><mo>&#x2061;</mo><mi mathvariant="normal">%</mi></mrow></mfrac></mfenced><mo>&#xD7;</mo><mi>S</mi><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mn>100</mn><mn>80</mn></mfrac></mfenced><mo>&#xD7;</mo><mn>2000</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>2500</mn></math>

Now, CP = $ 2500 and gain = 20%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, Required SP&#xA0;</mtext><mo>=</mo><mn>2500</mn><mo>&#xD7;</mo><mfenced separators="|"><mfrac><mn>120</mn><mn>100</mn></mfrac></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mn>3000</mn></math>

3.     A woman bought eggs at $ 30 per dozen. The selling price per hundred so as to gain 12% will be (in $)

Sol. Given 12 eggs cost = $ 30

 <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Then, cost price of&#xA0;</mtext><mn>1</mn><mtext>&#xA0;egg&#xA0;</mtext><mo>=</mo><mfrac><mn>30</mn><mn>12</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>2.5</mn></math>

Therefore, Cost Price of 100 eggs = 2.5 x 100 = $ 250

Now, let the SP of 100 eggs be $ u.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Then,&#xA0;</mtext><mfrac><mrow><mi>S</mi><mi>P</mi><mo>&#x2212;</mo><mi>C</mi><mi>P</mi></mrow><mrow><mi>C</mi><mi>P</mi></mrow></mfrac><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><mtext>&#xA0;Profit&#xA0;</mtext><mi mathvariant="normal">%</mi><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi mathvariant="normal">u</mi><mo>&#x2212;</mo><mn>250</mn></mrow><mn>250</mn></mfrac><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><mn>12</mn></mtd></mtr><mtr><mtd><mfrac><mrow><mi mathvariant="normal">u</mi><mo>&#x2212;</mo><mn>250</mn></mrow><mn>5</mn></mfrac><mo>&#xD7;</mo><mn>2</mn><mo>=</mo><mn>12</mn></mtd></mtr><mtr><mtd><mi mathvariant="normal">u</mi><mo>&#x2212;</mo><mn>250</mn><mo>=</mo><mfrac><mrow><mn>12</mn><mo>&#xD7;</mo><mn>5</mn></mrow><mn>2</mn></mfrac><mo>=</mo><mn>30</mn></mtd></mtr><mtr><mtd><mi mathvariant="normal">u</mi><mo>=</mo><mn>250</mn><mo>+</mo><mn>30</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>280</mn></mtd></mtr></mtable></math>

The selling price per hundred so as to gain 12% will be $280.

4.     A person sold his watch for $ 75 and got a percentage profit equal to the cost price. The cost price of the watch is 

Sol: Let CP of the watch be $ v.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>According to the question,&#xA0;</mtext><mfrac><mrow><mn>75</mn><mo>&#x2212;</mo><mi>v</mi></mrow><mi>v</mi></mfrac><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><mi>v</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mn>75</mn><mo>&#x2212;</mo><mi>v</mi><mo>)</mo><mo>&#xD7;</mo><mn>100</mn><mo>=</mo><msup><mi>v</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mi>v</mi><mn>2</mn></msup><mo>+</mo><mn>100</mn><mi>v</mi><mo>&#x2212;</mo><mn>7500</mn><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mi>v</mi><mn>2</mn></msup><mo>+</mo><mn>150</mn><mi>v</mi><mo>&#x2212;</mo><mn>50</mn><mi>v</mi><mo>&#x2212;</mo><mn>7500</mn><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>v</mi><mo>(</mo><mi>v</mi><mo>+</mo><mn>150</mn><mo>)</mo><mo>&#x2212;</mo><mn>50</mn><mo>(</mo><mi>v</mi><mo>+</mo><mn>150</mn><mo>)</mo><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mi>v</mi><mo>+</mo><mn>150</mn><mo>)</mo><mo>(</mo><mi>v</mi><mo>&#x2212;</mo><mn>50</mn><mo>)</mo><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>v</mi><mo>=</mo><mn>50</mn></mtd></mtr></mtable></math>

Therefore, CP of the watch = $ 50

5.     Two boxes of onions with equal quantity, one costing $ 10 per kg and the other costing $ 15 per kg, are mixed together in to a bag and whole bag is sold at $ 15 per kg. What is the profit or loss?

Sol: Let each box contains x kg onion,then total cost price of these two boxes together(bag) = 10x + 15x = 25x

Selling price of whole bag = 15(x + x) = 15(2x) = 30x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Profit percentage&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>30</mn><mi>x</mi><mo>&#x2212;</mo><mn>25</mn><mi>x</mi></mrow><mrow><mn>25</mn><mi>x</mi></mrow></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mrow><mn>5</mn><mi>x</mi></mrow><mrow><mn>25</mn><mi>x</mi></mrow></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mn>20</mn><mi mathvariant="normal">%</mi></math>

Therefore, profit percentage = 20%


15 Discount
N/A

Discount is defined as the amount of rebate given on a fixed price (called as marked price) of an article. It is given by merchants/ shopkeepers to increase their sales by attracting customers.

Discount = Marked Price - Selling Price

Marked Price (List Price):

The price on the label of an article/product is called the marked price or list price.

This is the price at which product is intended to be sold.

However, there can be some discount given on this price and actual selling price of the product may be less than the marked price.

 It is generally denoted by MP.

Selling Price:

Selling price = Marked price - Discount

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Selling price&#xA0;</mtext><mo>=</mo><mtext>&#xA0;Marked price&#xA0;</mtext><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mi>r</mi><mn>100</mn></mfrac></mrow></mfenced></math>

where, r% is the rate of discount allowed

Note: Discount is always calculated with respect to marked price of an article

Successive Discount:

When a series of discounts (one after the other) are allowed on marked price of an article, then these discounts are called successive discounts.

Let r1%, r2%, r3%......be the series of discounts on an article with marked price of $ P, then the selling price of the article after all the discounts is given as

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mi>P</mi><mo>=</mo><mi>P</mi><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mrow><mi>r</mi><mn>1</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mrow><mi>r</mi><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mrow><mi>r</mi><mn>3</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>&#xD7;</mo><mo>&#x2026;</mo><mo>&#x2026;</mo></math>

Basic Formulae Related to Discount:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Discount&#xA0;</mtext><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mtext>&#xA0;Marked price-Selling price&#xA0;</mtext><mtext>&#xA0;Marked price&#xA0;</mtext></mfrac><mo>&#xD7;</mo><mn>100</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Discount&#xA0;</mtext><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mtext>&#xA0;Discount&#xA0;</mtext><mtext>&#xA0;Marked price&#xA0;</mtext></mfrac><mo>&#xD7;</mo><mn>100</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Selling price&#xA0;</mtext><mo>=</mo><mtext>&#xA0;Marked price&#xA0;</mtext><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mi>r</mi><mn>100</mn></mfrac></mrow></mfenced></math>

Examples:

  1. What will be a single equivalent discount for successive discounts of 10% and 5% on marked price of an article?

Sol: Given r1 = 10% and r2 = 5%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>Single equivalent discount&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><msub><mi>r</mi><mn>1</mn></msub><mo>+</mo><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2212;</mo><mfrac><mrow><mi>r</mi><mn>1</mn><mo>&#xD7;</mo><mi>r</mi><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mfenced separators="|"><mrow><mn>10</mn><mo>+</mo><mn>5</mn><mo>&#x2212;</mo><mfrac><mrow><mn>10</mn><mo>&#xD7;</mo><mn>5</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mfenced separators="|"><mrow><mn>15</mn><mo>&#x2212;</mo><mfrac><mn>50</mn><mn>100</mn></mfrac></mrow></mfenced><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mo>(</mo><mn>15</mn><mo>&#x2212;</mo><mn>0.5</mn><mo>)</mo><mi mathvariant="normal">%</mi></mtd></mtr><mtr><mtd><mo>=</mo><mn>14.5</mn><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

Therefore, single equivalent discount = 14.5%

  2.  A man bought an article listed at $ 1500 with a discount of 20% offered on the list price. What additional discount must be offered to the man to bring the net price to $ 1104?

Sol: Listed price of an article = $ 1500

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, price after first discount&#xA0;</mtext><mo>=</mo><mn>1500</mn><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>20</mn><mn>100</mn></mfrac></mrow></mfenced><mo>=</mo><mn>1500</mn><mo>&#xD7;</mo><mfrac><mn>4</mn><mn>5</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>1200</mn></math>

Now, second discount = 1200 – 1104 = $ 96

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, required percentage&#xA0;</mtext><mo>=</mo><mfrac><mn>96</mn><mn>1200</mn></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mn>8</mn><mi mathvariant="normal">%</mi></math>

  3.  A man purchased a shirt and pant with a discount of 25% on its marked price. He sold them at a price 40% more than the price at which he bought them. How much per cent the new selling price to its marked price?

Sol:  Let the original price of pant and shirt to be $ p

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, Cost price of pant and shirt&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>p</mi><mo>&#xD7;</mo><mo>(</mo><mn>100</mn><mo>&#x2212;</mo><mn>25</mn><mo>)</mo></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mrow><mn>3</mn><mi>p</mi></mrow><mn>4</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>And selling price of shirt and pant&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>3</mn><mi>p</mi></mrow><mn>4</mn></mfrac><mo>&#xD7;</mo><mfenced separators="|"><mfrac><mrow><mn>100</mn><mo>+</mo><mn>40</mn></mrow><mn>100</mn></mfrac></mfenced><mo>=</mo><mfrac><mrow><mn>3</mn><mi>p</mi></mrow><mn>4</mn></mfrac><mo>&#xD7;</mo><mfenced separators="|"><mfrac><mn>140</mn><mn>100</mn></mfrac></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mn>21</mn><mn>20</mn></mfrac><mi>p</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, required percentage&#xA0;</mtext><mo>=</mo><mfrac><mrow><mfrac><mn>21</mn><mn>20</mn></mfrac><mi>p</mi><mo>&#x2212;</mo><mi>p</mi></mrow><mi>p</mi></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mn>100</mn><mn>20</mn></mfrac><mi mathvariant="normal">%</mi><mo>=</mo><mn>5</mn><mi mathvariant="normal">%</mi></math>

  4.  A dozen pair of socks quoted at $ 80 are available at a discount of 10%. How many pair of socks can be bought for $ 24?

Sol:  Since MP of one dozen of pairs of socks = $ 80

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, SP of one dozen of pairs of socks&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>80</mn><mo>&#xD7;</mo><mo>(</mo><mn>100</mn><mo>&#x2212;</mo><mn>10</mn><mo>)</mo></mrow><mn>100</mn></mfrac><mo>=</mo><mfrac><mrow><mn>80</mn><mo>&#xD7;</mo><mo>(</mo><mn>90</mn><mo>)</mo></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>72</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, required number of pairs of socks purchased for&#xA0;</mtext><mi mathvariant="normal">$</mi><mn>24</mn><mo>=</mo><mfrac><mrow><mo>(</mo><mn>12</mn><mo>&#xD7;</mo><mn>24</mn><mo>)</mo></mrow><mn>72</mn></mfrac><mo>=</mo><mn>4</mn></math>

  5.  A seller marks his goods 30% above their cost price but allows 15% discount for cash payment. His percentage of profit when sold in cash, is

Sol: Let CP of the goods = $ p

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, Marked price of the goods&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>p</mi><mi>x</mi><mo>(</mo><mn>100</mn><mo>+</mo><mn>30</mn><mo>)</mo></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mrow><mo>(</mo><mn>13</mn><mi>p</mi><mo>)</mo></mrow><mn>10</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;Now, SP of the goods&#xA0;</mtext><mo>=</mo><mfrac><mrow><mo>(</mo><mn>13</mn><mi mathvariant="normal">p</mi><mo>)</mo></mrow><mn>10</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>85</mn><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mn>221</mn><mn>200</mn></mfrac><mi mathvariant="normal">p</mi><mspace linebreak="newline"/><mtext>&#xA0;Profit&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mfrac><mn>221</mn><mn>200</mn></mfrac><mi mathvariant="normal">p</mi><mo>&#x2212;</mo><mi mathvariant="normal">p</mi></mrow></mfenced><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mn>21</mn><mn>200</mn></mfrac><mi mathvariant="normal">p</mi><mspace linebreak="newline"/><mtext>&#xA0;Hence, profit per cent&#xA0;</mtext><mo>=</mo><mfrac><mrow><mfrac><mn>21</mn><mn>200</mn></mfrac><mi mathvariant="normal">p</mi></mrow><mi mathvariant="normal">p</mi></mfrac><mo>&#xD7;</mo><mn>100</mn><mi mathvariant="normal">%</mi><mo>=</mo><mfrac><mn>2100</mn><mn>200</mn></mfrac><mo>=</mo><mn>10.5</mn><mi mathvariant="normal">%</mi></math>


16 Simple Interest
N/A

When a person borrows some amount of money from another person or organisation (bank), then the person borrowing money (borrower) pays some extra money during repayment, that extra money during repayment is called interest

Principal (P) Principal is the money borrowed or deposited for a certain time.

Amount (A) The sum of principal and interest is called amount

Amount = Principal + Simple Interest

Rate of Interest (R) It is the rate at which the interest is charged on principal. It is always specified in percentage terms.

Time (T) The period, for which the money is borrowed or deposited, is called time.

Simple Interest (SI):

If the interest is calculated on the original principal for any length of time, then it is called simple interest

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Simple interest&#xA0;</mtext><mo>=</mo><mfrac><mrow><mtext>&#xA0;Principal&#xA0;</mtext><mo>&#xD7;</mo><mtext>&#xA0;Rate&#xA0;</mtext><mo>&#xD7;</mo><mtext>&#xA0;Interest&#xA0;</mtext></mrow><mn>100</mn></mfrac><mfrac><mrow><mi>P</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mtext>T&#xA0;</mtext></mrow><mn>100</mn></mfrac></math>

Basic Formulae Related to Simple Interest:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>P</mi><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#xD7;</mo><mi>A</mi></mrow><mrow><mn>100</mn><mo>+</mo><mi>R</mi><mi>T</mi></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>S</mi><mi>l</mi><mo>=</mo><mfrac><mrow><mi>A</mi><mi>R</mi><mi>T</mi></mrow><mrow><mn>100</mn><mo>+</mo><mi>R</mi><mi>T</mi></mrow></mfrac><mo>,</mo></mtd></mtr></mtable></math>

Where, SI = Simple Interest, P = Principal, R = Rate of interest, T = Time, A = Amount.

Things to remember:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;If rate of interest is half-yearly, then rate&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mfrac><mi>R</mi><mn>2</mn></mfrac></mfenced><mi mathvariant="normal">%</mi><mtext>&#xA0;and time&#xA0;</mtext><mo>=</mo><mn>2</mn><mi>T</mi><mspace linebreak="newline"/><mtext>&#x2219;&#xA0;If rate of interest is quarterly, then rate&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mfrac><mi>R</mi><mn>4</mn></mfrac></mfenced><mi mathvariant="normal">%</mi><mtext>&#xA0;and time&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>T</mi><mspace linebreak="newline"/><mtext>&#x2219;&#xA0;If rate of interest is monthly, then rate&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mfrac><mi>R</mi><mn>12</mn></mfrac></mfenced><mi mathvariant="normal">%</mi><mtext>&#xA0;and time&#xA0;</mtext><mo>=</mo><mn>12</mn><mi>T</mi></math>

  • To calculate interest, the day on which amount is deposited, is not counted but the day on which amount is withdrawn is counted.

     Instalments:

When a borrower paid the total money in some equal parts (i.e., not in a single amount), then we say that he/she is paying in instalments

The important point is that borrower has to also pay the interest for using the borrowed sum or purchased article.

In general, the value of each instalment is kept constant even when the interest charged on each instalment vary for each instalment for n equal instalments we only calculate up to (n -1) term.

For simple interest

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>A</mi><mo>=</mo><mfenced open="[" close="]" separators="|"><mrow><mi>x</mi><mo>+</mo><mfenced separators="|"><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><mi>x</mi><mo>&#x2217;</mo><mi>R</mi><mo>&#x2217;</mo><mn>1</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>+</mo><mfenced separators="|"><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><mi>x</mi><mo>&#x2217;</mo><mi>R</mi><mo>&#x2217;</mo><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>+</mo><mfenced separators="|"><mrow><mi>x</mi><mo>+</mo><mfrac><mrow><mi>x</mi><mo>&#x2217;</mo><mi>R</mi><mo>&#x2217;</mo><mn>3</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mo>+</mo><mo>&#x2026;</mo><mo>&#x2026;</mo></mrow></mfenced></math>

Where, A = Total amount paid;

x = Value of each instalment.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Also,&#xA0;</mtext><mi>A</mi><mo>=</mo><mi>P</mi><mo>+</mo><mfrac><mrow><mi>P</mi><mo>&#x2217;</mo><mi>n</mi><mo>&#x2217;</mo><mi>R</mi></mrow><mn>100</mn></mfrac></math>

Where, P is the principal n is the number of instalments R is the rate of interest

Examples:

  1. A private finance company A claims to be lending money at simple interest. But the company includes the interest every 6 months for calculating principal. If company A is charging an interest of 10%, the effective rate of interest after 1 yr becomes

Sol: Let the sum be $100

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>SI for first&#xA0;</mtext><mn>6</mn><mtext>&#xA0;months&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#x2217;</mo><mn>10</mn><mo>&#x2217;</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></mrow><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>5</mn></math>

Now, principal becomes 100 + 5 = 105

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>SI for last&#xA0;</mtext><mn>6</mn><mtext>&#xA0;months&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>105</mn><mo>&#x2217;</mo><mn>10</mn><mo>&#x2217;</mo><mn>1</mn></mrow><mrow><mn>100</mn><mo>&#x2217;</mo><mn>2</mn></mrow></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>5.25</mn></math>

Hence, amount at the end of 1 year = 105 + 5.25 = $ 110.25

Therefore, Effective SI = 110.25 – 100 = $ 10.25%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Effective rate&#xA0;</mtext><mo>(</mo><mi>R</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#x2217;</mo><mi>S</mi><mi>I</mi></mrow><mrow><mi>P</mi><mo>&#xD7;</mo><mi>T</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#x2217;</mo><mn>10.25</mn></mrow><mrow><mn>100</mn><mo>&#x2217;</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>10.25</mn><mi mathvariant="normal">%</mi></math>

  1. Jack lent out $ 8750 at 7% annual interest. Find the simple interest in 3 yr.

Sol: Given, t = 3 yr, r = 7%, P = $ 8750

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mi>i</mi><mi>n</mi><mi>c</mi><mi>e</mi><mo>,</mo><mi>S</mi><mi>I</mi><mo>=</mo><mfrac><mrow><mo>&#xA0;</mo><mi>P</mi><mi>R</mi><mi>T</mi></mrow><mn>100</mn></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><mi>S</mi><mi>I</mi><mo>=</mo><mo>(</mo><mfrac><mrow><mn>8750</mn><mo>&#xA0;</mo><mi>x</mi><mo>&#xA0;</mo><mn>7</mn><mo>&#xA0;</mo><mi>x</mi><mo>&#xA0;</mo><mn>3</mn></mrow><mn>100</mn></mfrac><mo>)</mo></math>

 SI = $ 1837.50

  1. Simple interest for the sum of $ 1230 for 2 yr is $ 10 more than the simple interest for $ 1130 for the same duration. Find the rate of interest

Sol: According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mn>1230</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mi>R</mi></mrow><mn>100</mn></mfrac><mo>&#x2212;</mo><mfrac><mrow><mn>1130</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mi>R</mi></mrow><mn>100</mn></mfrac><mo>=</mo><mn>10</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>200</mn><mn>100</mn></mfrac><mi>R</mi><mo>=</mo><mn>10</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>R</mi><mo>=</mo><mn>5</mn><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

  4.  A principal amount to $ 944 in 3 yr and to $ 1040 in 5 yr, each sum being invested at the same simple interest. The principal was

Sol: Let the principal be $ p

Rate of interest = R%

Case I:

P = $ p, T = 3 yr

R = R%, SI = $(944 – p)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>S</mi><mi>I</mi><mo>=</mo><mfrac><mrow><mi>P</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>=</mo><mfrac><mrow><mi>p</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mn>3</mn></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mn>100</mn><mo>(</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mrow><mn>3</mn><mi>p</mi></mrow></mfrac><mo>=</mo><mi>R</mi><mo>&#x2212;</mo><mo>&gt;</mo><mo>(</mo><mn>1</mn><mo>)</mo></mtd></mtr></mtable></math>

Case II:

P = $ p, T = 5 yr

R = R%, SI = $(1040 – p)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>S</mi><mi>I</mi><mo>=</mo><mfrac><mrow><mi>P</mi><mi>x</mi><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>=</mo><mfrac><mrow><mi>p</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mn>5</mn></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mn>100</mn><mo>(</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mrow><mn>5</mn><mi>p</mi></mrow></mfrac><mo>=</mo><mi>R</mi><mo>&#x2212;</mo><mo>&gt;</mo><mo>(</mo><mn>2</mn><mo>)</mo></mtd></mtr></mtable></math>

From Eq (1) and (2), we get

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mn>100</mn><mo>(</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mrow><mn>3</mn><mi>p</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>100</mn><mo>(</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mrow><mn>5</mn><mi>p</mi></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mo>(</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mn>3</mn></mfrac><mo>=</mo><mfrac><mrow><mo>(</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mn>5</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mo>(</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mn>3</mn></mfrac><mo>=</mo><mfrac><mrow><mo>(</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow><mn>5</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mn>944</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo><mo>&#xD7;</mo><mn>5</mn><mo>=</mo><mo>(</mo><mn>1040</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo><mo>&#xD7;</mo><mn>3</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>4720</mn><mo>&#x2212;</mo><mn>5</mn><mi>p</mi><mo>=</mo><mn>3120</mn><mo>&#x2212;</mo><mn>3</mn><mi>p</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>4720</mn><mo>&#x2212;</mo><mn>3120</mn><mo>=</mo><mn>2</mn><mi>p</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>1600</mn><mo>=</mo><mn>2</mn><mi>p</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>p</mi><mo>=</mo><mi mathvariant="normal">$</mi><mn>800</mn></mtd></mtr></mtable></math>

Therefore, p = $ 800

  1. A person invests $ 12000 as fixed deposit at a bank at the rate of 10% per annum simple interest. But due to some pressing needs, he has to withdraw the entire money after 3 yr, for which the bank allowed him a lower rate of interest. If he gets $ 3320 less than, what he would have got at the end of 5 yr, the rate of interest allowed by bank is

Sol: Let the rate of interest allowed by bank be r%

According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mn>12000</mn><mo>&#xD7;</mo><mn>5</mn><mo>&#xD7;</mo><mn>10</mn></mrow><mn>100</mn></mfrac><mo>&#x2212;</mo><mfrac><mrow><mn>12000</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mi>r</mi></mrow><mn>100</mn></mfrac><mo>=</mo><mn>3320</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>6000</mn><mo>&#x2212;</mo><mn>360</mn><mi>r</mi><mo>=</mo><mn>3320</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>360</mn><mi>r</mi><mo>=</mo><mn>6000</mn><mo>&#x2212;</mo><mn>3320</mn><mo>=</mo><mn>2680</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>r</mi><mo>=</mo><mfrac><mn>2680</mn><mn>360</mn></mfrac><mo>=</mo><mn>7</mn><mfrac><mn>4</mn><mn>9</mn></mfrac><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>The rate of interest allowed by bank&#xA0;</mtext><mo>=</mo><mn>7</mn><mfrac><mn>4</mn><mn>9</mn></mfrac><mi mathvariant="normal">%</mi></math>


17 Compound Interest
N/A

As we know that when we borrow some money from bank or any person, then we have to pay some extra money at the time of repaying. This extra money is known as interest.

If interest accrued on principal, it is known as simple interest.

Sometimes it happens that we repay the borrow money some late.

After the completion of specific period, interest accrued on principal as well as interest due of the principal. Then, it is known as compound interest.

Compound Interest = Amount – Principal

Basic Formulae Related to Compound Interest

Let principal = P, rate = R% pa and time = n yr

  1.  If interest is compounded annually,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>then Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup></math>

Compound interest = Amount – Principal

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Compound interest&#xA0;</mtext><mo>=</mo><mi>P</mi><mfenced open="[" close="]" separators="|"><mrow><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup><mo>&#x2212;</mo><mn>1</mn></mrow></mfenced></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;2. If interest is compounded half-yearly, then&#xA0;</mtext><mi>R</mi><mo>=</mo><mfrac><mi>R</mi><mn>2</mn></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>n</mi><mo>=</mo><mn>2</mn><mi>n</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mrow><mn>2</mn><mo>&#xD7;</mo><mn>100</mn></mrow></mfrac></mrow></mfenced><mrow><mn>2</mn><mi>n</mi></mrow></msup></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;3. If the amount is compounded quarterly, then&#xA0;</mtext><mi>R</mi><mo>=</mo><mfrac><mi>R</mi><mn>4</mn></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>n</mi><mo>=</mo><mn>4</mn><mi>n</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mrow><mn>4</mn><mo>&#xD7;</mo><mn>100</mn></mrow></mfrac></mrow></mfenced><mrow><mn>4</mn><mi>n</mi></mrow></msup></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;4. If interest is compounded annually but time is in fraction (suppose time&#xA0;</mtext><mo>=</mo><mi>n</mi><mfrac><mi>a</mi><mi>b</mi></mfrac><mi>y</mi><mi>r</mi><mtext>&#xA0;),&#xA0;</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;then Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mfrac><mi>a</mi><mi>b</mi></mfrac><mi>R</mi></mrow><mn>100</mn></mfrac></mrow></mfenced></math>

  5. If rates of interest are R1%, R2% and R3% for 1st, 2nd and 3rd years respectively, then

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>3</mn></mrow><mn>100</mn></mfrac></mrow></mfenced></math>

Instalments:

When a borrower pays the sum in parts, then we say that he/she is paying in instalments.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>=</mo><mfenced open="[" close="]" separators="|"><mrow><mfrac><mi>x</mi><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced></mfrac><mo>+</mo><mfrac><mi>x</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mfrac><mo>+</mo><mfrac><mi>x</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mn>3</mn></msup></mfrac><mo>+</mo><mo>&#x2026;</mo><mo>&#x2026;</mo><mo>+</mo><mfrac><mi>x</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup></mfrac></mrow></mfenced></math>

x = value of each instalment

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Total amount paid in instalments,&#xA0;</mtext><mi>A</mi><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup></math>

n = Number of instalments

Examples:

  1. What will be the present worth of $ 169 due in 2 yr at 4% pa compound interest?

Sol: Given, R = 4%, n = 2 years and A = $ 169 and P =?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;According to the formula, Amount&#xA0;</mtext><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>n</mi></msup><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mn>169</mn><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>4</mn><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>169</mn><mo>=</mo><mi>P</mi><msup><mfenced separators="|"><mfrac><mn>26</mn><mn>25</mn></mfrac></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mrow><mn>169</mn><mo>&#xD7;</mo><mn>25</mn><mo>&#xD7;</mo><mn>25</mn></mrow><mrow><mn>26</mn><mo>&#xD7;</mo><mn>26</mn></mrow></mfrac></mfenced></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>P</mi><mo>=</mo><mfenced separators="|"><mfrac><mn>105625</mn><mn>676</mn></mfrac></mfenced></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>P</mi><mo>=</mo><mi mathvariant="normal">$</mi><mn>156.25</mn></mtd></mtr></mtable></math>

  2. A sum of $ 400 amounts to $ 441 in 2 yr. What will be its amount, if the rate of interest is increased by 5%?

Sol: According to the given condition,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mn>441</mn><mo>=</mo><mn>400</mn><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x2192;</mo><mfrac><mn>441</mn><mn>400</mn></mfrac><mo>=</mo><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x2192;</mo><msup><mfenced separators="|"><mfrac><mn>21</mn><mn>20</mn></mfrac></mfenced><mn>2</mn></msup><mo>=</mo><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>21</mn><mn>20</mn></mfrac><mo>=</mo><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x2192;</mo><mfrac><mn>21</mn><mn>20</mn></mfrac><mo>&#x2212;</mo><mn>1</mn><mo>=</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x2192;</mo><mfrac><mi>R</mi><mn>100</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>20</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x2192;</mo><mi>R</mi><mo>=</mo><mn>5</mn><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

New rate = 5 + 5 = 10%

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, amount&#xA0;</mtext><mo>=</mo><mn>400</mn><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>10</mn><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup><mspace linebreak="newline"/><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mn>400</mn><mo>&#xD7;</mo><mfrac><mn>11</mn><mn>10</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>11</mn><mn>10</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mi mathvariant="normal">$</mi><mn>484</mn></mtd></mtr></mtable></math>

  3. A sum, at the compound rate of interest, becomes 5/2 A times in 6 yr. The same sum becomes what times in 18 yr?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Sol: If sum is&#xA0;</mtext><mi>x</mi><mtext>, then&#xA0;</mtext><mi>x</mi><mtext>&#xA0;becomes&#xA0;</mtext><mfrac><mn>5</mn><mn>2</mn></mfrac><mtext>&#xD7;&#xA0;in&#xA0;</mtext><mn>6</mn><mtext>&#xA0;years&#xA0;</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x2192;</mo><mfrac><mn>5</mn><mn>2</mn></mfrac><mi>x</mi><mtext>&#xA0;becomes&#xA0;</mtext><mfrac><mn>25</mn><mn>4</mn></mfrac><mi>x</mi><mtext>&#xA0;in&#xA0;</mtext><mn>12</mn><mtext>&#xA0;years</mtext><mspace linebreak="newline"/><mspace linebreak="newline"/><mo stretchy="false">&#x2192;</mo><mfrac><mn>25</mn><mn>4</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;becomes&#xA0;</mtext><mfrac><mn>125</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;in&#xA0;</mtext><mn>18</mn><mtext>&#xA0;years</mtext><mspace linebreak="newline"/><mspace linebreak="newline"/><mtext>Thus, the sum becomes&#xA0;</mtext><mfrac><mn>125</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;in&#xA0;</mtext><mn>18</mn><mtext>&#xA0;years&#xA0;</mtext></math>

  1.  A borrowed sum was paid in the two annual instalments of $ 121 each. If the rate of compound interest is 10% pa, what sum was borrowed?

Sol: According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;Borrowed sum&#xA0;</mtext><mo>=</mo><mfrac><mn>121</mn><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>10</mn><mn>100</mn></mfrac></mrow></mfenced></mfrac><mo>+</mo><mfrac><mn>121</mn><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>10</mn><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mfrac><mn>121</mn><mfenced separators="|"><mfrac><mn>11</mn><mn>10</mn></mfrac></mfenced></mfrac><mo>+</mo><mfrac><mn>121</mn><mfenced separators="|"><mrow><mfrac><mn>11</mn><mn>10</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>11</mn><mn>10</mn></mfrac></mrow></mfenced></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>110</mn><mo>+</mo><mn>100</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mi mathvariant="normal">$</mi><mn>210</mn></mtd></mtr></mtable></math>

  5.  The population of a particular area A of a city is 5000. It increases by 10% in 1st yr. It decreases by 20% in the 2nd yr because of some reason. In the 3rd yr, the population increases by 30%. What will be the population of area A at the end of 3 yr?

Sol: Given that, P = 5000, R1 = 10%, R2 = -20%(decrease) and R3 = 30%

Therefore, Population at the end of 3rd year

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mi>P</mi><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>1</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>2</mn></mrow><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mrow><mi>R</mi><mn>3</mn></mrow><mn>100</mn></mfrac></mrow></mfenced></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>5000</mn><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>10</mn><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>20</mn><mn>100</mn></mfrac></mrow></mfenced><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>30</mn><mn>100</mn></mfrac></mrow></mfenced></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>5000</mn><mo>&#xD7;</mo><mfrac><mn>11</mn><mn>10</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>4</mn><mn>5</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>13</mn><mn>10</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>10</mn><mo>&#xD7;</mo><mn>11</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>13</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mi mathvariant="normal">$</mi><mn>5720</mn></mtd></mtr></mtable></math>

Population at the end of 3rd year = $ 5720


18 True Discount
N/A

True Discount If a person borrows certain money from another person for a certain period and the borrower wants to clear off the debt right now, then for paying back the debt, the borrower gets certain discount which is called True Discount (TD)

Present Worth The money to be paid back is called the Present Worth (PW).

 Amount Sum due is called Amount (A). Amount (A) = PW + TD

 True Discount It is the difference between the Amount (A) and the Present Worth (PW).

Discount (TD) = A-PW

Things to remember:

1. True discount is the interest on Present Worth (PW).

2. Interest is reckoned on PW and TD is reckoned on amount.

 According to the definition, we have TD = A – PW

Formulae:

  1. If rate of interest is R% per annum, time is T yr and present worth is PW, then True

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Discount&#xA0;</mtext><mo>(</mo><mi>T</mi><mi>D</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mi>P</mi><mi>W</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow><mn>100</mn></mfrac></math>

  2. If rate of interest is R% and money due for amount is A after T yr, then

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Present Worth&#xA0;</mtext><mo>(</mo><mi>P</mi><mi>W</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mn>100</mn><mo>&#xD7;</mo><mi>A</mi></mrow><mrow><mn>100</mn><mo>+</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow></mfrac></math>

  3. If money due A, rate of interest R% and time T are given, then

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>True Discount&#xA0;</mtext><mo>(</mo><mi>T</mi><mi>D</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mi>A</mi><mo>&#xD7;</mo><mi>T</mi><mo>&#xD7;</mo><mi>R</mi></mrow><mrow><mn>100</mn><mo>+</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow></mfrac></math>

  4. If the true discount on a certain sum of money due certain year hence and the simple interest on the same sum for the same time and at the same rate is given, then

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Money due&#xA0;</mtext><mo>(</mo><mi>A</mi><mo>)</mo><mo>=</mo><mfrac><mtext>&#xA0;SI&#xA0;&#xD7;&#xA0;TD&#xA0;</mtext><mtext>&#xA0;SI&#xA0;-&#xA0;TD&#xA0;</mtext></mfrac></math>

  5. If the true discount on a certain sum of money due T yr hence and the simple interest on the same sum for the same time and at the same rate of interest R% per annum are given, then

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mi>I</mi><mo>&#x2212;</mo><mi>T</mi><mi>D</mi><mo>=</mo><mfrac><mrow><mi>T</mi><mi>D</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow><mn>100</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;6. When the sum is put at compound Interest, then&#xA0;</mtext><mi>P</mi><mi>W</mi><mo>=</mo><mfrac><mi>A</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>T</mi></msup></mfrac></math>

Examples:

  1. What will be the true discount for the present worth of $ 6000 for a period of 9 months at 12% per annum?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Sol: Given that,&#xA0;</mtext><mi>T</mi><mo>=</mo><mn>9</mn><mtext>&#xA0;months&#xA0;</mtext><mo>=</mo><mfrac><mn>9</mn><mn>12</mn></mfrac><mi>y</mi><mi>r</mi><mtext>&#xA0;and&#xA0;</mtext><mi>R</mi><mo>=</mo><mn>12</mn><mi mathvariant="normal">%</mi><mo>,</mo><mi>P</mi><mi>W</mi><mo>=</mo><mi mathvariant="normal">$</mi><mn>6000</mn><mtext>;</mtext></math>

  2. The true discount on a certain sum of money due 4 yr hence is X 75 and the simple interest on the same sum for the same time and at the same rate of interest is X 225. Find the rate per cent.

Sol: Given that, SI = $ 225, TD = $ 75, T = 4 yr and R =?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;According to the formula,&#xA0;</mtext><mi>S</mi><mi>I</mi><mo>&#x2212;</mo><mi>T</mi><mi>D</mi><mo>=</mo><mfrac><mrow><mi>T</mi><mi>D</mi><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mi>T</mi></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mspace width="2em"/><mn>225</mn><mo>&#x2212;</mo><mn>75</mn><mo>=</mo><mfrac><mrow><mn>75</mn><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mn>4</mn></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>150</mn><mo>=</mo><mfrac><mrow><mn>75</mn><mo>&#xD7;</mo><mi>R</mi><mo>&#xD7;</mo><mn>4</mn></mrow><mn>100</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>3</mn><mi>R</mi><mo>=</mo><mn>150</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>R</mi><mo>=</mo><mn>50</mn><mi mathvariant="normal">%</mi></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Sol: PW of&#xA0;</mtext><mi mathvariant="normal">$</mi><mn>6440</mn><mtext>&#xA0;due&#xA0;</mtext><mn>8</mn><mtext>&#xA0;months hence&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>6440</mn><mo>&#xD7;</mo><mn>100</mn></mrow><mrow><mn>100</mn><mo>+</mo><mn>18</mn><mo>&#xD7;</mo><mfrac><mn>8</mn><mn>12</mn></mfrac></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>6440</mn><mo>&#xD7;</mo><mn>100</mn></mrow><mn>112</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>5750</mn></math>

From PW we can say that $10000 cash is better offer

  4. The true discount on a certain sum of money due 10 yr hence is $ 68 and the simple interest on the same sum for the same time and at the same rate of interest is $ 102. Find the sum due.

Sol: Given that, TD = $ 68 and SI = $ 102

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, Sum due&#xA0;</mtext><mo>(</mo><mi>A</mi><mo>)</mo><mo>=</mo><mfrac><mtext>&#xA0;SI x TD&#xA0;</mtext><mtext>&#xA0;SI-TD&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mrow><mn>102</mn><mo>&#xD7;</mo><mn>68</mn></mrow><mrow><mn>102</mn><mo>&#x2212;</mo><mn>68</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>102</mn><mo>&#xD7;</mo><mn>68</mn></mrow><mn>34</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>204</mn></math>

  5. What will be the present worth of $ 4840 due 2 yr hence, when the interest is compounded at 10% per annum? Also, find true discount.

Sol: Given that, A = $4840, T = 2 years, R = 10%, PW =? and TD =?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>According to the formula,&#xA0;</mtext><mi>P</mi><mi>W</mi><mo>=</mo><mfrac><mi>A</mi><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mi>R</mi><mn>100</mn></mfrac></mrow></mfenced><mi>T</mi></msup></mfrac><mo>=</mo><mfrac><mn>4840</mn><msup><mfenced separators="|"><mrow><mn>1</mn><mo>+</mo><mfrac><mn>10</mn><mn>100</mn></mfrac></mrow></mfenced><mn>2</mn></msup></mfrac><mo>=</mo><mfrac><mn>4840</mn><msup><mfenced separators="|"><mfrac><mn>11</mn><mn>10</mn></mfrac></mfenced><mn>2</mn></msup></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mn>4840</mn><mo>&#xD7;</mo><mn>10</mn><mo>&#xD7;</mo><mn>10</mn></mrow><mrow><mn>11</mn><mo>&#xD7;</mo><mn>11</mn></mrow></mfrac><mo>=</mo><mn>40</mn><mo>&#xD7;</mo><mn>10</mn><mo>&#xD7;</mo><mn>10</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>4000</mn></mtd></mtr></mtable></math>

                                                    TD = A – PW = 4840 – 4000 = $ 840


19 Ratio and Proportion
N/A

Ratio

When two or more similar quantities are compared, then to represent this comparison, ratios are used.

or

Ratio of two quantities is the number of times one quantity contains another quantity of same kind.

The ratio between x and y can be represented as x: y, where x is called antecedent and y is called consequent.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>x</mi><mi>y</mi></mfrac><mtext>&#xA0;or&#xA0;</mtext><mi>x</mi><mo>:</mo><mi>y</mi></math>

Type of Ratio:

The different types of ratio are explained as under

  1. Duplicate Ratio If two numbers are in ratio, then the ratio of their squares is called duplicate ratio. If x and y are two numbers, then the duplicate ratio of x and y would be xr: jr.

 For example: Duplicate ratio of 3 :4 = 3 :4 =9:16

  1. Sub-duplicate Ratio If two numbers are in ratio, then the ratio of their square roots is called sub-duplicate ratio. If x and y are two numbers, then the sub-duplicate ratio of x and y would be Vx: Jy.
  2. Triplicate Ratio If two numbers are in ratio, then the ratio of their cubes is called triplicate ratio. If x and y are two numbers, then the triplicate ratio of x and y would be x3: y3.

For example: Triplicate ratio of 2 :3 = 23 :33 = 8 :27

  4. Sub-triplicate Ratio If two numbers are in ratio, then the ratio of their cube roots is called sub-triplicate ratio. If x and y are two numbers, then the sub-triplicate ratio of x and y would be <math xmlns="http://www.w3.org/1998/Math/MathML"><mroot><mi>x</mi><mn>3</mn></mroot><mo>:</mo><mroot><mi>y</mi><mn>3</mn></mroot><mo>.</mo></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>For example, Sub-triplicate ratio of&#xA0;</mtext><mn>1</mn><mo>:</mo><mn>125</mn><mo>=</mo><mroot><mrow><mn>1</mn><mo>:</mo></mrow><mn>3</mn></mroot><mroot><mn>125</mn><mn>3</mn></mroot><mo>=</mo><mn>1</mn><mo>:</mo><mn>5</mn></math>

  5. Inverse Ratio If two numbers are in ratio, then their antecedent and consequent are interchanged and the ratio obtained is called inverse ratio, If x and y are two numbers and their ratio is x: y, then its inverse ratio will be y: x.

For example, Inverse ratio of 4: 5 is 5: 4

  6. Compound Ratio If two or more ratios are given, then the antecedent of one is multiplied with antecedent of other and respective consequents are also multiplied.

If a: b, c: d and e: f are three ratios, then their compound ratio will be ace: bdf.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>For example: The compound ratio of&#xA0;</mtext><mn>2</mn><mo>:</mo><mn>5</mn><mo>,</mo><mn>6</mn><mo>:</mo><mn>7</mn><mtext>&#xA0;and&#xA0;</mtext><mn>9</mn><mo>:</mo><mn>13</mn><mo>=</mo><mfrac><mrow><mn>2</mn><mo>&#xD7;</mo><mn>6</mn><mo>&#xD7;</mo><mn>9</mn></mrow><mrow><mn>5</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>13</mn></mrow></mfrac><mo>=</mo><mfrac><mn>108</mn><mn>455</mn></mfrac></math>

Note:

  1. If the antecedent is greater than the consequent, then the ratio is known as the ratio of greater inequality, such as 7: 5.
  2. If the antecedent is less than the consequent, then the ratio is called the ratio of less inequality, such as 5: 7.

Comparison of Ratios:

Rules used to compare different ratios are as follows

  1. If the given ratios are a: b and c: d, then (i) a: b> c: d, if ad > be (ii) a: b< c: d, if ad < be (iii) a: b = c: d, if ad = bc.
  2. if two ratios are given for comparison, convert each ratio in such a way that both ratios have same denominator, then compare their numerators, the fraction with greater numerator will be greater.
  3. If two ratios are given for comparison, convert each ratio in such a way that both ratios have same numerator, then compare their denominators, the fraction with lesser denominator will be greater.

Proportion:

An equality of two ratios is called the proportion.

If a/b = c/d  or a: b = c: d, then we can say that a, b, c and d are in proportion and can be written as a: b::c: d, where symbol '::' represents proportion and it is read as 'a is to b' as 'c is to d'.

Here, a and d are called 'Extremes' and b and c are called as 'Means'.

Basic Rules of Proportion:

  1. if a:b::b:c, then c is called third proportional to a and b, which are in continued proportion, c will be calculated as a:b::b:c => a: b = b: c => a x c = b x b => b2 = ac => c = (b2/a)
  2. if a: b :: c: d, then d is called the 4th proportional to a, b and c, d will be calculated as

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>:</mo><mi>b</mi><mo>::</mo><mi>c</mi><mo>:</mo><mi>d</mi><mo>=&gt;</mo><mi>a</mi><mo>:</mo><mi>b</mi><mo>=</mo><mi>c</mi><mo>:</mo><mi>d</mi><mo>=&gt;</mo><mi>a</mi><mo>&#xD7;</mo><mi>d</mi><mo>=</mo><mi>c</mi><mo>&#xD7;</mo><mi>b</mi><mo>=&gt;</mo><mi>d</mi><mo>=</mo><mfrac><mrow><mi>b</mi><mi>c</mi></mrow><mi>a</mi></mfrac></math>

  3. Mean proportional between a and b is <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mi>a</mi><mi>b</mi></msqrt></math>. If mean proportional is x, then a: x :: x: b

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=&gt;</mo><mi>a</mi><mi>x</mi><mi>b</mi><mo>=</mo><msup><mi>x</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=&gt;</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mi>a</mi><mi>b</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=&gt;</mo><mi>x</mi><mo>=</mo><msqrt><mi>a</mi><mi>b</mi></msqrt></mtd></mtr></mtable></math>

Examples:

  1. If P: Q = 8:15 and Q: R = 3:2, then find P: Q: R.

Sol: Given that, P: Q = 8: 15, Q: R = 3: 2

P: Q: R = (8 x 3): (15 x 3): (15 x 2) = 24: 45: 30

Therefore, P: Q: R = 8: 15: 10

Here, consequent of first ratio should be equal to the antecedent of second ratio.

  1. Salary of Mr. X is 80% of the salary of Mr. Y and the salary of Mr. Z is 120% of the salary of Mr. X. What is the ratio between the salaries of X, Y and Z, respectively?

Sol: Let Y’s salary = 100

Therefore, X’s salary = 80

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>And Z's salary&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>80</mn><mo>&#xD7;</mo><mn>120</mn></mrow><mn>100</mn></mfrac><mo>=</mo><mn>96</mn></math>

Therefore, required ratio = 80: 100: 96 = 20: 25: 24

  1. $ 710 were divided among A, B and C in such a way that A had $ 40 more than B and C had $ 30 more than A. How much was C’s share?

Sol: Let B gets x.

Then, A gets (x + 40) and C gets (x + 70).

According to the question,

x + 40 + x + x + 70 = 710

3x = 710 – 110

3x = 600

x = 200

C’s share = 200 + 70 = $ 270

  4. What will be the mean proportional between 4 and 25?

Sol: Let mean proportional be x.

Then, 4: x :: x: 25 => 4: x :: x: 25 => 4 x 25 = x2

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><msqrt><mn>4</mn><mo>&#xD7;</mo><mn>25</mn></msqrt><mo>=</mo><mn>10</mn></math>

  5. The ratio between the number of passengers travelling by 1st and 2nd class between the two railway stations is 1: 50, whereas the ratio of 1st and 2nd class fares between the same stations is 3: 1. If on a particular day, $ 1325 were collected from the passengers travelling between these stations, then what was the amount collected from the 2nd class passengers?

Sol: Let the number of passengers in 1st class be x and number of passengers in 2nd class be 50x.

Then, total amount of 1st class = 3x and total amount of 2nd class = 50x.

Ratio of the amounts collected from 1st class and the 2nd class passengers = 3: 50

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, amount collected from the&#xA0;</mtext><msup><mn>2</mn><mtext>nd&#xA0;</mtext></msup><mtext>&#xA0;class passengers&#xA0;</mtext><mo>=</mo><mfrac><mi>b</mi><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow></mfrac><mo>&#x2217;</mo><mi>x</mi></math>

Where, x = total amount a = 3, b = 50

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mfrac><mn>50</mn><mn>53</mn></mfrac><mo>&#x2217;</mo><mn>1325</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>1250</mn></math>


20 Mixture or Allegation
N/A

Mixture

The new product obtained by mixing two or more ingredients in a certain ratio is called a mixture. or Combination of two or more quantities is known as mixture.

Mean Price

The cost price of a unit quantity of the mixture is called the mean price. It will always be higher than cost price of cheaper quantity and lower than cost price of dearer quantity.

Rule of Mixture or Alligation:

 It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>According to this rule,&#xA0;</mtext><mfrac><msub><mi>n</mi><mn>1</mn></msub><msub><mi>n</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><mrow><msub><mi>A</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>A</mi><mi>w</mi></msub></mrow><mrow><msub><mi>A</mi><mi>w</mi></msub><mo>&#x2212;</mo><msub><mi>A</mi><mn>1</mn></msub></mrow></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Where,&#xA0;</mtext><mfrac><msub><mi>n</mi><mn>1</mn></msub><msub><mi>n</mi><mn>2</mn></msub></mfrac></math> is the ratio, in which two quantities should be mixed, while A1, A2 and Aw are the cheaper price, dearer price and mean price, respectively.

Remember: A1 < Aw < A2

The rule is also applicable for solving questions based on average i.e., speed, percentage, price, ratio etc., and not for absolute values. In other words, we can use this method whenever per cent, per hour, per kg etc., are being compared.

Examples:

  1. 600g of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?

Solution: 40% sugar is in 600g of sugar solution.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, quantity of sugar&#xA0;</mtext><mo>=</mo><mfrac><mrow><mn>600</mn><mo>&#xD7;</mo><mn>40</mn></mrow><mn>100</mn></mfrac><mo>=</mo><mn>240</mn><mi>g</mi></math>

Let x g sugar be added.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>According to the question,&#xA0;</mtext><mn>240</mn><mo>+</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mo>(</mo><mn>600</mn><mo>+</mo><mi>x</mi><mo>)</mo><mo>&#xD7;</mo><mn>50</mn></mrow><mn>100</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mn>240</mn><mo>+</mo><mi>x</mi></mrow><mrow><mn>600</mn><mo>+</mo><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>600</mn><mo>+</mo><mi>x</mi><mo>=</mo><mn>480</mn><mo>+</mo><mn>2</mn><mi>x</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>x</mi><mo>=</mo><mn>120</mn><mi>g</mi></mtd></mtr></mtable></math>

  2.  A container is filled with liquid, 6 parts of which are water and 10-part milk. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half milk?

Solution: Let the container initially contains 16 L of liquid.

Let a L of liquid be compressing water.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Quantity of water in the new mixture&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mn>6</mn><mo>&#x2212;</mo><mfrac><mrow><mn>6</mn><mi>a</mi></mrow><mn>16</mn></mfrac><mo>+</mo><mi>a</mi></mrow></mfenced><mi>L</mi><mspace linebreak="newline"/><mtext>&#xA0;Quantity of milk in the new mixture&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mn>10</mn><mo>&#x2212;</mo><mfrac><mrow><mn>10</mn><mi>a</mi></mrow><mn>16</mn></mfrac></mrow></mfenced><mi>L</mi><mspace linebreak="newline"/><mtext>&#xA0;According to the question,&#xA0;</mtext><mn>6</mn><mo>&#x2212;</mo><mfrac><mrow><mn>6</mn><mi>a</mi></mrow><mn>16</mn></mfrac><mo>+</mo><mi>a</mi><mo>=</mo><mn>10</mn><mo>&#x2212;</mo><mfrac><mrow><mn>10</mn><mi>a</mi></mrow><mn>16</mn></mfrac><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>96</mn><mo>&#x2212;</mo><mn>6</mn><mi>a</mi><mo>+</mo><mn>16</mn><mi>a</mi><mo>=</mo><mn>160</mn><mo>&#x2212;</mo><mn>10</mn><mi>a</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>96</mn><mo>+</mo><mn>10</mn><mi>a</mi><mo>=</mo><mn>160</mn><mo>&#x2212;</mo><mn>10</mn><mi>a</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>20</mn><mi>a</mi><mo>=</mo><mn>64</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>a</mi><mo>=</mo><mfrac><mn>64</mn><mn>20</mn></mfrac><mo>=</mo><mfrac><mn>16</mn><mn>5</mn></mfrac></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, part of mixture replaced&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>16</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>16</mn><mn>5</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>5</mn></mfrac></math>

  3.  How many kilograms of tea worth $ 25 per kg must be blended with 30 kg of tea worth $ 30 per kg, so that by selling the blended variety at $ 30 per kg, there should be a gain of 10%?

Solution: Let the quantity of tea worth $ 25 be x kg.

According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>(</mo><mn>25</mn><mi>x</mi><mo>+</mo><mn>30</mn><mo>&#xD7;</mo><mn>30</mn><mo>)</mo><mo>&#xD7;</mo><mfrac><mn>110</mn><mn>100</mn></mfrac><mo>=</mo><mn>30</mn><mo>(</mo><mn>30</mn><mo>+</mo><mi>x</mi><mo>)</mo></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mn>275</mn><mi>x</mi><mo>+</mo><mn>9900</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>9000</mn><mo>+</mo><mn>300</mn><mi>x</mi><mo>)</mo></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>300</mn><mi>x</mi><mo>&#x2212;</mo><mn>275</mn><mi>x</mi><mo>=</mo><mn>900</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>900</mn><mn>25</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>36</mn><mi>k</mi><mi>g</mi></mtd></mtr></mtable></math>

  4.  In a mixture of 60 L the ratio of acid and water is 2: 1. If the ratio of acid and water is to be 1: 2, then the amount of water (in litres) to be added to the mixture is

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Solution: Quantity of acid in the mixture&#xA0;</mtext><mo>=</mo><mfrac><mn>2</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mn>60</mn><mo>=</mo><mn>40</mn><mi>L</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Quantity of water in the mixture&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mn>60</mn><mo>=</mo><mn>20</mn><mi>L</mi><mspace linebreak="newline"/><mi>L</mi><mi>e</mi><mi>t</mi><mo>&#xA0;</mo><mi>r</mi><mi>e</mi><mi>q</mi><mi>u</mi><mi>i</mi><mi>r</mi><mi>e</mi><mi>d</mi><mo>&#xA0;</mo><mi>q</mi><mi>u</mi><mi>a</mi><mi>n</mi><mi>t</mi><mi>i</mi><mi>t</mi><mi>y</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>w</mi><mi>a</mi><mi>t</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mi>b</mi><mi>e</mi><mo>&#xA0;</mo><mi>x</mi><mo>&#xA0;</mo><mi>L</mi><mo>.</mo><mspace linebreak="newline"/><mtext>According to the question,&#xA0;</mtext><mfrac><mn>40</mn><mrow><mn>20</mn><mo>+</mo><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo stretchy="false">&#x21D2;</mo><mn>80</mn><mo>=</mo><mn>20</mn><mo>+</mo><mi>x</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x21D2;</mo></math> x = 60 L

  1. In a container, milk and water are present in the ratio 7: 5. If 15 L water is added to this mixture, the ratio of milk and water becomes 7: 8. Find the quantity of water in the new mixture.

Sol. Let the quantity of milk and water in initial mixture be 7x and 5x L.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Then, according to the question,&#xA0;</mtext><mfrac><mrow><mn>7</mn><mi>x</mi></mrow><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mn>7</mn><mn>6</mn></mfrac><mo stretchy="false">&#x21D2;</mo><mn>7</mn><mi>x</mi><mo>&#x2217;</mo><mn>8</mn><mo>=</mo><mn>7</mn><mo>(</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>15</mn><mo>)</mo><mspace linebreak="newline"/><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>56</mn><mi>x</mi><mo>=</mo><mn>35</mn><mi>x</mi><mo>+</mo><mn>105</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>56</mn><mi>x</mi><mo>&#x2212;</mo><mn>35</mn><mi>x</mi><mo>=</mo><mn>105</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>21</mn><mi>x</mi><mo>=</mo><mn>105</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>105</mn><mn>21</mn></mfrac><mo>=</mo><mn>5</mn></mtd></mtr></mtable></math>

Therefore, quantity of water in initial mixture = 5 x 5 = 25 L

And quantity of water in new mixture = 25 + 15 = 40 L


21 Partnership
N/A

When two or more persons make an association and invest money for running a certain business and after certain time receive profit in the ratio of their invested money and time period of investment, then such an association is called partnership and the persons involved in the partnership are called partners.

Partnership is of Two Types

 Simple Partnership

If all partners invest their different capitals (money) for the same time period or same capital for different time period then their profit or loss is in the ratio of their investments or time period of investment then such a partnership is called simple partnership

Compound Partnership

If all partners invest their different capitals (money) for different time period, then their profit not only depends on their investments but also on the time period of their investment, then such a partnership is called compound partnership

Partners are of Two Types

 Active or Working Partner

A partner who not only invests money, but also take part in the business activities for which he draws a defined salary or gets some share from profit before its division is called an active partner.

Sleeping Partner

A partner who only invests money and does not take part in business activities is called sleeping partner.

Ratio of division of gains:

  • When the investments made by all the partners are for the same time period, then gain or loss is distributed amongst them in the ratio of their investments,

Suppose A and B invest $ x and $ y respectively for a year in a business, then at the end of the year: (A’s share of profit): (B’s share of profit) = x: y

  • When investments are of different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now, gain or loss is divided in the ratio of these capitals.

Suppose A invests $ x for p months and B invests $ y for q months, then (A’s share of profit): (B’s share of profit) = xp: yq.

Examples:

  1. P and Q entered a partnership for 3 yr. At the start of the business, they invested $ 13000 and $ 25000, respectively. At the end of 3 yr their total profit was $ 76000. What will be share of Q out of this profit?

Solution: P’s share: Q’s share = Ratio of their investments

                                    = 13000: 25000

                                    = 13: 25

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore,&#xA0;</mtext><msup><mi>Q</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s share&#xA0;</mtext><mo>=</mo><mfrac><mn>25</mn><mrow><mn>13</mn><mo>+</mo><mn>25</mn></mrow></mfrac><mo>&#xD7;</mo><mn>76000</mn><mo>=</mo><mfrac><mn>25</mn><mn>38</mn></mfrac><mo>&#xD7;</mo><mn>76000</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>50000</mn></math>

  2. A, B and C invested their capitals in the ratio of 5: 6: 8. At the end of the business, they received the profits in the ratio of 5: 3: 1. Find the ratio of time for which they contributed their capitals.

Solution: Here P1: P2: P3 = 5: 3: 1 and x1: x2: x3 = 5: 6: 8

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;According to the formula, Required ratio&#xA0;</mtext><mo>=</mo><mfrac><msub><mi>P</mi><mn>1</mn></msub><msub><mi>x</mi><mn>1</mn></msub></mfrac><mo>:</mo><mfrac><msub><mi>P</mi><mn>2</mn></msub><msub><mi>x</mi><mn>2</mn></msub></mfrac><mo>:</mo><mfrac><msub><mi>P</mi><mn>3</mn></msub><msub><mi>x</mi><mn>3</mn></msub></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mfrac><mn>5</mn><mn>5</mn></mfrac><mo>:</mo><mfrac><mn>3</mn><mn>6</mn></mfrac><mo>:</mo><mfrac><mn>1</mn><mn>8</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>1</mn><mo>:</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>:</mo><mfrac><mn>1</mn><mn>8</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>8</mn><mo>:</mo><mfrac><mn>8</mn><mn>2</mn></mfrac><mo>:</mo><mfrac><mn>8</mn><mn>8</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>8</mn><mo>:</mo><mn>4</mn><mo>:</mo><mn>1</mn></mtd></mtr></mtable></math>

  3. A and B together start a business by investing in the ratio of 4: 3. If 9% of the total profit goes to charity and A's share is $ 1196, find the total profit.

Solution: Let total profit = x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Paid to charity&#xA0;</mtext><mo>=</mo><mn>9</mn><mi mathvariant="normal">%</mi><mtext>&#xA0;of&#xA0;</mtext><mi>x</mi><mo>=</mo><mfrac><mrow><mn>9</mn><mi>x</mi></mrow><mn>100</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, balance profit&#xA0;</mtext><mo>=</mo><mi>x</mi><mo>&#x2212;</mo><mfrac><mrow><mn>9</mn><mi>x</mi></mrow><mn>100</mn></mfrac><mo>=</mo><mfrac><mrow><mn>91</mn><mi>x</mi></mrow><mn>100</mn></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s share&#xA0;</mtext><mo>=</mo><mfrac><mn>4</mn><mrow><mn>4</mn><mo>+</mo><mn>3</mn></mrow></mfrac><mo>&#xD7;</mo><mfrac><mrow><mn>91</mn><mi>x</mi></mrow><mn>100</mn></mfrac><mo>=</mo><mfrac><mn>4</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mfrac><mrow><mn>9</mn><mi>x</mi></mrow><mn>100</mn></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><mtext>&#xA0;According to the question,&#xA0;</mtext><mfrac><mn>4</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mfrac><mrow><mn>91</mn><mi>x</mi></mrow><mn>100</mn></mfrac><mo>=</mo><mn>1196</mn><mspace linebreak="newline"/><mspace linebreak="newline"/><mtext>&#xA0;Therefore,&#xA0;</mtext><mi>x</mi><mo>=</mo><mfrac><mrow><mn>1196</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>100</mn></mrow><mrow><mn>4</mn><mo>&#xD7;</mo><mn>91</mn></mrow></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>2300</mn></math>

  4. A and B started a business with $ 20000 and $ 35000 respectively. They agreed to share the profit in the ratio of their capital. C joins the partnership with the condition that A, B and C will share profit equally and pays $ 220000 as premium for this, to be shared between A and B. This is to be divided between A and B in the ratio of

Solution: Ratio of total capital of A and B = 20000 x 12: 35000 x 12

                                                                   = 240000: 420000

Now, C gives $220000 to both to make the capital equal.

If A takes $ 200000 and B takes $ 20000 from C, then both have the equal capital

Therefore, required ratio of divided amount = 200000: 20000 = 20: 2 =10: 1

  4. P, Q and R hire a meadow for $ 2920. P puts 10 cows for 20 days; Q puts 30 cows for 8 days and R puts 16 cows for 9 days. Find the rent paid by R.

Solution: Ratio of rents to be paid by P, Q and R

Ratio of monthly equivalent = (10 x 20): (30 x 8): (16 x 9)

                                                = 200: 240: 144

                                                = 25: 30: 18

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Hence,&#xA0;</mtext><msup><mi>R</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s share&#xA0;</mtext><mo>=</mo><mfrac><mn>18</mn><mrow><mn>25</mn><mo>+</mo><mn>30</mn><mo>+</mo><mn>18</mn></mrow></mfrac><mo>&#xD7;</mo><mn>2920</mn><mspace linebreak="newline"/><mspace linebreak="newline"/><mo>=</mo><mfrac><mn>18</mn><mn>73</mn></mfrac><mo>&#xD7;</mo><mn>2920</mn><mo>=</mo><mn>18</mn><mo>&#xD7;</mo><mn>40</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>720</mn></math>


22 Problem Based on Ages
N/A

Age is defined as a period of time that a person has lived or a thing has existed. Age is measured in months, years, decades and so on.

Problem based on ages generally consists of information of ages of two or more persons and a relation between their ages in present/future/past.

Using the information, it is asked to calculate the ages of one 01 more persons in present/future/ past.

Important Rules for Problem Based on Ages:

Rule1:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>If ratio of present ages of&#xA0;</mtext><mi>A</mi><mtext>&#xA0;and&#xA0;</mtext><mi>B</mi><mtext>&#xA0;is&#xA0;</mtext><mi>x</mi><mo>:</mo><mi>y</mi><mtext>&#xA0;and&#xA0;</mtext><mi>n</mi><mtext>&#xA0;yr ago, the ratio of their ages was&#xA0;</mtext><mi>p</mi><mo>:</mo><mi>q</mi><mtext>&#xA0;then&#xA0;</mtext><mfrac><mrow><mi>x</mi><mo>&#x2212;</mo><mi>n</mi></mrow><mrow><mi>y</mi><mo>&#x2212;</mo><mi>n</mi></mrow></mfrac><mo>=</mo><mfrac><mi>p</mi><mi>q</mi></mfrac></math>

Rule2:

If ratio of present ages of A and B is x: y and after n yr, the ratio of their ages will be p: q,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mi>h</mi><mi>e</mi><mi>n</mi><mo>&#xA0;</mo><mfrac><mrow><mi>x</mi><mo>+</mo><mi>n</mi></mrow><mrow><mi>y</mi><mo>+</mo><mi>n</mi></mrow></mfrac><mo>=</mo><mfrac><mi>p</mi><mi>q</mi></mfrac></math>

Note:

Mostly questions on ages can be solved with the use of linear equations.

 

Examples:

  1. Present age of Kevin is 5 times the age of Steve. After 10 yr, Kevin will be 3 times as old as Steve. What are the present ages of Kevin and Steve?

Solution: Let the present age of Steve be x yr. Then, present age of Kevin = 5x yr

After 10 yr, the ratio of ages will be 3: 1.

According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>10</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>10</mn></mrow></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>1</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>10</mn><mo>=</mo><mn>3</mn><mo>(</mo><mi>x</mi><mo>+</mo><mn>10</mn><mo>)</mo></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>10</mn><mo>=</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>30</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>5</mn><mi>x</mi><mo>&#x2212;</mo><mn>3</mn><mi>x</mi><mo>=</mo><mn>30</mn><mo>&#x2212;</mo><mn>10</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>2</mn><mi>x</mi><mo>=</mo><mn>20</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>20</mn><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>10</mn></mtd></mtr></mtable></math>

Therefore, Kevin’s present age = 5 x 10 = 50 yr and Steve’s present age = 10 yr

  1. The average age of a family of five members is 24. If the present age of the youngest member is 8 yr, what was the average age of the family at the time of the birth of the youngest member?

Solution: Total age of five members of a family = 24 x 5 = 120

Therefore, total age of four members at the time of birth of youngest = 120 – (8 x 5)

                                                                                                            = 120 – 40 = 80 yr

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, required average age&#xA0;</mtext><mo>=</mo><mfrac><mn>80</mn><mn>4</mn></mfrac><mo>=</mo><mn>20</mn><mi>y</mi><mi>r</mi></math>

  1. Before 7 yr, the ratio of ages of A and B was 3: After 9 yr, ratio of their ages will be 7: 8. The present age of B will be

Solution: Let the ages of A and B before 7 yr were 3x yr and 4x yr, respectively.

Therefore, present age of A = 3x + 7 and present age of B = 4x + 7

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Now, according to the question,&#xA0;</mtext><mfrac><mrow><mn>3</mn><mi>x</mi><mo>+</mo><mn>7</mn><mo>+</mo><mn>9</mn></mrow><mrow><mn>4</mn><mi>x</mi><mo>+</mo><mn>7</mn><mo>+</mo><mn>9</mn></mrow></mfrac><mo>=</mo><mfrac><mn>7</mn><mn>8</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>24</mn><mi>x</mi><mo>+</mo><mn>128</mn><mo>=</mo><mn>28</mn><mi>x</mi><mo>+</mo><mn>112</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>4</mn><mi>x</mi><mo>=</mo><mn>16</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>4</mn></mtd></mtr></mtable></math>

Hence, present age of B = 4 x 4 + 7 = 16 + 7 = 23 yr

 

  4.  The present ages of two persons are 36 and 50 yr, respectively. If after n yr the ratio of their ages will be 3: 4, then the value of n is

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: According to the question,&#xA0;</mtext><mfrac><mrow><mn>36</mn><mo>+</mo><mi>n</mi></mrow><mrow><mn>50</mn><mo>+</mo><mi>n</mi></mrow></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>4</mn></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>144</mn><mo>+</mo><mn>4</mn><mi>n</mi><mo>=</mo><mn>150</mn><mo>+</mo><mn>3</mn><mi>n</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>n</mi><mo>=</mo><mn>6</mn></mtd></mtr></mtable></math>

  5.  The present age of Peter's father is four times Peter's present age. Five years back, Peter's father was seven times as old as Peter was at that time. What is the present age of Peter's father?

Solution: Let present age of Peter be x.

Then, present age of Peter’s father = 4x

Now, 5 yr ago, Peter's father's age = 7x

Peter's age => 4x - 5 = 7(x - 5)

 => 4x - 5 = 7x - 35

=> 3x = 30

=> x = 10

Peter's present age = x = 10 yr

Peter's father's present age 4x = 4 * 10 = 40 yr


23 Unitary Method
N/A

Unitary method is a fundamental tool to solve arithmetic problems based on variation in quantities.

The method endorses a simple technique to find the amount related to unit quantity.

This method can be applied in questions based on time and work, speed and distance, work and wages etc.

Direct proportion:

Two quantities are said to be in direct proportion to each other, if on increasing (decreasing) a quantity, the other quantity also increases (decreases) to the same extent

i.e., (Quantity 1) ∞ (Quantity 2)

For example: Number of men ∞ Volume of work done (time constant)

i.e., if number of men increases, then the volume of work done also increases.

Similarly, if volume of work increases, then number of men required to finish the work also increases.

 

Indirect Proportion:

Two quantities are said to be in indirect proportion to each other, if on increasing (or decreasing) a quantity, the other quantity decreases (or increases) to the same extent

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;i.e., (Quantity 1)&#xA0;</mtext><mi mathvariant="normal">&#x221E;</mi><mfrac><mn>1</mn><mrow><mo>(</mo><mtext>&#xA0;Quantity 2)&#xA0;</mtext></mrow></mfrac></math>

For example:

The time taken by a vehicle in covering a certain distance is inversely proportional to the speed of the vehicle.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;i.e., (Speed)&#xA0;</mtext><mi mathvariant="normal">&#x221E;</mi><mfrac><mn>1</mn><mrow><mo>(</mo><mtext>&#xA0;Time&#xA0;</mtext><mo>)</mo></mrow></mfrac></math>

Note: If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days, then we have a general formula, M1 W2 D1 = M2 W1 D2

  1. If 30 men working 18 h per day can reap a field in 32 days, in how many days can 36 men reap the field working 16 h per day?

Solution: Let the required number of days be ‘d’.

More men, less days (Indirect proportion)

Less hours, more days (Indirect proportion)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>36</mn><mo>&#xD7;</mo><mn>16</mn><mo>&#xD7;</mo><mi>d</mi><mo>=</mo><mn>30</mn><mo>&#xD7;</mo><mn>18</mn><mo>&#xD7;</mo><mn>32</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>d</mi><mo>=</mo><mfrac><mrow><mn>30</mn><mo>&#xD7;</mo><mn>18</mn><mo>&#xD7;</mo><mn>32</mn></mrow><mrow><mn>36</mn><mo>&#xD7;</mo><mn>16</mn></mrow></mfrac><mo>=</mo><mn>30</mn></mtd></mtr></mtable></math>

Therefore, Required number of days = 30

  1. Steve completes 5/8 of a job in 20 days. At this rate, how many more days will he take to finish the job?

Solution: Let the required number of days be ‘d’.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Remaining work&#xA0;</mtext><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>5</mn><mn>8</mn></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>8</mn></mfrac></math>

Less work, Less days (Direct proportion)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>5</mn><mn>8</mn></mfrac><mo>:</mo><mfrac><mn>3</mn><mn>8</mn></mfrac><mo>:</mo><mn>20</mn><mo>:</mo><mi>x</mi></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>3</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mn>20</mn><mo>&#xD7;</mo><mfrac><mn>8</mn><mn>5</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>12</mn><mtext>&#xA0;days&#xA0;</mtext></mtd></mtr></mtable></math>

  1. If in a hostel, food is available for 45 days for 50 students. For how many days will this food be sufficient for 75 students?

Solution: For 50, food is sufficient for 45 days.

Therefore, for 1 student food is sufficient for 45 x 50 days

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Hence, for&#xA0;</mtext><mn>75</mn><mtext>&#xA0;students, food is sufficient for&#xA0;</mtext><mfrac><mrow><mn>45</mn><mo>&#xD7;</mo><mn>50</mn></mrow><mn>75</mn></mfrac><mtext>&#xA0;days&#xA0;</mtext><mo>=</mo><mn>30</mn><mtext>&#xA0;days&#xA0;</mtext></math>

  4. If 12 engines consume 30 metric tonnes of coal when each is running 18 h per day, how much coal will be required for 16 engines, each running 24 h per day, it being given that 6 engines of former type consume as much as 8 engines of latter type?

Solution:

Let the required quantity of coal consumed be x.

More engines, More coal consumption (Direct proportion)

More hours, More coal consumption (Direct proportion)

Less rate of consumption, Less coal consumption (Direct proportion)

Engines                        12: 16

Working hours             18: 24

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Rate of consumption&#xA0;</mtext><mfrac><mn>1</mn><mn>6</mn></mfrac><mo>:</mo><mfrac><mn>1</mn><mn>8</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;Therefore,&#xA0;</mtext><mn>12</mn><mo>&#xD7;</mo><mn>18</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mo>(</mo><mi>x</mi><mo>)</mo><mo>=</mo><mn>16</mn><mo>&#xD7;</mo><mn>24</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mn>30</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>36</mn><mi>x</mi><mo>=</mo><mn>14400</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>1440</mn><mn>36</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>40</mn></mtd></mtr></mtable></math>

  1. A worker makes a toy in every 2 h. If he works for 80 h, then how many toys will he make?

Solution: Let number of toys be x.

More hours, more toys (Direct proportion)

2: 80:: 1: x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi><mo>=</mo><mfrac><mn>80</mn><mn>2</mn></mfrac><mo>=</mo><mn>40</mn><mtext>&#xA0;toys</mtext></math>


24 Work and Time
N/A

In this, we will study techniques to solve problems based on work and its completion time as well as number of persons required to finish the given work in stipulated time.

Suppose that you are a contractor and you got a contract to construct a flyover in a certain time. For this, you need to calculate the number of men required to finish the work according to their work efficiency.

Important Relations:

 1. Work and Person Directly proportional (more work, more men and conversely more men, more work).

2. Time and Person Inversely proportional (more men, less time and conversely more time, less men).

3. Work and Time Directly proportional (more work, more time and conversely more time, more work).

Basic Rules Related to Work and Time:

Rule1: If a person can do a piece of work in n days, then that person’s 1 day’s (hour’s) work = (1/n)

Rule2: If a person’s 1 day’s(hour’s) work = (1/n), then the person will complete the work in n days(hours).

Rule3: If a person is n times efficient than the second person, then work done by

First person: Second person = n: 1 and time taken to complete a work by

First person: Second person = 1: n

Rule4: If ratio of numbers of men required to complete a work is m: n, then the ratio of time taken by them will be n: m

Examples:

  1. A and B together can do a piece of work in 12 days, while B alone can finish it in 30 days. A alone can finish the work in

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution:&#xA0;</mtext><mo>(</mo><mi>A</mi><mo>+</mo><mi>B</mi><msup><mo>)</mo><mi mathvariant="normal">&#x2032;</mi></msup><mtext>s&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>12</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;B's&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>30</mn></mfrac><mspace linebreak="newline"/><msup><mtext>&#xA0;A's&#xA0;</mtext><mi mathvariant="normal">&#x2032;</mi></msup><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>12</mn></mfrac><mo>&#x2212;</mo><mfrac><mn>1</mn><mn>30</mn></mfrac><mo>=</mo><mfrac><mrow><mn>5</mn><mo>&#x2212;</mo><mn>2</mn></mrow><mn>60</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>20</mn></mfrac></math>

  2. A can do a piece of work in 10 days and B in 20 days. They begin together but A leaves 2 days before the completion of the work. The whole work will be done in

Solution: Let the required days be x.

A works for (x-2) days, while B works for x days.

According to the question,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi>x</mi><mo>&#x2212;</mo><mn>2</mn></mrow><mn>10</mn></mfrac><mo>+</mo><mfrac><mi>x</mi><mn>20</mn></mfrac><mo>=</mo><mn>1</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>2</mn><mi>x</mi><mo>&#x2212;</mo><mn>4</mn><mo>+</mo><mi>x</mi><mo>=</mo><mn>20</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mn>3</mn><mi>x</mi><mo>=</mo><mn>24</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>x</mi><mo>=</mo><mn>8</mn><mtext>&#xA0;days</mtext></mtd></mtr></mtable></math>

  3. 6 boys can complete a piece of work in 16 h. In how many hours will 8 boys complete the same work?

Solution: Given, M1 = 120, D1 = 45, M2 = 120 + 30 = 150 and D2 = x

Then, using M1 D1 = M2 D2

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>120</mn><mo>&#xD7;</mo><mn>45</mn><mo>=</mo><mn>150</mn><mo>(</mo><mi>x</mi><mo>)</mo></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mspace width="1em"/><mi>x</mi><mo>=</mo><mfrac><mrow><mn>120</mn><mo>&#xD7;</mo><mn>45</mn></mrow><mn>150</mn></mfrac><mo>=</mo><mn>36</mn></mtd></mtr></mtable></math>

  4. If 3 men or 4 women can build a wall in 43 days, in how many days can 7 men and 5 women build this wall?

Solution: 3 men = 4 women

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mn>1</mn><mtext>&#xA0;man&#xA0;</mtext><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mtext>&#xA0;women</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore&#xA0;</mtext><mn>7</mn><mtext>&#xA0;men&#xA0;</mtext><mo>+</mo><mn>5</mn><mtext>&#xA0;women&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mn>7</mn><mo>&#xD7;</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mo>+</mo><mn>5</mn></mrow></mfenced><mo>=</mo><mfrac><mn>43</mn><mn>3</mn></mfrac><mtext>&#xA0;women&#xA0;</mtext><mspace linebreak="newline"/><mtext>&#xA0;Therefore,&#xA0;</mtext><msub><mi>M</mi><mn>1</mn></msub><mo>=</mo><mn>4</mn><mo>,</mo><msub><mi>D</mi><mn>1</mn></msub><mo>=</mo><mn>43</mn><mo>,</mo><msub><mi>M</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>43</mn><mn>3</mn></mfrac><mo>,</mo><msub><mi>D</mi><mn>2</mn></msub><mo>=</mo><mo>?</mo><mtext>&#xA0;and&#xA0;</mtext><msub><mi>W</mi><mn>1</mn></msub><mo>=</mo><msub><mi>W</mi><mn>2</mn></msub><mo>=</mo><mn>1</mn></math>

According to the formula, M1 D1 W2 = M2 D2 W1

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mn>4</mn><mo>&#xD7;</mo><mn>43</mn><mo>&#xD7;</mo><mn>1</mn><mo>=</mo><mfrac><mn>43</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><msub><mi>D</mi><mn>2</mn></msub><mo>&#xD7;</mo><mn>1</mn></math>

Therefore, D2 = 3 x 4 = 12 days

  5. A, B and C can do a piece of work individually in 8, 12 and 15 days, respectively. A and B start working but A quits after working for 2 days. After this, C joins B till the completion of work. In how many days will the work be completed?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Work done by&#xA0;</mtext><mi>A</mi><mtext>&#xA0;and&#xA0;</mtext><mi>B</mi><mtext>&#xA0;in&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>8</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>12</mn></mfrac><mo>=</mo><mfrac><mn>5</mn><mn>24</mn></mfrac></math>

After 2 day’s A left the work

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, Remaining work&#xA0;</mtext><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>10</mn><mn>24</mn></mfrac><mo>=</mo><mfrac><mn>14</mn><mn>24</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;One day work of&#xA0;</mtext><mi>B</mi><mtext>&#xA0;and&#xA0;</mtext><mi>C</mi><mtext>&#xA0;together&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>12</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>15</mn></mfrac><mo>=</mo><mfrac><mn>9</mn><mn>60</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;So, the number of days required by&#xA0;</mtext><mi>B</mi><mtext>&#xA0;and&#xA0;</mtext><mi>C</mi><mtext>&#xA0;to finish work&#xA0;</mtext><mo>=</mo><mfrac><mfrac><mn>14</mn><mn>24</mn></mfrac><mfrac><mn>2</mn><mn>60</mn></mfrac></mfrac><mo>=</mo><mfrac><mn>14</mn><mn>24</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>60</mn><mn>9</mn></mfrac><mo>=</mo><mfrac><mn>35</mn><mn>9</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;Therefore, total days to complete the work&#xA0;</mtext><mo>=</mo><mn>2</mn><mo>+</mo><mfrac><mn>35</mn><mn>9</mn></mfrac><mo>=</mo><mfrac><mn>53</mn><mn>9</mn></mfrac><mo>=</mo><mn>5</mn><mfrac><mn>8</mn><mn>9</mn></mfrac><mtext>&#xA0;days&#xA0;</mtext></math>


25 Work and Wages
N/A

Activity involving physical efforts, done in order to achieve a result is known as work.

Money received by a person for a certain work is called the wages of the person for that particular work,

In other words, we can find the entire wages of any person by the following formula

Entire wages = Total number of days x Wages of 1 day of any person

 

For example:

 If Arjun's monthly wages$ 4200 and he worked for all 30 days, then his daily wages will be calculated as Total wages = Number of days x Daily wages 4200 = 30 x Daily wages

Important Points:

  • Wages is directly proportional to the work done. It means, more money will be received for more work and less money will be received for less work.
  • Wages is indirectly proportional to the time taken by the individual.
  • Wages is directly proportional to 1 day work of each individual.

Examples:

  1. Alex can do a piece of work in 6 days, while Kevin can do the same work in 5 days. If the total amount to be given for this work is $ 660, then what will be the share of Kevin, if both work together

Solution: Time taken by Alex = 6 days

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore,&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mtext>&#xA0;and&#xA0;</mtext></math>

time taken by Kevin = 5 days

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>5</mn></mfrac></math>

Total amount earned = $660

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, Ratio of their incomes&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mo>:</mo><mfrac><mn>1</mn><mn>5</mn></mfrac><mo>=</mo><mn>5</mn><mo>:</mo><mn>6</mn><mspace linebreak="newline"/><mtext>&#xA0;Therefore, Kevin's share&#xA0;</mtext><mo>=</mo><mfrac><mn>660</mn><mrow><mn>5</mn><mo>+</mo><mn>6</mn></mrow></mfrac><mo>&#xD7;</mo><mn>6</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>360</mn></math>

  2. Wages of 45 women for 48 days amount to $ 31050. How many men must work for 16 days to receive $ 11500, if the daily wages of a man being double those of a woman?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: 1-day wages of a woman&#xA0;</mtext><mo>=</mo><mfrac><mn>31050</mn><mrow><mn>45</mn><mo>&#xD7;</mo><mn>48</mn></mrow></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mn>115</mn><mn>8</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;1-day wages of a man&#xA0;</mtext><mo>=</mo><mfrac><mn>115</mn><mn>8</mn></mfrac><mo>&#xD7;</mo><mn>2</mn><mo>=</mo><mi mathvariant="normal">$</mi><mfrac><mn>115</mn><mn>4</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;Therefore, Required number of men&#xA0;</mtext><mo>=</mo><mfrac><mn>11500</mn><mrow><mn>16</mn><mo>&#xD7;</mo><mfrac><mn>115</mn><mn>4</mn></mfrac></mrow></mfrac><mo>=</mo><mn>25</mn><mtext>&#xA0;men</mtext></math>

  3. Men, women and children are employed to do a work in the proportion of 3: 2: 1 and their wages as 5:3:2. When 90 men are employed, total daily wages of all amounts to $ 10350. Find the daily wages of a man.

Solution:

Let the numbers of men, women and children are 3y, 2y and y, respectively.

Given, 3y = 90 => y = 30

Number of women = 60 and number of children = 30

Let the men's, women's and children's wages be $ 5x, $ 3x and $ 2x, respectively.

 According to the question,

Total daily wages = $ 10350

=> 90 x (5x) + 60 x (3x) + 30 x (2x) = 10350

=> x (450 + 180 + 60) = 10350

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore,&#xA0;</mtext><mi>x</mi><mo>=</mo><mfrac><mn>10350</mn><mn>690</mn></mfrac><mo>=</mo><mn>15</mn></math>

Therefore, Daily wages of a man = 15 x 5 = $ 75

  4. A alone can finish a work in 2 days, while B alone can finish it in 3 days. If they work together to finish it, then out of total wages of $ 6000, what will be the 20% of A's share?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: A's&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;B's&#xA0;</mtext><mn>1</mn><mtext>&#xA0;day's work&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mspace linebreak="newline"/><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>s share:&#xA0;</mtext><msup><mi>B</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>s share&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>:</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>6</mn></mfrac><mo>:</mo><mfrac><mn>2</mn><mn>6</mn></mfrac><mo>=</mo><mn>3</mn><mo>:</mo><mn>2</mn><mspace linebreak="newline"/><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>s share&#xA0;</mtext><mo>=</mo><mfrac><mn>3</mn><mn>5</mn></mfrac><mo>&#xD7;</mo><mn>6000</mn><mo>=</mo><mn>3</mn><mo>&#xD7;</mo><mn>1200</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>3600</mn><mspace linebreak="newline"/><mtext>&#xA0;Therefore,&#xA0;</mtext><mn>20</mn><mi mathvariant="normal">%</mi><mtext>&#xA0;of&#xA0;</mtext><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mtext>s share&#xA0;</mtext><mo>=</mo><mn>3600</mn><mo>&#xD7;</mo><mfrac><mn>20</mn><mn>100</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>720</mn></math>

  5. A man and a boy received $ 1400 as wages for 10 days for the work they did together. The man's efficiency in the work was six times that of the boy. What is the daily wages of the boy?

Solution: The ratio of efficiency of man to boy = 6: 1

Efficiency ∞ Wages

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, Boy's share&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><mn>6</mn></mrow></mfrac><mo>&#xD7;</mo><mn>1400</mn><mo>=</mo><mfrac><mn>1</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>1400</mn><mo>=</mo><mi mathvariant="normal">$</mi><mn>200</mn></math>

Now, they worked for 10 days.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, Daily wages of a boy&#xA0;</mtext><mo>=</mo><mfrac><mn>200</mn><mn>10</mn></mfrac><mo>=</mo><mi mathvariant="normal">$</mi><mn>20</mn></math>


26 Pipes and Cisterns
N/A

Problems on Pipes and Cisterns are based on the basic concept of time and work

Pipes are connected to a tank or cistern and are used to fill or empty the tank or cistern.

In pipe and cistern, the work is done in form of filling or emptying a cistern/tank

Inlet pipe It fills a tank/cistern/reservoir. Outlet pipe It empties a tank/cistern/reservoir

 

Important Points:

  • If a pipe can fill/empty a tank in m h, then the pat of tank filled/emptied in 1 hr = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi>m</mi></mfrac></math>
  • If a pipe can fill/empty <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>'</mo><mfrac><mn>1</mn><mi>m</mi></mfrac><mo>'</mo></math> part of a tank in 1 h, then it can fill/empty the whole tank in ‘m’ h
  • Generally, time taken to fill a tank is taken(+ve) and time taken to empty a tank is taken negative(-ve).
  • If a pipe fills a tank in m h and another pipe fills in n h. Then, part filled by both pipes in <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mi>h</mi><mi>r</mi><mo>=</mo><mfrac><mn>1</mn><mi>m</mi></mfrac><mo>+</mo><mfrac><mn>1</mn><mi>n</mi></mfrac></math>

Examples:

  1. If a pipe can fill a tank in 2 h and another pipe can fill the same tank in 6 h, then what part of a tank will be filled by both the pipes in 1 h, if they are opened simultaneously?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: In&#xA0;</mtext><mn>1</mn><mi>h</mi><mi>r</mi><mtext>, part filled by&#xA0;</mtext><msup><mn>1</mn><mtext>st&#xA0;</mtext></msup><mtext>&#xA0;pipe&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mi>m</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>;</mo><mspace linebreak="newline"/><mtext>&#xA0;In&#xA0;</mtext><mn>1</mn><mi>h</mi><mi>r</mi><mtext>, part filled by&#xA0;</mtext><msup><mn>2</mn><mtext>nd&#xA0;</mtext></msup><mtext>&#xA0;pipe&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mi>n</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mtext>;</mtext><mspace linebreak="newline"/><mtext>&#xA0;Therefore, in&#xA0;</mtext><mn>1</mn><mi>h</mi><mi>r</mi><mtext>, part filled by both the pipes together&#xA0;</mtext><mo>=</mo><mfenced separators="|"><mrow><mfrac><mn>1</mn><mi>m</mi></mfrac><mo>+</mo><mfrac><mn>1</mn><mi>n</mi></mfrac></mrow></mfenced><mo>=</mo><mfenced separators="|"><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mn>6</mn></mfrac></mrow></mfenced><mo>=</mo><mfrac><mrow><mn>3</mn><mo>+</mo><mn>1</mn></mrow><mn>6</mn></mfrac><mo>=</mo><mfrac><mn>4</mn><mn>6</mn></mfrac><mo>=</mo><mfrac><mn>2</mn><mn>3</mn></mfrac><mtext>&#xA0;part</mtext></math>

  1. A pipe can fill a tank in 10 h. Due to a leak in the bottom, it fills the tank in 20 h. If the tank is full, how much time will the leak take to empty it?

Solution: Let the leak empties full tank in x h, then part emptied in 1 h by leak = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mi>x</mi></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Also, part filled by u=inlet pipe in&#xA0;</mtext><mn>1</mn><mi>h</mi><mi>r</mi><mo>=</mo><mfrac><mn>1</mn><mn>10</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;According to the question,&#xA0;</mtext><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mi>x</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>20</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;Therefore,&#xA0;</mtext><mfrac><mn>1</mn><mi>x</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>20</mn></mfrac><mo>&#x2212;</mo><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn><mo>&#x2212;</mo><mn>2</mn></mrow><mn>20</mn></mfrac><mo>=</mo><mo>&#x2212;</mo><mfrac><mn>1</mn><mn>20</mn></mfrac><mo>[</mo><mtext>-ve sign means leak empties tank]</mtext></math>

Therefore, leak will empty the full tank in 20 h.

  1. There are two tanks A and B to fill up a water tank. The tank can be filled in 40 min, if both taps are on. The same tank can be filled in 60 min, if tap A alone is on. How much time will tap B alone take, to fill up the same tank?

Solution:  Part filled by tap A in 1 min = 1/60

Let tap B fills the tank in x min

Then, part filled by tap, B in 1 min = 1/x

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;According to the question,&#xA0;</mtext><mfrac><mn>1</mn><mn>60</mn></mfrac><mo>+</mo><mfrac><mn>1</mn><mi>x</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>40</mn></mfrac><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mfrac><mn>1</mn><mi>x</mi></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>40</mn></mfrac><mo>&#x2212;</mo><mfrac><mn>1</mn><mn>60</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>120</mn></mfrac></math>

Therefore, Tap B can fill the tank in 120 min.

  1. Inlet A is four times faster than inlet B to fill a tank. If A alone can fill it in 15 min, how long will it take if both the pipes are opened together?

Solution: Time taken by A to fill the tank, m = 15 min

Therefore, time taken by B to fill the tank, n = 15 x 4 = 60 min

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, required time taken&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>m</mi><mi>x</mi><mi>n</mi></mrow><mrow><mi>m</mi><mo>+</mo><mi>n</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>15</mn><mo>&#xD7;</mo><mn>60</mn></mrow><mrow><mn>15</mn><mo>+</mo><mn>60</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>15</mn><mo>&#xD7;</mo><mn>60</mn></mrow><mn>75</mn></mfrac><mo>=</mo><mn>12</mn><mi>m</mi><mi>i</mi><mi>n</mi></math>

  5. A tap having diameter 'd' can empty a tank in 40 min. How long another tap having diameter '2d' take to empty the same tank?

Solution: Area of tap ∞ work done by pipe

When diameter is doubled, area will be four times. So, it will work four times faster.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, required time taken to empty the tank&#xA0;</mtext><mo>=</mo><mn>40</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>4</mn></mfrac><mo>=</mo><mn>10</mn><mi>m</mi><mi>i</mi><mi>n</mi><mtext>.&#xA0;</mtext></math>


27 Clock
N/A

Clock

A clock is an instrument which displays time divided into hours, minutes and seconds.

 A clock mainly consists of four components.

Dial

A clock is a circular dial. The periphery of the dial is numbered 1 through 12 indicating the hours in a 12 h cycle. The circumference of a dial is divided into 60 equal spaces.

Every clock has mainly two hands, one is smaller and other is bigger. The smaller hand is slower the and the bigger hand is faster.

Hour Hand

The smaller or slower hand of a clock is called the hour hand. It makes two revolutions in a day. Minute Hand

The bigger or faster hand of a clock is called the minute hand. It makes one revolution in every hour.

 

Second Hand

Second hand indicates seconds on a circular dial. It makes one revolution per minute.

 Note:

In 1 h minute hand covers 60 min spaces whereas the hour hand covers 5 min spaces Therefore, minute hand gains (60 - 5) = 55 min in 1 h

Important Points Related to Clock

  1. In 1 h, both hands coincide once (i.e., 0° apart)

For example: Between 3 and 4'o clock, hands are together as shown in adjacent figure

  2. In 12 h, both hands coincide 11 times (between 11 and 1'o clock they coincide once) and in a day both hands coincide 22 times.

For example, between 11 and 1'o clock, hands are together as shown in adjacent figure.

  3. If two hands are at 90 ° they are 15 min spaces apart. This happens twice in 1 h. In a period of 12 h, the hands are at right angle 22 times (2 common positions) and in a day both hands are at right angle 44 times.

  4. If two hands are in opposite direction. (i.e., 180°apart), then they are 30 min spaces apart. This happens once in 1 h. In a period of 12 h both hands are in opposite direction 11 times and in a day both hands are in opposite direction 22 times.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;5. Angle covered by minute hand in&#xA0;</mtext><mn>1</mn><mi>m</mi><mi>i</mi><mi>n</mi><mo>=</mo><mfrac><mtext>&#xA0;Total angle&#xA0;</mtext><mtext>&#xA0;Number of spaces&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mn>360</mn><mn>60</mn></mfrac><mo>=</mo><msup><mn>6</mn><mn>0</mn></msup><mtext>&#xA0;in&#xA0;</mtext><mn>1</mn><mi>m</mi><mi>i</mi><mi>n</mi></math>

  6. Angle covered by hour hand in 1 min.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;As hour hand covers&#xA0;</mtext><msup><mn>360</mn><mo>&#x2218;</mo></msup><mtext>&#xA0;in&#xA0;</mtext><mn>12</mn><mi>h</mi><mtext>. Hence, hour hand covers&#xA0;</mtext><mfrac><mn>360</mn><mn>12</mn></mfrac><mo>=</mo><msup><mn>30</mn><mo>&#x2218;</mo></msup><mtext>&#xA0;in&#xA0;</mtext><mn>1</mn><mi>h</mi><mtext>.</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Hence, hour hand covers&#xA0;</mtext><mfrac><mn>30</mn><mn>60</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mtext>&#xA0;in&#xA0;</mtext><mn>1</mn><mi>m</mi><mi>i</mi><mi>n</mi></math>

  7. From point 5 and 6, we can say that the minute hand goes ahead by <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>5</mn><mfrac><mn>1</mn><mn>2</mn></mfrac></math>  in comparison to hour band.

Concept of Slow or Fast Clocks

If a watch/clock indicates 9: 15, when the correct time is 9, then it is said to be 15 min too fast.

On the other hand, if the watch/clock indicates 6: 45, when the correct time is 7, then it is said to be 15 min too slow.

Examples:

  1.  A clock gains 10 s in every 3 h. If the clock was set right at 4:00 am on Monday morning, then the time it will indicate on Tuesday evening at 7:00 pm.

Solution: Difference of time between 4.00 am on Monday to 7.00 pm Tuesday

                                              24 + 12 + 3 = 39 h

  1. At what time between 4 o'clock and 5 o'clock, will the hands of a clock be together?

Solution: At 4 o'clock, the hour hand is at 4 and the minute hand is at 12.

 It means that they are 20 min spaces apart.

To be together, the minute hand must gain 20 min over the hour hand.

As we know, 55 min is gained by minute hand in 60 min

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore,&#xA0;</mtext><mn>20</mn><mtext>&#xA0;min will be gained in&#xA0;</mtext><mfenced separators="|"><mrow><mfrac><mn>60</mn><mn>55</mn></mfrac><mo>&#xD7;</mo><mn>20</mn></mrow></mfenced><mo>min</mo><mo>=</mo><mfrac><mrow><mn>60</mn><mo>&#xD7;</mo><mn>4</mn></mrow><mn>11</mn></mfrac><mo>=</mo><mfrac><mn>240</mn><mn>11</mn></mfrac><mi>m</mi><mi>i</mi><mi>n</mi><mo>=</mo><mn>21</mn><mfrac><mn>9</mn><mn>11</mn></mfrac><mi>m</mi><mi>i</mi><mi>n</mi></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Hence, the hands will coincide at&#xA0;</mtext><mn>21</mn><mfrac><mn>9</mn><mn>11</mn></mfrac><mo>min</mo><mtext>&#xA0;past&#xA0;</mtext><mn>4</mn><mo>.</mo></math>

  3. At what time between 3 o'clock and 4 o'clock, will the hands of a clock be in opposite directions?

 Solution: At 3 o'clock, the hour hand is at 3 and the minute hand is at 12.

It means that the two hands are 15 min spaces apart. But to be in opposite directions, the hands must be 30 min spaces apart.

Therefore, the minute hand will have to gain (30 + 15) = 45 min spaces over the hour hand.

Therefore 55 min spaces are gained in 60 min

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore&#xA0;</mtext><mn>45</mn><mi>m</mi><mi>i</mi><mi>n</mi><mtext>&#xA0;spaces are gained in&#xA0;</mtext><mfenced separators="|"><mrow><mfrac><mn>60</mn><mn>55</mn></mfrac><mo>&#xD7;</mo><mn>45</mn></mrow></mfenced><mo>min</mo><mo>=</mo><mfrac><mrow><mn>60</mn><mo>&#xD7;</mo><mn>9</mn></mrow><mn>11</mn></mfrac><mo>min</mo><mo>=</mo><mfrac><mn>540</mn><mn>11</mn></mfrac><mo>=</mo><mn>49</mn><mfrac><mn>1</mn><mn>11</mn></mfrac><mi>m</mi><mi>i</mi><mi>n</mi><mspace linebreak="newline"/><mtext>&#xA0;Hence, required time&#xA0;</mtext><mo>=</mo><mn>49</mn><mfrac><mn>1</mn><mn>11</mn></mfrac><mo>min</mo><mtext>&#xA0;past&#xA0;</mtext><mn>3</mn></math>

  4. At what point of time after 3 O' clock, hour hand and the minute hand are at right angles for the first time?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Clock will make right angle at&#xA0;</mtext><mo>(</mo><mn>5</mn><mi>n</mi><mo>+</mo><mn>15</mn><mo>)</mo><mo>&#xD7;</mo><mfrac><mn>12</mn><mn>11</mn></mfrac><mtext>&#xA0;min past&#xA0;</mtext><mi>n</mi><mtext>.&#xA0;</mtext></math>

Given that, n = 3

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore&#xA0;</mtext><mo>(</mo><mn>5</mn><mo>&#xD7;</mo><mn>3</mn><mo>+</mo><mn>15</mn><mo>)</mo><mo>&#xD7;</mo><mfrac><mn>12</mn><mn>11</mn></mfrac><mtext>&#xA0;min past&#xA0;</mtext><mn>3</mn><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mn>30</mn><mo>&#xD7;</mo><mfrac><mn>12</mn><mn>11</mn></mfrac><mo>min</mo><mtext>&#xA0;past&#xA0;</mtext><mn>3</mn><mo>.</mo></mtd></mtr><mtr><mtd><mo>=</mo><mn>32</mn><mfrac><mn>8</mn><mn>11</mn></mfrac><mtext>&#xA0;min past&#xA0;</mtext><mn>3</mn><mtext>&#xA0;i.e.,&#xA0;</mtext><mn>3</mn><mi>h</mi><mtext>&#xA0;and&#xA0;</mtext><mn>32</mn><mfrac><mn>8</mn><mn>11</mn></mfrac><mi>m</mi><mi>i</mi><mo>.</mo></mtd></mtr></mtable></math>

  5. At what time between 9 o'clock and 10 o'clock, will the hands of a clock be in the same straight line but not together?

Solution: At 9 o' clock, the hour hand is at 9 and the minute hand is at 12.

 It means that the two hands are 15 min spaces apart.

To be in the same straight line (but not together), they will be 30 min space apart.

... The minute hand will have to gain (30-15) = 15 min spaces over the hour hand.

As we know, 55 min spaces are gained in 60 min.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2234;</mo><mn>15</mn><mtext>&#xA0;min will be gained in&#xA0;</mtext><mfenced separators="|"><mrow><mn>15</mn><mo>&#xD7;</mo><mfrac><mn>60</mn><mn>55</mn></mfrac></mrow></mfenced><mo>min</mo><mo>=</mo><mfrac><mn>180</mn><mn>11</mn></mfrac><mo>min</mo><mo>=</mo><mn>16</mn><mfrac><mn>4</mn><mn>11</mn></mfrac><mi>m</mi><mi>i</mi><mi>n</mi><mspace linebreak="newline"/><mtext>&#xA0;Hence, the hands will be in the same straight line but not together at&#xA0;</mtext><mn>16</mn><mfrac><mn>4</mn><mn>11</mn></mfrac><mtext>&#xA0;min past&#xA0;</mtext><mn>9</mn><mo>.</mo></math>


28 Calendar
N/A

Calendar

 A calendar is chart or series of pages showing the days, weeks and months of a particular year. A calendar consists of 365 or 366 days divided into 12 months.

Ordinary Year

A year having 365 days is called an ordinary year (52 complete weeks + 1 extra day = 365 days)

Leap Year

 A leap year has 366 days (the extra day is 29th of February) (52 complete weeks + 2 extra days =366 days.)

 A leap year is divisible by 4 except for a century. For a century to be a leap year it must be divisible by 400. e.g.,

  • Years like 1988, 2008 are leap year (divisible by 4).
  •  Centuries like 2000, 2400 are leap year (divisible by 400).
  •  Years like 1999, 2003 are not leap year (not divisible by 4).
  •  Centuries like 1700, 1800 are not leap year (not divisible by 400).
  • In a century, there is 76 ordinary year and 24 leap years.

Odd Days

  • Extra days, apart from the complete weeks in a given period are called odd days.
  • Number of days in an ordinary year = 365 = (52 x 7) + 1 = 52 weeks + 1 odd day
  •  An ordinary year has 1 odd day
  • Number of days in a leap year = 366 = (52 x 7) + 2 = 52 weeks + 2 days
  • while a leap year has 2 odd days.
  •  Number of days in a century (100 yr) = 76 ordinary years + 24 leap years = 76x1+ 24x2) =124 =17x7+5 =17 week+ 5 odd days
  • Therefore, 100yrs has 5 odd days.
  • Number of odd days in 200 yr = 5x2=10 days = 1 week + 3 days = 3 odd days
  • Number of odd days in 300 yr = 5 x 3 = 15 days = 2 weeks + 1 day = 1 odd day
  • Number of odd days in 400 yr = (5 x 4 +1) days =21 days = 3 weeks = 0 odd days
  • As 400th is a leap year, therefore 1 more day has been taken
  •  Similarly, each one of 800 yr, 1200 yr, 1600 yr, 2000 yr, 2400 yr etc., has no odd days.
  • Remember the adjacent table for the number of odd days in different months of an year.
  • In an ordinary year, February has no odd days, but in a leap year, February has one odd day.
  •  The 1 st day of a century must be Tuesday, Thursday or Saturday.
  • The last day of a century cannot be Tuesday, Thursday or Saturday.

Day Gain/Loss

Ordinary Year (± 1 day)

  • When we proceed forward by 1 yr, then 1 day is gained.

 For example: 9th August 2013 is Friday, then 9th August 2014 has to be Friday +1 = Saturday.

  • When we move backward by 1 yr, then 1 day is lost.
  •  For example: 24th December 2013 is Tuesday, then 24th December 2012 has to be Tuesday -1 = Monday.

Leap Year (+ 2 days)

  • When we proceed forward by 1 leap year, then 2 days are gained.

For example: If it is Wednesday on 25th December 2011, then it would be Friday on 25th December 2012 [Wednesday + 2] because 2012 is a leap year.

  • When we move backward by 1 leap year, then 2 days are lost.

For example: If it is Wednesday on 18th December 2012, then it would be Monday on 18th December 2011. [Wednesday -2] because 2012 is a leap year.

Exception

• The day must have crossed 29th February for adding 2 days otherwise 1 day.

For example: If 26th January 2011 is Wednesday, 26th January 2012 would be Wednesday + 1 = Thursday (even if 2012 is leap year, we have added + 1 day because 29 February is not crossed).

If 23rd March 2011 is Wednesday, then 23rd March 2012 would be Wednesday + 2 = Friday (+ 2 days 29th February of leap year is crossed)

To Find a Particular Day on the Basis of Given Day and Date

 Following steps are taken into consideration to solve such questions

 Step I: Firstly, you have to find the number of odd days between the given date and the date for which the day is to be determined.

 Step II: The day (for a particular date) to be determined, will be that day of the week which is equal to the total number of odd days and this number is counted forward from the given day, in case the given day comes before the day to be determined.

 But, if the given day comes after the day to be determined, then the same counting is done backward from the given day.

To Find a Particular Day without Given Date and Day

 Following steps are taken into consideration to solve such questions

 Step I: Firstly, you have to find the number of odd days up to the date for which the day is to be determined.

Step II: Your required day will be according to the following conditions

(a) If the number of odd days = 0, then required day is Sunday.

 (b) If the number of odd days = 1, then required day is Monday.

(c) If the number of odd days = 2, then required day is Tuesday.

 (d) If the number of odd days = 3, then required day is Wednesday.

(e) If the number of odd days = 4, then required day is Thursday.

(f) If the number of odd days = 5, then required day is Friday. (g) If the number of odd days = 6, then required day is Saturday.

Examples:

  1. 14 How many days are there in x weeks x days?

Solution: Number of days in x weeks = 7x + x

... Total number of days is x week x days = 7x + x = 8x days

  1. Find the day of the week on 26th January 1950.

Solution: Number of odd days up to 26th January, 1950

= Odd days for 1600 yr + Odd days for 300 yr + Odd days for 49 yr + Odd days of 26 days of January 1950 = 0 + 1 + (12 X2 +37) + 5 = 0 + 1 + 61 + 5 = 67 days = 9 weeks + 4 days = 4 odd days

Therefore, It was Thursday on 26th January 1950.

  1. January 3, 2007 was Wednesday. What day of the week fell on January 3, 2008?

Solution: The year 2007 is an ordinary year, so it has 1 odd day.

 3rd day of the year 2007 was Wednesday.

... 3rd day of the year 2008 will be one day beyond the Wednesday.

Hence, it will be Thursday.

  1. What was the day of the week on 17th August, 2010?

Solution: Period up to 17th August, 2010 = (2009yr + Period from 1.1.2010 to 17.8.2010)

Counting of odd days odd days in 1600 yr = 0 Odd days in 400 yr = 0

9 yr = (2 leap years + 7 ordinary years) = (2x2+7xl) = l week + 4 days = 4 odd days

Number of days between 1.1.2010 to 17.8.2010

January + February + March + April + May + June + July + August

 = (31 +28 + 31+30 + 31 + 30 + 31 + 17) days

 = 229 days = 32 weeks + 5 odd days

Total number of odd days = (0 + 0 + 4 + 5) days = 9 days = 1 week + 2 odd days

 Hence, the required day is Tuesday

  1. If 5th March, 1999 was Friday, what day of the week was it on 9th March 2000?

Solution: 5th March, 1999 is Friday.

Then, 5th March 2000 = Friday + 2 = Sunday.

{2000 is leap year and it crosses 29th Feb 2000, so 2 is taken as odd day}

...  5th March 2000 = Sunday. Then, 9th March 2000 = Thursday.


29 Boats and Streams
N/A

Boats and streams

It is an application of concepts of speed, time and distance. Speed of river flowing either aides a swimmer (boat), while travelling with the direction of river or it opposes when travelling against the direction of river.

Still Water:

 If the speed of water of a river is zero, then water is considered to be still water.

 Stream Water

 If the water of a river is moving at a certain speed, then it is called as stream water.

Speed of Boat:

 Speed of boat means speed of boat (swimmer) in still water. In other words, if the speed of a boat (swimmer) is given, then that particular speed is the speed in still water.

 Downstream Motion:

 If the motion of a boat (swimmer) is along the direction of stream, then such motion is called downstream motion.

Upstream Motion:

 If the motion of a boat (swimmer) is against the direction of stream, then such motion is called upstream motion.

Formulae related to Boats and Strems:

  • If the speed of a boat in still water is a miles/h and speed of the stream is b miles/h, then
  • Speed downstream = (a + b) miles/h
  • Speed upstream = (a – b) miles/h

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Speed of a boat in still water&#xA0;</mtext><mo>(</mo><mi>a</mi><mo>)</mo><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mi>S</mi><mi>p</mi><mi>e</mi><mi>e</mi><mi>d</mi><mtext>&#xA0;downstream&#xA0;</mtext><mo>+</mo><mtext>&#xA0;Speed upstream&#xA0;</mtext><mo>)</mo><mspace linebreak="newline"/><mo>&#x2219;</mo><mo>&#xA0;</mo><mtext>&#xA0;Speed of stream&#xA0;</mtext><mo>(</mo><mi>b</mi><mo>)</mo><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mtext>&#xA0;Speed downstream - Speed upstream)&#xA0;</mtext></math>

Examples:

  1. If the speed of a boat in still water is 10 miles/h and the rate of stream is 5 miles/h, then find upstream speed of the boat.

Sol. Given,

speed of a boat = a = 10 miles/h Speed of stream = b = 5 miles/h

 Hence, Speed upstream = a – b = 10 - 5 = 5 miles/h.

 

 

  1. Kevin can row a certain distance downstream in 24 h and can return the same distance in 36 h. If the stream flows at the rate of 12 miles/h, then find the speed of Kevin in still water.

Solution:

  • Let the Kevin’s speed in still water = a
  •  Then, Kevin's speed downstream = (a + 12)
  • Kevin's speed upstream = (a - 12)

Given that,

  • Distance travelled downstream = Distance travelled upstream
  • 24 (a + 12) = 36 (a - 12)
  • 2 (a + 12) = 3 (a – 12)
  •  2a + 24 = 3a – 36
  •  3a – 2a = 36 + 24
  •  a = 60 miles/h
  1. A boat's speed in still water is 12 miles/h, while river is flowing with a speed of 4 miles/h and time taken to cover a certain distance upstream is 6 h more than time taken to river the same distance downstream. Find the distance.

Solution: Let the distance = d

Boat’s rate downstream = 12 + 4 = 16 miles/h

Boat’s rate upstream = 12 – 4 = 8 miles/h

Difference between the time = Time taken by boat to travel upstream - Time taken by boat to travel downstream

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mi>d</mi><mn>8</mn></mfrac><mo>&#x2212;</mo><mfrac><mi>d</mi><mn>16</mn></mfrac><mo>=</mo><mn>6</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mn>2</mn><mi>d</mi><mo>&#x2212;</mo><mi>d</mi></mrow><mn>16</mn></mfrac><mo>=</mo><mn>6</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mi>d</mi><mn>16</mn></mfrac><mo>=</mo><mn>6</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>d</mi><mo>=</mo><mn>96</mn><mtext>&#xA0;miles</mtext></mtd></mtr></mtable></math>

  4.  Andrew can row 36 miles/h in still water. It takes him twice as long to row up as to row down the river. Find the rate of stream.

Solution:

  • Let rate of stream be a miles/h.

According to the question,

  • 36 + a = 2(36 - a)
  • 36 + a = 72 - 2a
  •  3a = 72 - 36 = 36
  •  a = 36/3 = 12 miles/h

  5. A boatman takes twice as long to row a distance against the stream as to row the same distance with the stream. Find the ratio of speeds of the boat in still water and the stream.

Solution:

  • Let boatman's speed upstream p
  •  His speed downstream = 2p

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Speed in still water&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mtext>&#xA0;Speed downstream&#xA0;</mtext><mo>+</mo><mtext>&#xA0;Speed upstream&#xA0;</mtext><mo>)</mo><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mn>2</mn><mi>p</mi><mo>+</mo><mi>p</mi><mo>)</mo><mo>=</mo><mfrac><mrow><mn>3</mn><mi>p</mi></mrow><mn>2</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Speed of stream&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mtext>&#xA0;Speed downstream&#xA0;</mtext><mo>&#x2212;</mo><mtext>&#xA0;Speed upstream&#xA0;</mtext><mo>)</mo><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mn>2</mn><mi>p</mi><mo>&#x2212;</mo><mi>p</mi><mo>)</mo><mo>=</mo><mfrac><mi>p</mi><mn>2</mn></mfrac><mspace linebreak="newline"/><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mtext>Ratio&#xA0;</mtext><mo>=</mo><mo>(</mo><mtext>&#xA0;Speed in still water&#xA0;</mtext><mo>)</mo><mo>:</mo><mo>(</mo><mtext>&#xA0;Speed of stream&#xA0;</mtext><mo>)</mo><mo>=</mo><mfrac><mrow><mn>3</mn><mi>p</mi></mrow><mn>2</mn></mfrac><mo>:</mo><mfrac><mi>p</mi><mn>2</mn></mfrac><mo>=</mo><mn>3</mn><mo>:</mo><mn>1</mn></math>


1 Inequalities
N/A

The mathematical expressions in which both sides are not equal are called inequalities.

In inequality, unlike in equations, we compare two values where the equal to sign in between is replaced by <, > and ≠ sign.

  • a < b → a is less a than b
  • a > b → a is greater than b
  • a ≠ b → a is not equal to b
  • a ≤ b → a is less than or equal to b
  • a ≥ b → a is greater than or equal to b

Rules of inequalities:

Rule1:

  • If, a < b and b < c, then a < b
  • If, a > b and b > c, then a > b

Rule2:

  • If, a > b, then b < a
  • If, a < b, then b > a

Rule3:

  • If a < b, then a − c < b – c and a + c < b + c
  • If a > b, then a + c > b + c and a − c > b – c

Rule4:

  • If a < b, and c is positive, then ac < bc
  • If a < b, and c is negative, then ac > bc (inequality swaps)

Rule5:

Having minus in front of a and b changes the direction of the inequality.

  • If a < b then −a > −b
  • If a > b, then −a < −b 

Rule6:

The reciprocal of both a and b changes the direction of the inequality.

When a and b are both positive or both negative.

  • If a < b, then 1/a > 1/b
  • If a > b, then 1/a < 1/b

Rule7:

Taking a square root will not change the inequality.

  • If a ≤ b, then 

Solving inequalities:

To solve an inequality, the following steps:

  • Eliminate fractions
  • Simplify
  • Add or subtract quantities

Important points:

  • Inequalities can be solved by adding, subtracting, multiplying, or dividing both sides by the same number.
  • Do not multiply or divide by a variable.
  • Dividing or multiplying both sides by negative numbers will alter the inequality's direction.

 

  1. Absolute value inequalities:

It is an inequality with an absolute value symbol in it and can be solved by two methods.

  • Using number line
  • Using formulae

The absolute value inequalities are of two types:

  • one with < or ≤
  • one with > or ≥

 Formulae

  1. When the inequality is in the form |x|< a or |x|≤ a

Formulae to solve the inequality:

If |x|< a => -a < x < a

If |x|≤ a => -a ≤ x ≤ a

  1. When the inequality is in the form |x|> a or |x|≥ a

Formulae to solve the inequality:

If |x|> a => x < - a or x > a

If |x|≥ a => x ≤ - a or x ≥ a

  1. When the inequality is in the form |x|< - a or |x|≤ - a

|x|< −a or |x|≤ −a ⇒ No Solution

  1. When the inequality is in the form |x|> - a or |x|≥ - a

|x|> −a or |x|≥ −a ⇒ Set of all Real Numbers, R

 2. Quadratic Inequalities:

The standard form of quadratic inequalities in one variable is almost the same as the standard form of a quadratic equation.

The only difference is that the quadratic equation has an "equal to" sign in it while a quadratic inequality has a "greater than" or "less than" sign (> or <).

The quadratic inequality is represented as: ax2 + bx + c > 0 or ax2 + bx + c < 0

Quadratic inequality can have infinite values of x which satisfy the condition ax2+ bx +c<0 ax2+ bx + c <0.

Now consider a quadratic expression x2 + bx + c. We can write the quadratic expression in the form of (x−α) (x−β) and α < β.

By number line method

It means that if x2 + bx + c, then x can take values between -  to α and β to + 

  • If x2 + bx + c, then x ? (- , α) ∪ (β, + )
  • If x2 + bx + c, then x can take values between α and β.
  • If x2 + bx + c, then x ? (α, β)

Symbols used inequalities:

(-1, 1) -> x cannot take value -1 and 1.

[-1, 1) -> x can take value -1 and but not 1.

(-1,1] -> x cannot take value -1 but it can take value 1.

[-1, 1] -> x can take both -1 and 1 values

  1. Linear Inequalities:

Linear inequalities are defined as expressions in which two linear expressions are compared using the inequality symbols.

Rules of Linear Inequalities:

Four types of operations that are done on linear inequalities are addition, subtraction, multiplication, and division.

Addition Rule:

If a > b, then a + c > b + c

if a < b, then a + c < b + c.

Subtraction Rule:

If a > b, then a − c > b – c

if a < b, then a – c < b − c.

Multiplication Rule:

If a > b and c > 0, then a × c > b × c

If a < b and c > 0, then a × c < b × c,

If a > b and c < 0, then a × c < b × c

If a < b and c < 0, then a × c > b × c.

Division Rule:

If a > b and c > 0, then (a/c) > (b/c)

If a < b and c > 0, then (a/c) < (b/c)

If a > b and c < 0, then (a/c) < (b/c)

If a < b and c < 0, then (a/c) > (b/c)

Examples:

    1. What is the scope of (x -1) (x + 5)/ (x + 3) < x?

Solution:

For inequalities, if you multiply both sides by (x + 3)2 the squared number is always positive, so the inequality sign doesn’t change.

Thus, you get (x + 3)2 ((x -1) (x + 5)/ (x + 3)) < x (x + 3)2

(x + 3) (x - 1) (x + 5) = (x + 3)2 x

(x + 3) (x - 1) (x + 5) - (x + 3)2 x < 0

(x + 3) {(x - 1) (x + 5) - (x + 3) x} < 0

(x + 3) {(x2 + 4x – 5) - (x2 + 3x)} < 0

(x + 3) {x - 5} < 0

- 3 < x < 5

 2.  If  is not a real number, what is the scope of x?

Solution:

  • If a certain polynomial is “not a real number” or “not defined”, the denominator must be 0 or the number inside the” √” root must be negative.
  • Then, this question also has to be -4x2 + 16 ≤ 0, and if you divide both sides by -4, from x2 – 4 ≥ 0, (x – 2) (x + 2) ≥ 0,
  •  x ≤ -2 or 2 ≤ x.

 3. Solve the linear inequality in one variable: 10x + 5 < 8x + 25

Solution:

Given inequality 10x + 5 < 8x + 25

10x + 5 – 8x < 8x + 25 – 8x

2x + 5 < 25

2x < 25 - 5

2x < 20

x < 10

    1. A gardener is shaping a plant. In order to get the exact shape, it needs 350 millimetres wide, allowing for a margin of error of 5.5 millimetres. Find the range of width of the plant using absolute value inequality.

Solution: Let the width of the plant to be x.

Then the absolute value inequality is |x−350|≤ 5.5

−5.5 ≤ x − 350 ≤ 5.5

344.5 ≤ x ≤ 355.5

The range of the plant’s width [344.5, 355.5]

 5. Find the solution of the inequalities 10x + 12 >52 and 6x + 18 < 72.

Solution: Given 10x + 12 >52 and 6x + 18 < 72

10x > 52 – 12 and 6x < 72 – 18

10x > 40 and 6x < 54

x > 4 and x < 9

 


2 Exponent
N/A

Exponent:

It is a numerical notation that indicates the number of times a number is to be multiplied by itself.

It is also called as power or index.

It is used to write a very big number in the simplest form.

Example: 10000 = 104

Suppose, a number ‘p’ is multiplied by itself k-times, then it is represented as pk where p is the base and k are the exponent.

In the number 64, 6 is called as base and 4 is called as exponent or power.

It is read as 6 raised to the power 4

Comparison using exponents:

  • When two numbers in the standard form have same power of 10, then number with larger factor is greater

7.54 x 1022 < 8.432 x 1024

  • When two numbers in the standard form have same factor, then number with larger power of 10 is greater

7.54 x 1012 > 7.54 x 1010

  • When two numbers in the standard form have different factors and the different powers of 10, then the number with larger power of 10 is larger number

7.54 x 1012 > 7.54 x 1010

Important points to remember:

  • A positive number raised to an even or odd power always remains positive.
  • A negative number raised to an odd power always remains negative.
  • A negative number raised to an even power always turns to be positive.

In simple we can say that any number raised to an even power either remains positive or becomes positive and any number raised to odd power keeps the sign it starts with.

Exponents Rules:

  1. Product Rule:

same base different power

          au x av = au + v

different base same power

          au x bu = (a x b) u

Example: 25 x 26 = 25+6 = 211

                 42 x 52 = (4 x 5)2 = 202

  1. Quotient Rule:

same base different power

(au/ av) = au – v

different base same power

(au/ bu) = (a/b) u

Example: 35/ 32 = 35 – 2 = 33 = 27

               : 123/ 63 = 23 = 8

  1. Power Rule:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><msup><mfenced separators="|"><msup><mi>a</mi><mi>u</mi></msup></mfenced><mi>v</mi></msup><mo>=</mo><msup><mi>a</mi><mrow><mi>u</mi><mi>v</mi></mrow></msup></mtd></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><msup><mi>a</mi><msup><mi>u</mi><mi>v</mi></msup></msup><mo>=</mo><msup><mi>a</mi><mfenced separators="|"><msup><mi>u</mi><mi>v</mi></msup></mfenced></msup></mtd></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><msup><mi>a</mi><mrow><mn>1</mn><mo>/</mo><mi>u</mi></mrow></msup><mo>=</mo><mroot><mi>a</mi><mi>u</mi></mroot></mtd></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mroot><msup><mi>a</mi><mi>v</mi></msup><mi>u</mi></mroot><mo>=</mo><msup><mi>a</mi><mrow><mi>v</mi><mo>/</mo><mi>u</mi></mrow></msup></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>E</mi><mi>x</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mo>:</mo><msup><mfenced separators="|"><msup><mn>5</mn><mn>2</mn></msup></mfenced><mn>3</mn></msup><mo>=</mo><msup><mn>5</mn><mrow><mn>2</mn><mo>&#xD7;</mo><mn>3</mn></mrow></msup><mo>=</mo><msup><mn>5</mn><mn>6</mn></msup></mtd></mtr><mtr><mtd><msup><mn>5</mn><msup><mn>2</mn><mn>3</mn></msup></msup><mo>=</mo><msup><mn>5</mn><mfenced separators="|"><msup><mn>2</mn><mn>3</mn></msup></mfenced></msup><mo>=</mo><msup><mn>5</mn><mn>8</mn></msup></mtd></mtr><mtr><mtd><msup><mn>5</mn><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup><mo>=</mo><mroot><mn>5</mn><mn>2</mn></mroot></mtd></mtr><mtr><mtd><mroot><msup><mn>5</mn><mn>3</mn></msup><mn>2</mn></mroot><mo>=</mo><msup><mn>5</mn><mrow><mn>3</mn><mo>/</mo><mn>2</mn></mrow></msup></mtd></mtr></mtable></math>

  4. Negative exponent rule:

a-u = 1/au

Example: 3-2 = 1/32

  5. Zero rule:

a0 = 1

0u = 0, for n > 0

Example: 50 = 1

                  03 = 0

  6. One rule:

b1 = b

1u = 1

Example: 51 = 5

                  15 = 1

  7. Minus one rule:


3 Linear Equations
N/A

A linear equation is an equation for a straight line. So, the equation which has degree 1, i.e., which has linear power of the variables, is called a linear equation.

It is written as ax + by + c = 0, where a, b and c are real numbers and a and b both are not zero

For example, y = 2x +1 is a linear equation. The different values of x and y are

All these values of (x, y) as (1,3), (2,5), (0,1) etc., are the solutions of the given linear equation.

If we are given two equations in x and y, then we are to find those values of x and y which satisfy both the given equations.

Linear Equation in One Variable

A linear equation in which number of unknown variables is one, is known as linear equation in one variable. For example, 3x + 5=10, y + 3=5 etc

Linear Equation in Two Variables

A linear equation in which number of unknown variables are two, is known as linear equation in two variables. For example, 2x + 5y = 10, x + 4y = 8 etc

Linear Equation in Three Variables

A linear equation in which number of unknown variables are three, is known as linear equation in three variables. For example, 4x + 6y + 7z = 20, x + y + 2z = 5 etc.

Note:

1. Linear equation in one variable represents a point in number line.

2. Linear equation in two variable represents a line in XY-plane (cartesian plane).

3. Linear equation in three variables represents a plane in XVZ-coordinate system.

Methods of Solving Linear Equations

There are following methods which are useful to solve the linear equations

Substitution Method

In this method, first the value of one variable must be represented in the form of another variable and put this value in another equation and solve it. Thus, a value of one variable is obtained and this value is used to find the value of another variable.

Elimination Method

In this method, the coefficients of one of the variables of each equation become same by multiplying a proper multiple. Solve these equations and by which we get the value of another variable and thus with the help of this value, we can find the value of another variable.

Cross Multiplication Method

Let a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are two equations

Therefore, By cross multiplication method,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mi>x</mi><mrow><msub><mi>b</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>b</mi><mn>2</mn></msub><msub><mi>c</mi><mn>1</mn></msub></mrow></mfrac><mo>=</mo><mfrac><mi>y</mi><mrow><msub><mi>c</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>c</mi><mn>2</mn></msub><msub><mi>a</mi><mn>1</mn></msub></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mrow><msub><mi>a</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>a</mi><mn>2</mn></msub><msub><mi>b</mi><mn>1</mn></msub></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mi>x</mi><mo>=</mo><mfrac><mrow><msub><mi>b</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>b</mi><mn>2</mn></msub><msub><mi>c</mi><mn>1</mn></msub></mrow><mrow><msub><mi>a</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>a</mi><mn>2</mn></msub><msub><mi>b</mi><mn>1</mn></msub></mrow></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>y</mi><mo>=</mo><mfrac><mrow><msub><mi>c</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>c</mi><mn>2</mn></msub><msub><mi>a</mi><mn>1</mn></msub></mrow><mrow><msub><mi>a</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub><mo>&#x2212;</mo><msub><mi>a</mi><mn>2</mn></msub><msub><mi>b</mi><mn>1</mn></msub></mrow></mfrac></mtd></mtr></mtable></math>

Consistency of the System of Linear Equations

A set of linear equations is said to be consistent, if there exists at least one solution for this equation.

A set of linear equations is said to be inconsistent, if there are no solution for this equation. Let us consider a system of two linear equations as shown, a1x+b1y+c1=0 and a2x+b2y+c2=0

Consistent System:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;The above system will be consistent, if&#xA0;</mtext><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>&#x2260;</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mtext>&#xA0;or&#xA0;</mtext><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>c</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub></mfrac><mspace linebreak="newline"/><mtext>&#xA0;If&#xA0;</mtext><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>&#x2260;</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mtext>, then system has unique solution and represents a pair of intersecting lines.</mtext><mspace linebreak="newline"/><mtext>&#xA0;If&#xA0;</mtext><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>c</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub></mfrac><mtext>, then system has infinite solutions and represents overlapping lines.</mtext></math>

Inconsistent System:

The above system will be inconsistent, if <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mo>&#x2260;</mo><mfrac><msub><mi>c</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub></mfrac></math> and do not have any solution. It represents a pair of parallel lines.

Examples:

  1. Solve the following equations with cross multiplication method.

2x – 3y + 1 = 0, 3x + 4y – 5 = 0

Solution: Given 2x – 3y + 1 = 0, 3x + 4y – 5 = 0

By cross multiplication method,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mi>x</mi><mrow><mo>(</mo><mo>&#x2212;</mo><mn>3</mn><mo>)</mo><mo>(</mo><mo>&#x2212;</mo><mn>5</mn><mo>)</mo><mo>&#x2212;</mo><mo>(</mo><mn>4</mn><mo>&#xD7;</mo><mn>1</mn><mo>)</mo></mrow></mfrac><mo>=</mo><mfrac><mi>y</mi><mrow><mo>(</mo><mn>1</mn><mo>&#xD7;</mo><mn>3</mn><mo>)</mo><mo>&#x2212;</mo><mo>(</mo><mo>(</mo><mo>&#x2212;</mo><mn>5</mn><mo>)</mo><mo>&#xD7;</mo><mn>2</mn><mo>)</mo></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mrow><mo>(</mo><mn>2</mn><mo>&#xD7;</mo><mn>4</mn><mo>)</mo><mo>&#x2212;</mo><mo>(</mo><mn>3</mn><mo>&#xD7;</mo><mo>(</mo><mo>&#x2212;</mo><mn>3</mn><mo>)</mo><mo>)</mo></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mi>x</mi><mrow><mn>15</mn><mo>&#x2212;</mo><mn>4</mn></mrow></mfrac><mo>=</mo><mfrac><mi>y</mi><mrow><mn>3</mn><mo>+</mo><mn>10</mn></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mrow><mn>8</mn><mo>+</mo><mn>9</mn></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mi>x</mi><mn>11</mn></mfrac><mo>=</mo><mfrac><mi>y</mi><mn>13</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>17</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>11</mn><mn>17</mn></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>y</mi><mo>=</mo><mfrac><mn>13</mn><mn>17</mn></mfrac></mtd></mtr></mtable></math>

  2. For what value of K, the system of equations 2x + 4y +16 = 0 and 3x + Ky + 24 = 0 has an infinite number of solutions.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: For infinite number of solutions, we have&#xA0;</mtext><mfrac><msub><mi>a</mi><mn>1</mn></msub><msub><mi>a</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>b</mi><mn>1</mn></msub><msub><mi>b</mi><mn>2</mn></msub></mfrac><mo>=</mo><mfrac><msub><mi>c</mi><mn>1</mn></msub><msub><mi>c</mi><mn>2</mn></msub></mfrac><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mfrac><mn>2</mn><mn>3</mn></mfrac><mo>=</mo><mfrac><mn>4</mn><mi>K</mi></mfrac><mo>=</mo><mfrac><mn>16</mn><mn>24</mn></mfrac></math>

Therefore, K = 6

  3. The cost of 21 pencils and 9 clippers is $ 819. What is the total cost of 7 pencils and 3 clippers together?

Solution:  Let the cost of 1 pencil and 1 clipper be p and c respectively.

Now, according to the question,

21p + 9c = $ 819

3(7p + 3c) = $ 819

7p + 3c = $ 273

Cost of 7 pencils and 3 clippers = $ 273

  4. David has some hens and some dogs. If the total number of animal heads is 100 and the total number of animal feet is 248, what is the total number of dogs David has?

Solution: Let hens = H, dogs = D

 According to the question, H + D = 100 ...(i)

2H + 4D = 248 ...(ii)

On multiplying Eq. (i) by 2 and subtracting from Eq. (ii), we get

2H + 2D = 200

  5. In an examination, a student scores 4 marks for every correct answer and losses 1 mark for every wrong answer. A student attempted all the 200 questions and scored 200 marks. Find the number of questions, he answered correctly.

Solution: Let the number of correct answers be x and number of wrong answers be y.

Then, 4x – y = 200 -> (1) and x + y = 200 -> (2)

On adding Equations (1) and (2), we get

Therefore, number of correct answers = 80


4 Square & Square root
N/A

Square

If a number is multiplied with itself, then the result of this multiplication is called the square of that number.

For example

(i) Square of 7 = 7 x 7 = 49

(ii) Square of 11 = 11 x 11 = 121

(iii) Square of 100 = 100 x 100 = 10000

Methods to Find Square:

Different methods to calculate the square of a number are as follows

Multiplication Method

In this method, the square of any 2-digit number can be calculated by the following given steps.

  • Step I: Square the unit's digit {If the square has two digits, then write ten's digit as carry.}
  • Step II: 2 x Ten's digit x Unit's digit + Carry
  • Step III: (Ten's digit)2 + Carry from step II
  • Step IV: Now arrange the numbers first write step III number, then unit’s place number of step II and at unit's place step I’s unit’s place number.

For example:  Find the square of 74:

Step I: (4)2 = 16 {Carry = 1}

Step II: 2 x 7 x 4 + 1 = 57 {Carry = 5}

Step III: (7)2 + 5 = 49 + 5 = 54

Step IV: (74)2 = 5476

Algebraic Method:

To calculate square by this method, two formulae are used.

(i) (a + b)2 = a2 + b2 + 2ab (ii) (a-b)2 = a2 + b2 -2ab

For example: The square of 34 is (34)2 = (30 + 4)2 = (30)2 + (4)2 + 2 x 30 x 4 = 900+ 16+240; (34)2 = 1156

Square of Decimal Numbers:

To find the square of any decimal number, write the square of the number ignoring the decimal and then place the decimal twice the place of the original number starting from unit's place.

For example: The square of 3.5 is as follows (35)2 = 1225

  • Here, the decimal is after one-digit in 3.5.
  • Hence, the decimal will be placed twice the place of original number in the result. (3.5)2 = 12.25

Square Root:

The square root of a number is that number, the square of which is equal to the given number.

There are two types of square roots of a number, positive and negative.

It is denoted by the sign '√'.

For example: 49 has two square roots 7 and - 7, because (7)2 = 49 and (- 7)2 = 49. Hence, we can write <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mn>49</mn></msqrt><mo>=</mo><mo>&#xB1;</mo><mn>7</mn></math>

Methods to Find Square Root:

Different methods to calculate the square root of a number are as follows

Prime Factorisation Method

This method has the following steps

Step I: Express the given number as the product of prime factors.

Step II: Arrange the factors in pairs of same prime numbers.

Step III: Take the product of these prime factors taking one out of every pair of the same primes. This product gives us the square root of the given number.

Division Method

If it is not easy to evaluate square root using prime factorisation method, then we use division method.

The steps of this method can be easily understood with the help of following examples.

Example: Find the square root of 18769.

Solution.

Step I: In the given number, mark off the digits in pairs starting from the unit digit. Each pair and the remaining one-digit (if any) are called a period.

Step II: Now, 1=1; On subtracting, we get 0 (zero) as remainder.

Step III: Bring down the next period, i.e., 87. Now, the trial divisor is 1 x 2 = 2 and trial dividend is 87. So, we take 23 as divisor and put 3 as quotient. The remainder is 18 now.

Step IV: Bring down the next period, which is 69. Now, trial divisor is 13 x 2 = 26 and trial dividend is 1869. So, we take 267 as dividend and 7 as quotient. The remainder is 0.

Step V The process (processes like III and IV) goes on till all the periods (pairs) come to an end and we get remainder as 0 (zero) now.

Hence, the required square root = 137

Properties of Squares and Square Roots

  • The difference of squares of two consecutive numbers will always be equal to the sum of the number i.e., (a2 -b2) = (a + b) (a-b). Here, a>b and (a, b) being consecutive (a — b) =1.

Example: If a = 12 and b =11, then (122 -112) = (12 + 11) (12-11) = 23

  • If the square of any number ends with 1, then its square root will end with 1 or 9.
  • If the square of any number ends with 4, then its square root will end with 2 or 8.
  • If the square of any number ends with 5, then its square root will end with 5.
  • If the square of any number ends with 6, then its square root will end with 7 or 6.
  • If the square of any number ends with 9, then its square root will end with 3 or 7.
  •  The square of any number always ends with 0,1,4, 5,6 or 9 but will never end with 2,3, 7 or 8,
  • Square root of negative number is imaginary

Important Relations:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><mo>?</mo></msqrt><mo>=</mo><mi>y</mi><mtext>. The required number&#xA0;</mtext><mo>=</mo><msup><mi>y</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>&#xD7;</mo><msup><mi>b</mi><mn>2</mn></msup></msqrt><mo>=</mo><mi>a</mi><mi>b</mi></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><msup><mi>a</mi><mn>3</mn></msup><mo>&#xD7;</mo><msup><mi>b</mi><mn>3</mn></msup></msqrt><mo>=</mo><mi>a</mi><mi>b</mi><msqrt><mi>a</mi><mi>b</mi></msqrt></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><msup><mi>a</mi><mn>4</mn></msup><mo>&#xD7;</mo><msup><mi>b</mi><mn>4</mn></msup><mo>&#xD7;</mo><msup><mi>c</mi><mn>4</mn></msup></msqrt><mo>=</mo><msup><mi>a</mi><mn>2</mn></msup><msup><mi>b</mi><mn>2</mn></msup><msup><mi>c</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><msup><mi>a</mi><mi>n</mi></msup><mo>&#xD7;</mo><msup><mi>b</mi><mi>m</mi></msup></msqrt><mo>=</mo><msup><mi>a</mi><mrow><mi>n</mi><mo>/</mo><mn>2</mn></mrow></msup><msup><mi>b</mi><mrow><mi>m</mi><mo>/</mo><mn>2</mn></mrow></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><mi>x</mi></msqrt><mo>&#xD7;</mo><msqrt><mi>y</mi></msqrt><mo>=</mo><msqrt><mi>x</mi><mi>y</mi></msqrt></mtd></mtr><mtr><mtd><mfrac><mrow><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><mi>x</mi></msqrt></mrow><mrow><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msqrt><mi>y</mi></msqrt></mrow></mfrac><mo>=</mo><msqrt><mfrac><mi>x</mi><mi>y</mi></mfrac></msqrt></mtd></mtr></mtable></math>

Square Root of Decimal Numbers:

If in a given decimal number, the number of digits after decimal are not even, then we put a 0 (zero) at the extreme right, So, that these are even number of digits after the decimal point. Now, periods are marked as marked in previous explanation starting from right hand side before the decimal point and from the left hand after the decimal digit.

For example: 156.694

There are odd number of digits after decimal.

So, we put a zero after the digit, so that there are even digits after the decimal 156.6940

Now, periods are marked as

After the periods are marked, then previous method is used to find the square root

Square Root of a Fraction:

To find square root of a fraction, we have to find the square roots of numerators and denominators, separately.

<math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><mfrac><mn>461</mn><mn>8</mn></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mrow><mn>461</mn><mo>&#xD7;</mo><mn>2</mn></mrow><mrow><mn>8</mn><mo>&#xD7;</mo><mn>2</mn></mrow></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mn>922</mn><mn>16</mn></mfrac></msqrt><mo>=</mo><mfrac><msqrt><mn>922</mn></msqrt><msqrt><mn>16</mn></msqrt></mfrac><mo>=</mo><mfrac><mn>30.3644</mn><mn>4</mn></mfrac><mo>=</mo><mn>7.5911</mn><mtext>&#xA0;(approx.)</mtext></math>

Note:

Sometimes, numerator and denominator are not a complete square. In these types of cases, it is better to convert the given fraction into decimal fraction to find the square root.

Examples:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;1. What is the value of&#xA0;</mtext><msqrt><mn>110</mn><mfrac><mn>1</mn><mn>4</mn></mfrac><mtext>&#xA0;?&#xA0;</mtext></msqrt><mspace linebreak="newline"/><mtext>&#xA0;Solution:&#xA0;</mtext><msqrt><mn>110</mn><mfrac><mn>1</mn><mn>4</mn></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mn>441</mn><mn>4</mn></mfrac></msqrt><mo>=</mo><msqrt><mfrac><mrow><mn>3</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>7</mn></mrow><mrow><mn>2</mn><mo>&#xD7;</mo><mn>2</mn></mrow></mfrac></msqrt><mo>=</mo><mfrac><mrow><mn>3</mn><mo>&#xD7;</mo><mn>7</mn></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mn>21</mn><mn>2</mn></mfrac><mo>=</mo><mn>10.5</mn><mspace linebreak="newline"/><mtext>&#xA0;2. Find the value of&#xA0;</mtext><msqrt><mfrac><mrow><mo>(</mo><mn>0.05</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.41</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mrow><mo>(</mo><mn>0.005</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.041</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.0066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow></mfrac></msqrt><mo>&#xF7;</mo><mn>5</mn><mspace linebreak="newline"/><mtext>&#xA0;Solution:&#xA0;</mtext><msqrt><mfrac><mrow><mo>(</mo><mn>0.05</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.41</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mrow><mo>(</mo><mn>0.005</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.041</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.0066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow></mfrac></msqrt><mo>&#xF7;</mo><mn>5</mn><mo>=</mo><msqrt><mfrac><mrow><mo>(</mo><mn>0.05</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.41</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mrow><msup><mfenced separators="|"><mfrac><mn>0.05</mn><mn>10</mn></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><msup><mfenced separators="|"><mfrac><mn>0.41</mn><mn>10</mn></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><msup><mfenced separators="|"><mfrac><mn>0.066</mn><mn>10</mn></mfrac></mfenced><mn>2</mn></msup></mrow></mfrac></msqrt><mo>&#xF7;</mo><mn>5</mn><mspace linebreak="newline"/><mo>=</mo><msqrt><mfrac><mrow><mn>100</mn><mfenced open="[" close="]" separators="|"><mrow><mo>(</mo><mn>0.05</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.41</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow></mfenced></mrow><mrow><mo>(</mo><mn>0.05</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.41</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>0.066</mn><msup><mo>)</mo><mn>2</mn></msup></mrow></mfrac></msqrt><mo>&#xF7;</mo><mn>5</mn><mspace linebreak="newline"/><mo>=</mo><msqrt><mn>100</mn></msqrt><mo>&#xF7;</mo><mn>5</mn><mspace linebreak="newline"/><mo>=</mo><mn>10</mn><mo>&#xF7;</mo><mn>5</mn><mspace linebreak="newline"/><mo>=</mo><mn>2</mn></math>

  3. A shop-keeper has 1000 boxes. He wants to keep them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of boxes that he needs more for this purpose?

Solution: Let the number of rows and columns be m.

Then, total boxes should be m x m.

Now, 1000 is not a square of any number.

Let m = 30 Then, m x m = 30 x 30 = 900 Which is less than total boxes.

Now, let m = 32 Then, m x m = 32 x 32 = 1024 Which is greater than 1000.

So, the minimum number of boxes that he needs for this purpose = 1024 - 1000 = 24 boxes

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;4. If&#xA0;</mtext><mi>x</mi><mo>=</mo><mn>3</mn><mo>+</mo><msqrt><mn>3</mn></msqrt><mtext>&#xA0;and&#xA0;</mtext><mi>y</mi><mo>=</mo><mn>3</mn><mo>&#x2212;</mo><msqrt><mn>3</mn></msqrt><mtext>, then find the value of&#xA0;</mtext><mfenced separators="|"><mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></mfenced><mspace linebreak="newline"/><mtext>&#xA0;Solution:&#xA0;</mtext><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><msup><mi>y</mi><mn>2</mn></msup><mo>=</mo><mo>(</mo><mn>3</mn><mo>+</mo><msqrt><mn>3</mn></msqrt><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>3</mn><mo>+</mo><msqrt><mn>3</mn></msqrt><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mo>(</mo><mn>9</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>6</mn><msqrt><mn>3</mn></msqrt><mo>)</mo><mo>+</mo><mo>(</mo><mn>9</mn><mo>+</mo><mn>3</mn><mo>&#x2212;</mo><mn>6</mn><msqrt><mn>3</mn></msqrt><mo>)</mo><mo>=</mo><mn>24</mn></math>

  5. What is the least number to be added to 8200 to make it a perfect square?

Solution: Given number = 8200

Now, lets find the nearest square values of given number

-> (90)2 = 8100 and (91)2 = 8281

-> (90)2 < 8200 > (91)2

Therefore, least number to be added to 8200 to make it a perfect square = 8281 – 8200 = 81


5 Cube & Cube root
N/A

If a number is multiplied two times with itself, then the result of this multiplication is called the cube of that number.

For example (i) Cube of 6 = 6 x 6 x 6=216 (ii) Cube of 8 = 8 x 8 x 8=512

Methods to Find Cube

Different methods to calculate the cube of a number are as follows

Algebraic Method

To calculate cube by this method, two formulae are used.

  1. (a + b)3 = a3 + 3ab (a +b) + b3
  2.  (a -b)3 = a3 - 3ab (a-b)- b3

For example:

The cube of 16 is (16)3 = (10 + 6)3 = (10)3 + 3 x 10 x 6 (10 + 6) + (6)3 = 1000 + 2880 + 216 = 4096

Shortcut Method:

Step I: The answer consists of 4 parts each of which has to be calculated separately,

Step II: First write down the cube of ten's digit to the extreme left. Write the next two terms to the right of it by creating; GP (Geometric Progression) having common ratio which is equal to <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msup><mtext>&#xA0;unit&#xA0;</mtext><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s digit&#xA0;</mtext></mrow><mrow><msup><mtext>&#xA0;ten&#xA0;</mtext><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s digit&#xA0;</mtext></mrow></mfrac></math> and the fourth number will be cube of unit’s digit.

Step III: Write the double of 2nd and 3rd number below them.

Step IV:  Now, add the number with numbers written below it and write the unit's place digit in a straight line and remaining number is carried forward to the next number.

Example: Find the cube of 35.

Sol. Here, unit's digit is 5 and ten's digit is 3.

Step I: Write the cube of ten's digit at extreme left i.e., (3)3 =27

Step II: Now, the next two terms on the right will be in a GP of common ratio equals to <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msup><mtext>&#xA0;unit&#xA0;</mtext><mi mathvariant="normal">&#x2032;</mi></msup><mtext>&#xA0;s digit&#xA0;</mtext></mrow><mrow><mi>t</mi><mi>e</mi><msup><mi>n</mi><mi mathvariant="normal">&#x2032;</mi></msup><mi>s</mi><mtext>&#xA0;digit&#xA0;</mtext></mrow></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>So, the two terms will be&#xA0;</mtext><mn>27</mn><mo>&#xD7;</mo><mfrac><mn>5</mn><mn>3</mn></mfrac><mo>=</mo><mn>45</mn><mo>;</mo><mn>45</mn><mo>&#xD7;</mo><mfrac><mn>5</mn><mn>3</mn></mfrac><mo>=</mo><mn>75</mn></math> and last term will be cube of unit’s digit i.e., (5)3 = 125, So, they are arranged as 27 45 75 125

Step III: Twice the second and third terms are written under it and are added.

(35)3 = 42875

Cube Root:

The cube root of a given number is the number whose cube is the given number. The cube root is denoted by the sign <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x201D;</mo><mroot><mrow/><mn>3</mn></mroot><mo>&#x201D;</mo><mo>.</mo></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>For example:&#xA0;</mtext><mroot><mn>8</mn><mn>3</mn></mroot><mo>=</mo><mroot><mrow><mn>2</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>2</mn></mrow><mn>3</mn></mroot><mo>=</mo><mn>2</mn></math>

Methods to Find Cube Root

Method to calculate the Cube root of a number is as follow

Prime Factorisation Method

This method has following steps.

Step I: Express the given number as the product of prime factors.

Step II: Arrange the factors in a group of three of same prime numbers.

Step III: Take the product of these prime factors picking one out of every group (group of three) of the same primes.

This product gives us the cube root of given number

Example: Find the cube root of 9261.

Sol: Prime factors of 9261 = (3 x 3 x 3) x (7 x 7 x 7)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mroot><mn>9261</mn><mn>3</mn></mroot><mo>=</mo><mroot><mrow><mn>3</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>7</mn></mrow><mn>3</mn></mroot></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Now, taking one number from each group of three, we get&#xA0;</mtext><mroot><mn>9261</mn><mn>3</mn></mroot><mo>=</mo><mn>3</mn><mo>&#xD7;</mo><mn>7</mn><mo>=</mo><mn>21</mn></math>

Properties of Cubes and Cube Roots

  • If the cube of a number is of 2 or 3-digits, then its cube root will be of 1-digit
  •  If the cube of a number is of 4, 5 or 6 digits, then its cube root will be of 2 digits.
  •  If the cube of a number has 0,1, 2,3,4, 5, 6, 7,8,9 in its unit's place, then its cube root will have 0, 1, 8, 7, 4, 5, 6, 3, 2 or 9 in their unit's place, respectively.

Example Problems:

  1. Find the cube root of -5832

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Solution:&#xA0;</mtext><mroot><mrow><mo>(</mo><mo>&#x2212;</mo><mn>5832</mn><mo>)</mo></mrow><mn>3</mn></mroot><mo>=</mo><mo>&#x2212;</mo><mroot><mn>5832</mn><mn>3</mn></mroot><mo>=</mo><mo>&#x2212;</mo><mroot><mrow><mn>18</mn><mo>&#xD7;</mo><mn>18</mn><mo>&#xD7;</mo><mn>18</mn></mrow><mn>3</mn></mroot><mo>=</mo><mo>&#x2212;</mo><mn>18</mn></math>

  2. What least number should be subtracted from 6862, so that 19 be the cube root of the result from this subtraction?

Solution: Given,

  • number = 6862
  • Let’s find out the nearest cube values of 6862
  • (18)3 = 5832
  • (19)3 = 6859
  • (18)3 < 6862 > (19)3
  • Therefore, required number = 6862 – 6859 = 3

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>3. Find the value of&#xA0;</mtext><mroot><mrow><mn>4</mn><mroot><mrow><mn>2</mn><mroot><mrow><mn>4</mn><msqrt><mn>2</mn><mroot><mrow><mn>4</mn><mo>&#x2026;</mo><mo>&#x2026;</mo></mrow><mn>3</mn></mroot></msqrt></mrow><mn>3</mn></mroot></mrow><mn>3</mn></mroot></mrow><mn>3</mn></mroot></math>

Solution:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;Let&#xA0;</mtext><mi>x</mi><mo>=</mo><msqrt><mn>2</mn><mroot><mrow><mn>4</mn><msqrt><mn>2</mn><mroot><mrow><mn>4</mn><mo>&#x2026;</mo><mo>.</mo></mrow><mn>3</mn></mroot></msqrt></mrow><mn>3</mn></mroot></msqrt></math>

  • On squaring both sides, we get

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2219;</mo><mo>&#xA0;</mo><msup><mi>x</mi><mn>2</mn></msup><mo>=</mo><mn>2</mn><mo>&#x22C5;</mo><mroot><mrow><mn>4</mn><mi>x</mi></mrow><mn>3</mn></mroot></math>

  • and now cubing both sides, we get
  • (x2)3 = 8*4x
  • x5 = 32
  • x5 = 25
  • x = 2

  4. Find the cube value of 102

Solution: Given number = 102

It can be written as (100 + 2)3 = 1003 + 23 + 3 x 100 x 2 (100 + 2)

                                                = 1000000 + 8 + 600(102)

                                                = 1000000 + 8 + 61,200

                                                = 1061208

  5. Find the cube root of 1259712

Solution: Given number = 1259712

Let’s write factors of the give number = 3 x 3 x 3 x 4 x 4 x 4 x 9 x 9 x 9

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mroot><mrow><mn>3</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>9</mn><mo>&#xD7;</mo><mn>9</mn><mo>&#xD7;</mo><mn>9</mn></mrow><mn>3</mn></mroot><mo>=</mo><mn>3</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>9</mn><mo>=</mo><mn>108</mn></math>

Therefore, cube root of 1259712 = 108


6 Indices and Surds
N/A

Indices:

When a number 'P' is multiplied by itself 'n' times, then the product is called nth power of 'P' and is written as Pn. Here, P is called the base and 'n' is known as the index of the power.

 Therefore, Pn is the exponential expression.

 Pn is read as 'P raised to the power n’ or ‘P to the power n’.

Rules:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msup><mi>P</mi><mi>m</mi></msup><mo>&#xD7;</mo><msup><mi>P</mi><mi>n</mi></msup><mo>=</mo><msup><mi>P</mi><mrow><mi>m</mi><mo>+</mo><mi>n</mi></mrow></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mfrac><msup><mi>P</mi><mi>m</mi></msup><msup><mi>P</mi><mi>n</mi></msup></mfrac><mo>=</mo><msup><mi>P</mi><mrow><mi>m</mi><mo>&#x2212;</mo><mi>n</mi></mrow></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msup><mfenced separators="|"><msup><mi>P</mi><mi>m</mi></msup></mfenced><mi>n</mi></msup><mo>=</mo><msup><mi>P</mi><mrow><mi>m</mi><mi>n</mi></mrow></msup></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mrow><mo>(</mo><mi>P</mi><mi>Q</mi><msup><mo>)</mo><mi>n</mi></msup><mo>=</mo><msup><mi>P</mi><mi>n</mi></msup><mo>&#xD7;</mo><msup><mi>Q</mi><mi>n</mi></msup></mrow></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msup><mfenced separators="|"><mfrac><mi>P</mi><mi>Q</mi></mfrac></mfenced><mi>n</mi></msup><mo>=</mo><mfrac><msup><mi>P</mi><mi>n</mi></msup><msup><mi>Q</mi><mi>n</mi></msup></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msup><mi>P</mi><mn>0</mn></msup><mo>=</mo><mn>1</mn></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><msup><mi>P</mi><mrow><mo>&#x2212;</mo><mi>n</mi></mrow></msup><mo>=</mo><mfrac><mn>1</mn><msup><mi>P</mi><mrow><mo>&#x2212;</mo><mi>n</mi></mrow></msup></mfrac></mtd></mtr></mtable></math>

Surds

Numbers which can be expressed in the form √p + √q, where p and q are natural numbers and not perfect squares.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Irrational numbers which contain the radical sign&#xA0;</mtext><mo>(</mo><mroot><mi>x</mi><mi>n</mi></mroot><mo>)</mo><mtext>&#xA0;are called as surds</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Hence, the numbers in the form of&#xA0;</mtext><msqrt><mn>3</mn></msqrt><mo>,</mo><mroot><mrow><mo>,</mo><mo>&#x2026;</mo><mo>&#x2026;</mo><mo>.</mo><mo>.</mo><mroot><mi>x</mi><mi>n</mi></mroot></mrow><mn>3</mn></mroot></math>

Note

 1. All surds are irrational numbers

2. All irrational numbers are not surds

 Order of Surds

  • The order of a surd indicates the index of root to be extracted.
  • <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>In&#xA0;</mtext><mroot><mi>p</mi><mi>n</mi></mroot><mo>,</mo><mi>n</mi><mtext>&#xA0;is called the order of the surd and&#xA0;</mtext><mi>p</mi><mtext>&#xA0;is called the radicand.&#xA0;</mtext></math>
  • <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;For example: The order of the surd&#xA0;</mtext><mroot><mi>x</mi><mn>7</mn></mroot><mtext>&#xA0;is&#xA0;</mtext><mn>7</mn><mo>.</mo></math>
  • A surd with index of root 2 is called a second order surd or quadratic surd.
  • For example: √2, √3, √5, √7, √x are the surds of order 2. Example: √2, √5, √10, √a, √m, √x, √ (x + 1) are second order surd or quadratic surd (since the indices of roots are 2).
  • A surd with index of root 3 is called a third order surd or cubic surd.
  • For example: ?2, ?3, ?10, ?17, ?x are the surds of order 3 or cubic surds. Example: ?2, ?5, ?7, ?15, ?100, ?a, ?m, ?x, ? (x - 1) are third order surd or cubic surd (since the indices of roots are 3).

Types of Surds

Pure Surds

Those surds which do not have factor other than 1, are known as pure surds

Mixed Surds

Those surds which have factor other than 1. are known as mixed surds

Like and Unlike Surds

  • When the radicands of two surds are same, then those are known as like surds.

Example: 2√2, 3√2, 4√2

  • When radicands are different, then they are called unlike surds

Example: 2√2, 2√3, 2√5

Properties of Surds

  • A quadratic surd cannot be equal to the sum and difference of a rational number and a quadratic surd.

Example:  a + b ≠ √c or √a – b ≠ √c

To Arrange the Surds in Increasing or Decreasing Order

  • <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Suppose given surds are&#xA0;</mtext><mi>p</mi><mo>=</mo><mroot><mi>x</mi><mi>a</mi></mroot><mo>,</mo><mi>q</mi><mo>=</mo><mroot><mi>y</mi><mi>b</mi></mroot><mo>,</mo><mi>r</mi><mo>=</mo><mroot><mi>z</mi><mi>c</mi></mroot></math>
  • First of all, take the LCM of a, b and c and use it to make the denominator of the powers the same.
  •  Then, easily we can find the required order.

Operations on Surds

Addition and Subtraction of Surds

Only like surds can be added or subtracted. Therefore, to add or subtract two or more surds, first simplify them and add or subtract them respectively like surds

Note:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><msqrt><mi>a</mi></msqrt><mo>+</mo><msqrt><mi>b</mi></msqrt><mo>&#x2260;</mo><msqrt><mi>a</mi><mo>+</mo><mi>b</mi></msqrt></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><msqrt><mi>a</mi></msqrt><mo>&#x2212;</mo><msqrt><mi>b</mi></msqrt><mo>&#x2260;</mo><msqrt><mi>a</mi><mo>&#x2212;</mo><mi>b</mi></msqrt></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mi>x</mi><msqrt><mi>a</mi></msqrt><mo>+</mo><mi>x</mi><msqrt><mi>b</mi></msqrt><mo>&#x2260;</mo><mi>x</mi><msqrt><mi>a</mi><mo>+</mo><mi>b</mi></msqrt></mtd></mtr></mtable></math>

Multiplication and Division of Surds

To multiply or divide the surds, we make the denominators of the powers equal to each other. Then, multiply or divide as usual.

Comparison of Surds

To compare two or more surds, first of all the denominators of the power of given surds are made equal to each other and then the radicand of the new surds is compared.

Rationalisation of Surds

The method of obtaining a rational number from a surd by multiplying it with another surd is known as rationalisation of surds. Both the surds are known as rationalising factor of each other

Examples:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>1. Value of&#xA0;</mtext><msqrt><mn>5</mn><mo>+</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt><mo>&#x2212;</mo><mfrac><mn>1</mn><msqrt><mn>5</mn><mo>&#x2212;</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt></mfrac><mtext>&#xA0;is</mtext><mspace linebreak="newline"/><mtext>Solution: Given&#xA0;</mtext><msqrt><mn>5</mn><mo>+</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt><mo>&#x2212;</mo><mfrac><mn>1</mn><msqrt><mn>5</mn><mo>&#x2212;</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt></mfrac><mo>=</mo><mfrac><mrow><msqrt><mo>(</mo><mn>5</mn><msup><mo>)</mo><mn>2</mn></msup><mo>&#x2212;</mo><mo>(</mo><mn>2</mn><msqrt><mn>6</mn></msqrt><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>&#x2212;</mo><mn>1</mn></mrow><msqrt><mn>5</mn><mo>&#x2212;</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt></mfrac><mo>=</mo><mfrac><mrow><msqrt><mn>25</mn><mo>&#x2212;</mo><mn>24</mn></msqrt><mo>&#x2212;</mo><mn>1</mn></mrow><msqrt><mn>5</mn><mo>&#x2212;</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn><mo>&#x2212;</mo><mn>1</mn></mrow><msqrt><mn>5</mn><mo>&#x2212;</mo><mn>2</mn><msqrt><mn>6</mn></msqrt></msqrt></mfrac><mo>=</mo><mn>0</mn></math>

  1. If ax = b, by = c and xyz = 1, then what is the value of cz?

Solution: Given, ax = b, by = c and xyz = 1

Let us take b = ax

  • by = axy
  • byz = axyz
  • cz = a

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>3. If&#xA0;</mtext><mi>p</mi><mtext>&#xA0;and&#xA0;</mtext><mi>q</mi><mtext>&#xA0;are natural, then&#xA0;</mtext><mroot><mi>q</mi><mi>p</mi></mroot><mtext>&#xA0;is</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: If&#xA0;</mtext><mi>p</mi><mtext>&#xA0;and&#xA0;</mtext><mi>q</mi><mtext>&#xA0;are natural numbers, then&#xA0;</mtext><mroot><mi>q</mi><mi>p</mi></mroot><mtext>&#xA0;is irrational unless&#xA0;</mtext><mi>q</mi><mtext>&#xA0;is&#xA0;</mtext><msup><mi>p</mi><mrow><mi>t</mi><mi>h</mi></mrow></msup><mtext>&#xA0;power of an integer</mtext></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>4. Arrange&#xA0;</mtext><mroot><mn>3</mn><mn>4</mn></mroot><mo>,</mo><mroot><mn>10</mn><mn>6</mn></mroot><mo>,</mo><mroot><mn>25</mn><mn>12</mn></mroot><mtext>&#xA0;in descending order.</mtext></math>

Solution: LCM of 4, 6, 12 = 12

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mroot><mn>3</mn><mn>4</mn></mroot><mo>=</mo><mroot><msup><mn>3</mn><mn>3</mn></msup><mn>12</mn></mroot><mo>=</mo><mroot><mn>27</mn><mn>12</mn></mroot></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mroot><mn>10</mn><mn>6</mn></mroot><mo>=</mo><mroot><msup><mn>10</mn><mn>2</mn></msup><mn>12</mn></mroot><mo>=</mo><mroot><mn>100</mn><mn>12</mn></mroot></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mroot><mn>25</mn><mn>12</mn></mroot><mo>=</mo><mroot><mn>25</mn><mn>12</mn></mroot></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>From this we can say that,&#xA0;</mtext><mroot><mn>100</mn><mn>12</mn></mroot><mo>&gt;</mo><mroot><mn>27</mn><mn>12</mn></mroot><mo>&gt;</mo><mroot><mn>25</mn><mn>12</mn></mroot><mo stretchy="false">&#x21D2;</mo><mroot><mn>10</mn><mn>6</mn></mroot><mo>&gt;</mo><mroot><mn>3</mn><mn>4</mn></mroot><mo>&gt;</mo><mroot><mn>25</mn><mn>12</mn></mroot></math>

  5. If 3x – 3x-1 = 18, then xx is equal to?

Solution: Given 3x – 3x-1 = 18

  • 3x-1(3 – 1) = 18
  • 3x-1 = 9
  • 3x-1 = 32 [Since bases are equal exponents are also equal]
  • x – 1 = 2
  • x = 3

Therefore, xx = 33 = 27


7 Quadratic Equations
N/A

A quadratic equation is an equation in which the highest power of an unknown quantity is a square that can be written as ax2 + bx + c = 0 where, a and b are coefficients of x2 and x, respectively and c is a constant.

The factor that identifies this expression as quadratic is the exponent 2. The coefficient of x2 i.e., a cannot be zero, (a≠0)

To check whether an equation is quadratic or not, following examples will help to understand it in a better way.

  1. 5x3 – 4x + 5 = 0 is not a quadratic equation because here the first term is raised to the 3rd power. It must be raised to the 2nd power in order to be quadratic.
  2. 7x2 – 6x + 2 = 0 is a quadratic equation, which is in the correct form i.e., ax2 + bx + c = 0
  3. 9x2 = 81 is a quadratic equation, here it can be written as 9x2 – 81 = 0 where a = 9, b = 0, c = - 81

Note:

  1. A quadratic equation has two and only two roots.
  2.  If a be the root of the quadratic equation ax2 + bx + c = 0, then (x - a) is a factor of ax2 + bx + c=0

Important Points Related to Quadratic Equations:

  1.  A quadratic equation ax2 + bx + c = 0 has
    • two distinct real roots, if D > 0.
    •   two equal real roots, if D = 0.
    •  no real roots, if D < 0.
    •  reciprocal roots, if a=c.
    • one root = 0, if c = 0.
    •  negative and reciprocal roots, if c = - a.
    •  both roots equal to 0, if fa = 0, c = 0 where, discriminant (D) = b2 - 4ac
  1. Formation of Quadratic Equation Let p and q be two roots. Then, we can form a quadratic equation as
      •  x2 – (sum of roots) x + (product of roots) = 0
      • x2 – (p + q) x + (pq) = 0

=> (x – p) (x – q) = 0

  • Here, for standard quadratic equation ax2 + bx + c = 0, <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>sum of roots&#xA0;</mtext><mo>=</mo><mi>p</mi><mo>+</mo><mi>q</mi><mo>=</mo><mo>&#x2212;</mo><mfrac><mi>b</mi><mi>a</mi></mfrac><mo>;</mo><mtext>&#xA0;product of roots&#xA0;</mtext><mo>=</mo><mi>p</mi><mi>q</mi><mo>=</mo><mfrac><mi>c</mi><mi>a</mi></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;If&#xA0;</mtext><mi>a</mi><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mi>b</mi><mi>x</mi><mo>+</mo><mi>c</mi><mo>=</mo><mn>0</mn><mtext>, where&#xA0;</mtext><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>c</mi><mtext>&#xA0;is rational, has one root&#xA0;</mtext><mi>u</mi><mo>+</mo><msqrt><mi>v</mi></msqrt><mtext>, then the other root will be&#xA0;</mtext><munder accentunder="false"><mi>u</mi><mo>_</mo></munder><mo>&#x2212;</mo><msqrt><mi>v</mi></msqrt></math>

Examples:

  1. In solving a problem, one student makes a mistake in the coefficient of the first-degree term and obtains - 9 and - 1 for the roots. Another student makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct equation was

Solution:

  • When mistake is done in first degree term, the roots of the equation are - 9 and - 1.
  • So, equation is (x + 1) (x + 9) = x2 + 10x + 9 ...(i)
  • When mistake is done in constant term, the roots of equation are 8 and 2.
  •  Therefore, Equation is (x - 2) (x - 8) = x2 - 10x + 16 ...(ii).
  •  Required equation from Eqs. (i) and (ii) is = x2 - 10x + 9
  • Also, we see in both the cases 1 st degree term is same with opposite sign i. e., in such questions we should take data from given conditions and find the correct equation
  1. Two numbers whose sum is 8 and difference is 4, are the roots of the equation, find the equation?

Solution: Let the roots be p and q.

Given p + q = 8 – (1)

p – q = 4 – (2)

By solving 1 and 2, we get p = 6 and q = 2

Therefore, required equation is x2 – (p + q) x + pq = 0

                                                    x2 – (8) x +12 = 0

  3. If one of the roots of quadratic equation 7x2 - 50x + k = 0 is 7, then what is the value of k?

Solution: Given equation is 7x2 – 50x + k = 0.

Here, a = 7, b = - 50, c = k

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Since,&#xA0;</mtext><mi>p</mi><mo>+</mo><mi>q</mi><mo>=</mo><mo>&#x2212;</mo><mfrac><mi>b</mi><mi>a</mi></mfrac><mo>=</mo><mfrac><mn>50</mn><mn>7</mn></mfrac><mspace linebreak="newline"/><mi>q</mi><mo>=</mo><mfrac><mn>50</mn><mn>7</mn></mfrac><mo>&#x2212;</mo><mn>7</mn><mo>=</mo><mfrac><mn>1</mn><mn>7</mn></mfrac><mspace linebreak="newline"/><mi>p</mi><mi>q</mi><mo>=</mo><mfrac><mi>c</mi><mi>a</mi></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mn>7</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>7</mn></mfrac><mo>=</mo><mfrac><mi>k</mi><mn>7</mn></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>k</mi><mo>=</mo><mn>7</mn></mtd></mtr></mtable></math>

  4. The sum of the roots of the equation 5x + (p + q + r) x + pqr = 0 is equal to zero. What is the value of (p3 + q3 + r3)?

Solution: Given equation, 5x + (p + q + r) x + pqr = 0

Here a = 5, b = p + q + r and c = pqr

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Sum of the roots&#xA0;</mtext><mo>=</mo><mo>&#x2212;</mo><mfrac><mi>b</mi><mi>a</mi></mfrac><mo>=</mo><mo>&#x2212;</mo><mfrac><mrow><mo>(</mo><mi>p</mi><mo>+</mo><mi>q</mi><mo>+</mo><mi>r</mi><mo>)</mo></mrow><mn>5</mn></mfrac><mo>=</mo><mn>0</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x21D2;</mo><mo>&#xA0;</mo><mo>(</mo><mi>p</mi><mo>&#xA0;</mo><mo>+</mo><mo>&#xA0;</mo><mi>q</mi><mo>&#xA0;</mo><mo>+</mo><mo>&#xA0;</mo><mi>r</mi><mo>)</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>0</mn></math>

According to the formula, a3 + b3 + c3 = 3abc, if a + b + c = 0

From this p3 + q3 + r3 = 3pqr

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;5. If&#xA0;</mtext><msup><mfenced separators="|"><mrow><mi>p</mi><mo>+</mo><mfrac><mn>1</mn><mi>p</mi></mfrac></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>3</mn><mtext>, what is the value of&#xA0;</mtext><msup><mi>p</mi><mn>3</mn></msup><mo>+</mo><mfrac><mn>1</mn><mrow><mi>p</mi><mn>3</mn></mrow></mfrac><mtext>&#xA0;?&#xA0;</mtext><mspace linebreak="newline"/><mtext>&#xA0;Solution: Given&#xA0;</mtext><msup><mfenced separators="|"><mrow><mi>p</mi><mo>+</mo><mfrac><mn>1</mn><mi>p</mi></mfrac></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>3</mn><mspace linebreak="newline"/><mtext>&#xA0;Taking square roots on both sides&#xA0;</mtext><mo>&#x21D2;&gt;</mo><mfenced separators="|"><mrow><mi>p</mi><mo>+</mo><mfrac><mn>1</mn><mi>p</mi></mfrac></mrow></mfenced><mo>=</mo><msqrt><mn>3</mn></msqrt><mo>&#x2212;</mo><mo>(</mo><mn>1</mn><mo>)</mo><mspace linebreak="newline"/><mi>C</mi><mi>u</mi><mi>b</mi><mi>i</mi><mi>n</mi><mi>g</mi><mo>&#xA0;</mo><mi>o</mi><mi>n</mi><mo>&#xA0;</mo><mi>b</mi><mi>o</mi><mi>t</mi><mi>h</mi><mo>&#xA0;</mo><mi>s</mi><mi>i</mi><mi>d</mi><mi>e</mi><mi>s</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>e</mi><mi>q</mi><mo>-</mo><mo>(</mo><mn>1</mn><mo>)</mo><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><msup><mfenced separators="|"><mrow><mi>p</mi><mo>+</mo><mfrac><mn>1</mn><mi>p</mi></mfrac></mrow></mfenced><mn>3</mn></msup><mo>=</mo><mo>(</mo><msqrt><mn>3</mn></msqrt><msup><mo>)</mo><mn>3</mn></msup></mtd></mtr><mtr><mtd><msup><mi>p</mi><mn>3</mn></msup><mo>+</mo><mfrac><mn>1</mn><mrow><mi>p</mi><mn>3</mn></mrow></mfrac><mo>+</mo><mn>3</mn><mo>&#xD7;</mo><mi>p</mi><mo>&#xD7;</mo><mfrac><mn>1</mn><mi>p</mi></mfrac><mfenced separators="|"><mrow><mi>p</mi><mo>+</mo><mfrac><mn>1</mn><mi>p</mi></mfrac></mrow></mfenced><mo>=</mo><mn>3</mn><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><msup><mi>p</mi><mn>3</mn></msup><mo>+</mo><mfrac><mn>1</mn><mrow><mi>p</mi><mn>3</mn></mrow></mfrac><mo>+</mo><mn>3</mn><mo>&#xD7;</mo><mo>(</mo><msqrt><mn>3</mn></msqrt><mo>)</mo><mo>=</mo><mn>3</mn><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><mtext>Therefore,&#xA0;</mtext><msup><mi>p</mi><mn>3</mn></msup><mo>+</mo><mfrac><mn>1</mn><mrow><mi>p</mi><mn>3</mn></mrow></mfrac><mo>=</mo><mn>0</mn></mtd></mtr></mtable></math>


1 Coordinate Geometry
N/A

 

It is a system of geometry, where the position of points on the plane is described by using an ordered pair of numbers.

 Rectangular Coordinate Axes

The lines XOX' and YOY' are mutually perpendicular to each other and they meet at point O which is called the origin.

 

Line XOX' represents X-axis and line YOY' represents Y-axis and together taken, they are called coordinate axes.

 

Any point in coordinate axis can be represented by specifying the position of x and y-coordinates

Quadrants

The X and Y-axes divide the cartesian plane into four regions referred to quadrants

  • The coordinates of point O (origin) are taken as (0,0).
  • The coordinates of any point on X-axis are of the form (x, 0).
  • The coordinates of any point on Y-axis are of the form (0, y)

Formulae:

Distance Formula

Distance between Two Points If A (x1, y1) and B (x2, y2) are two points, then 

Distance of a Point from the Origin

The distance of a point A (x, y) from the origin O (0, 0) is given by 

Area of triangle

If A (x1, y1) B (x2, y2) and C (x3, y3) are three vertices of a Triangle ABC, then its area is given by

 Area of triangle  (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))

Collinearity of Three Points

Three points A (x1, y1) B (x2, y2) and C (x3, y3) are collinear, if

(i)   Area of triangle ABC is 0

(ii)   Slope of AB = Slope of BC = Slope of AC

(iii)   Distance between A and B + Distance between B and C = Distance between A and C

Centroid of a Triangle

 

Centroid is the point of intersection of all the three medians of a triangle. If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of its centroid are

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mrow><mfrac><mn>1</mn><mn>3</mn></mfrac><mfenced separators="|"><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>+</mo><msub><mi>x</mi><mn>2</mn></msub><mo>+</mo><msub><mi>x</mi><mn>3</mn></msub></mrow></mfenced><mo>,</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mfenced separators="|"><mrow><msub><mi>y</mi><mn>1</mn></msub><mo>+</mo><msub><mi>y</mi><mn>2</mn></msub><mo>+</mo><msub><mi>y</mi><mn>3</mn></msub></mrow></mfenced></mrow></mfenced></math>

Circumcentre

The circumcentre of a triangle is the point of inter section of the perpendicular bisectors of its sides and is equidistance from all three vertices.

 

 If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of triangles and O (x, y) is the circumcentre of triangle ABC, then OA = OB= OC

Incentre

The centre of the circle, which touches the sides of a triangle, is called its incentre.

 

 Incentre is the point of intersection of internal angle bisectors of triangle.

If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC such that BC = a, CA = b and AB = c, then coordinates of its incentre I are 

Section formulae

  • Let A (x1, y1) B (x2, y2) be two points on the cartesian plane.
  • Let point P (x, y) divides the line AB in the ratio of m: n internally.

If P divides AB externally, then 

If P is the mid-point of AB, then  

Basic Points Related to Straight Lines

1. General form of equation of straight line is ax + by + c = 0. Where, a, b and c are real constants and x and y are two unknowns.

2. The equation of a line having slope m and intersects at c on x-axis is y = mx + c.

 

3. Slope (gradient) of a line ax + by + c = 0, by = - ax – c

Comparing with y = mx + c, where m is slope, therefore m = tan θ

Slope of the line is always measured in anti-clockwise direction.

 

4. Point slope form A line in terms of coordinates of any two points on it, if (x1, y1) and (x2, y2) are coordinates of any two points on a line, then its slope is 

5. Two-point form a line the equation of a line passing through the points A (x1, y1) and B (x2, y2) is 

 

6. Condition of parallel lines

If the slopes of two lines i.e., m1 and m2 are equal then lines are parallel.

Equation of line parallel to ax + by + c = 0 is ax + by + q =

 

7. Condition of perpendicular lines

If the multiplication of slopes of two lines i.e., m1 and m2 is equal to -1 then lines are perpendicular.

m1 x m2 = -1

Equation of line perpendicular to ax + by + c = 0 is bx - ay + q =0

 

 8. Angle between the two lines 

 

9. Intercept form Equation of line L intersects at a and b on x and y-axes, respectively is 

 

10. Condition of concurrency of three lines:

Let the equation of three lines are a1x + b1y + c1 = 0,

a2x + b2y + c2 = 0, and a3x + b3y + c3 = 0.

Then, three lines will be concurrent, if

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|" separators="|"><mtable columnspacing="1em" columnalign="left left left"><mtr><mtd><msub><mi>a</mi><mn>1</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>1</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>1</mn></msub></mtd></mtr><mtr><mtd><msub><mi>a</mi><mn>2</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>2</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>2</mn></msub></mtd></mtr><mtr><mtd><msub><mi>a</mi><mn>3</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>3</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>3</mn></msub></mtd></mtr></mtable></mfenced><mo>=</mo><mn>0</mn></math>

Distance of a point from the line:

Let ax + by + c = 0 be any equation of line and P (x1, y1) be any point in space. Then the perpendicular Distance(d) of a point P from a line is given by 

 

12. The length of the perpendicular from the origin to the line ax + by + c = 0, is 

 

 13. Area of triangle by straight line ax + by + c = 0 where a 0 and b 0 with coordinate axes is 

 

 14. Distance between parallel lines ax + by + c = 0 and ax + by + d = 0 is equal to 

 

15. Area of trapezium, between two parallel lines and axes,

Area of trapezium ABCD = Area of OCD

Examples:

Find the area of triangle ABC, whose vertices are A (8, - 4), B (3, 6) and C (- 2, 4).

Solution: Here, A (8, - 4) so, x1 = 8, y1 = - 4

                        B (3, 6) so, x2 = 3, y2 = 6

 

                           C (-2, 4) so, x3 = -2, y3 = 4

Therefore, area of triangle ABC (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))

(8(6 – 4) + 3(4 – (- 4)) + (-2) (-4-6))

 (16 + 24 + 20)

= 30 sq units

If A (-2,1), B (2, 3) and C (-2, -4) are three points, then find the angle between AB and BC.

 

Solution: Let m1 and m2 be the slopes of line AB and BC, respectively.

 

Let θ be the angle between AB and BC

 

 3. In what ratio, the line made by joining the points A (- 4, - 3) and B (5,2) intersects x-axis?

Solution: We know that y-coordinate is zero on x-axis,

 

Given, y1 = - 3, y2 = 2

Therefore, 

 

2m – 3n = 0

 

 4. Coordinates of a point is (0, 1) and ordinate of another point is - 3. If distance between both the points is 5, then abscissa of second point is

Solution: Let abscissa be x.

So, (x – 0)2 + (-3 -1)2 = 52

x2 + 16 = 25

x2 = 9

 5. Do the points (4, 3), (- 4, - 6) and (7, 9) form a triangle? If yes, then find the longest side of the triangle

Solution: Let P (4, 3), Q (-4, -6) and R (7, 9) are given points.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>Q</mi><mo>=</mo><msqrt><mo>(</mo><mo>&#x2212;</mo><mn>4</mn><mo>&#x2212;</mo><mn>4</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mo>&#x2212;</mo><mn>6</mn><mo>&#x2212;</mo><mn>3</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mo>(</mo><mo>&#x2212;</mo><mn>9</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mo>&#x2212;</mo><mn>9</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>64</mn><mo>+</mo><mn>81</mn></msqrt><mo>=</mo><mn>12.04</mn><mspace linebreak="newline"/><mi>Q</mi><mi>R</mi><mo>=</mo><msqrt><mo>(</mo><mn>7</mn><mo>&#x2212;</mo><mo>(</mo><mo>&#x2212;</mo><mn>4</mn><mo>)</mo><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>9</mn><mo>&#x2212;</mo><mo>(</mo><mo>&#x2212;</mo><mn>6</mn><mo>)</mo><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mo>(</mo><mn>11</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>15</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>121</mn><mo>+</mo><mn>225</mn></msqrt><mo>=</mo><mn>18.6</mn><mspace linebreak="newline"/><mi>P</mi><mi>R</mi><mo>=</mo><msqrt><mo>(</mo><mn>7</mn><mo>&#x2212;</mo><mn>4</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>9</mn><mo>&#x2212;</mo><mn>3</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>9</mn><mo>+</mo><mn>36</mn></msqrt><mo>=</mo><mn>6.7</mn></math>

Since, the sum of 12.04 and 6.7 is greater than 18.6.

 

So, it will form a triangle, whose longest side is 18.6

 


2 Coordinate Geometry
N/A

 

It is a system of geometry, where the position of points on the plane is described by using an ordered pair of numbers.

 Rectangular Coordinate Axes

The lines XOX' and YOY' are mutually perpendicular to each other and they meet at point O which is called the origin.

 

Line XOX' represents X-axis and line YOY' represents Y-axis and together taken, they are called coordinate axes.

 

Any point in coordinate axis can be represented by specifying the position of x and y-coordinates

Quadrants

The X and Y-axes divide the cartesian plane into four regions referred to quadrants

  • The coordinates of point O (origin) are taken as (0,0).
  • The coordinates of any point on X-axis are of the form (x, 0).
  • The coordinates of any point on Y-axis are of the form (0, y)

Formulae:

Distance Formula

Distance between Two Points If A (x1, y1) and B (x2, y2) are two points, then 

Distance of a Point from the Origin

The distance of a point A (x, y) from the origin O (0, 0) is given by 

Area of triangle

If A (x1, y1) B (x2, y2) and C (x3, y3) are three vertices of a Triangle ABC, then its area is given by

 Area of triangle  (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))

Collinearity of Three Points

Three points A (x1, y1) B (x2, y2) and C (x3, y3) are collinear, if

(i)   Area of triangle ABC is 0

(ii)   Slope of AB = Slope of BC = Slope of AC

(iii)   Distance between A and B + Distance between B and C = Distance between A and C

Centroid of a Triangle

 

Centroid is the point of intersection of all the three medians of a triangle. If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of its centroid are

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mrow><mfrac><mn>1</mn><mn>3</mn></mfrac><mfenced separators="|"><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>+</mo><msub><mi>x</mi><mn>2</mn></msub><mo>+</mo><msub><mi>x</mi><mn>3</mn></msub></mrow></mfenced><mo>,</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mfenced separators="|"><mrow><msub><mi>y</mi><mn>1</mn></msub><mo>+</mo><msub><mi>y</mi><mn>2</mn></msub><mo>+</mo><msub><mi>y</mi><mn>3</mn></msub></mrow></mfenced></mrow></mfenced></math>

Circumcentre

The circumcentre of a triangle is the point of inter section of the perpendicular bisectors of its sides and is equidistance from all three vertices.

 

 If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of triangles and O (x, y) is the circumcentre of triangle ABC, then OA = OB= OC

Incentre

The centre of the circle, which touches the sides of a triangle, is called its incentre.

 

 Incentre is the point of intersection of internal angle bisectors of triangle.

If A (x1, y1) B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC such that BC = a, CA = b and AB = c, then coordinates of its incentre I are 

Section formulae

  • Let A (x1, y1) B (x2, y2) be two points on the cartesian plane.
  • Let point P (x, y) divides the line AB in the ratio of m: n internally.

If P divides AB externally, then 

If P is the mid-point of AB, then  

Basic Points Related to Straight Lines

1. General form of equation of straight line is ax + by + c = 0. Where, a, b and c are real constants and x and y are two unknowns.

2. The equation of a line having slope m and intersects at c on x-axis is y = mx + c.

 

3. Slope (gradient) of a line ax + by + c = 0, by = - ax – c

Comparing with y = mx + c, where m is slope, therefore m = tan θ

Slope of the line is always measured in anti-clockwise direction.

 

4. Point slope form A line in terms of coordinates of any two points on it, if (x1, y1) and (x2, y2) are coordinates of any two points on a line, then its slope is 

5. Two-point form a line the equation of a line passing through the points A (x1, y1) and B (x2, y2) is 

 

6. Condition of parallel lines

If the slopes of two lines i.e., m1 and m2 are equal then lines are parallel.

Equation of line parallel to ax + by + c = 0 is ax + by + q =

 

7. Condition of perpendicular lines

If the multiplication of slopes of two lines i.e., m1 and m2 is equal to -1 then lines are perpendicular.

m1 x m2 = -1

Equation of line perpendicular to ax + by + c = 0 is bx - ay + q =0

 

 8. Angle between the two lines 

 

9. Intercept form Equation of line L intersects at a and b on x and y-axes, respectively is 

 

10. Condition of concurrency of three lines:

Let the equation of three lines are a1x + b1y + c1 = 0,

a2x + b2y + c2 = 0, and a3x + b3y + c3 = 0.

Then, three lines will be concurrent, if

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|" separators="|"><mtable columnspacing="1em" columnalign="left left left"><mtr><mtd><msub><mi>a</mi><mn>1</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>1</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>1</mn></msub></mtd></mtr><mtr><mtd><msub><mi>a</mi><mn>2</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>2</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>2</mn></msub></mtd></mtr><mtr><mtd><msub><mi>a</mi><mn>3</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>b</mi><mn>3</mn></msub><mo>&#x2005;&#x2005;&#x2005;&#x2005;</mo><msub><mi>c</mi><mn>3</mn></msub></mtd></mtr></mtable></mfenced><mo>=</mo><mn>0</mn></math>

Distance of a point from the line:

Let ax + by + c = 0 be any equation of line and P (x1, y1) be any point in space. Then the perpendicular Distance(d) of a point P from a line is given by 

 

12. The length of the perpendicular from the origin to the line ax + by + c = 0, is 

 

 13. Area of triangle by straight line ax + by + c = 0 where a 0 and b 0 with coordinate axes is 

 

 14. Distance between parallel lines ax + by + c = 0 and ax + by + d = 0 is equal to 

 

15. Area of trapezium, between two parallel lines and axes,

Area of trapezium ABCD = Area of OCD

Examples:

Find the area of triangle ABC, whose vertices are A (8, - 4), B (3, 6) and C (- 2, 4).

Solution: Here, A (8, - 4) so, x1 = 8, y1 = - 4

                        B (3, 6) so, x2 = 3, y2 = 6

 

                           C (-2, 4) so, x3 = -2, y3 = 4

Therefore, area of triangle ABC (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2))

(8(6 – 4) + 3(4 – (- 4)) + (-2) (-4-6))

 (16 + 24 + 20)

= 30 sq units

If A (-2,1), B (2, 3) and C (-2, -4) are three points, then find the angle between AB and BC.

 

Solution: Let m1 and m2 be the slopes of line AB and BC, respectively.

 

Let θ be the angle between AB and BC

 

 3. In what ratio, the line made by joining the points A (- 4, - 3) and B (5,2) intersects x-axis?

Solution: We know that y-coordinate is zero on x-axis,

 

Given, y1 = - 3, y2 = 2

Therefore, 

 

2m – 3n = 0

 

 4. Coordinates of a point is (0, 1) and ordinate of another point is - 3. If distance between both the points is 5, then abscissa of second point is

Solution: Let abscissa be x.

So, (x – 0)2 + (-3 -1)2 = 52

x2 + 16 = 25

x2 = 9

 5. Do the points (4, 3), (- 4, - 6) and (7, 9) form a triangle? If yes, then find the longest side of the triangle

Solution: Let P (4, 3), Q (-4, -6) and R (7, 9) are given points.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mi>Q</mi><mo>=</mo><msqrt><mo>(</mo><mo>&#x2212;</mo><mn>4</mn><mo>&#x2212;</mo><mn>4</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mo>&#x2212;</mo><mn>6</mn><mo>&#x2212;</mo><mn>3</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mo>(</mo><mo>&#x2212;</mo><mn>9</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mo>&#x2212;</mo><mn>9</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>64</mn><mo>+</mo><mn>81</mn></msqrt><mo>=</mo><mn>12.04</mn><mspace linebreak="newline"/><mi>Q</mi><mi>R</mi><mo>=</mo><msqrt><mo>(</mo><mn>7</mn><mo>&#x2212;</mo><mo>(</mo><mo>&#x2212;</mo><mn>4</mn><mo>)</mo><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>9</mn><mo>&#x2212;</mo><mo>(</mo><mo>&#x2212;</mo><mn>6</mn><mo>)</mo><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mo>(</mo><mn>11</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>15</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>121</mn><mo>+</mo><mn>225</mn></msqrt><mo>=</mo><mn>18.6</mn><mspace linebreak="newline"/><mi>P</mi><mi>R</mi><mo>=</mo><msqrt><mo>(</mo><mn>7</mn><mo>&#x2212;</mo><mn>4</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>9</mn><mo>&#x2212;</mo><mn>3</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>9</mn><mo>+</mo><mn>36</mn></msqrt><mo>=</mo><mn>6.7</mn></math>

Since, the sum of 12.04 and 6.7 is greater than 18.6.

 

So, it will form a triangle, whose longest side is 18.6

 


1 Volume and Surface Area of Cylinder
N/A

 

Cylinder

  • A cylinder is a solid or hollow body that is formed by keeping circles of equal radii one on another.

  • A part from this, a cylinder is formed by rolling a rectangular sheet also.

 

  • A cylinder has three surfaces
  • Curved surface
  • Bottom
  • Top

Solid Cylinder:

  • Volume of cylinder = Area of base x Height = <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>&#xD7;</mo><mi>h</mi></math>
  • Curved surface area = Perimeter of base x Height = <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi mathvariant="normal">&#x3C0;</mi></math>rh
  • Total surface area = Curved surface area + Area of both the circles (top and bottom surfaces) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mi mathvariant="normal">&#x3C0;</mi></math>r (h + r) where, r = Radius of base and h = Height

 

Hollow Cylinder:

 

If cylinder is hollow, then

  • Volume of hollow cylinder = Outer volume - Inner volume = π(R2 – r2) h cubic units.
  • If Curved surface area = Curved surface area of outer surface + Curved surface area of inner surface = 2πRh + 2πrh = 2πrh (R + r)
  • Total surface area of hollow cylinder = Curved surface area + Area of both top and bottom surface = 2πh (R + r) + 2π(R2 – r2)

 

 where, R = External radius of base, r = Internal radius of base and h = Height

 

Examples:

 

  1. How many iron rods each of length 14 m and diameter 4 cm can be made out of 0.88 m3 of iron?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Here,&#xA0;</mtext><mi>r</mi><mo>=</mo><mfrac><mn>2</mn><mn>100</mn></mfrac><mi>m</mi><mtext>&#xA0;and&#xA0;</mtext><mi>h</mi><mo>=</mo><mn>14</mn><mi>m</mi><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;Volume of&#xA0;</mtext><mn>1</mn><mtext>&#xA0;rod&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>&#xD7;</mo><mi>h</mi><mo>=</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>2</mn><mn>100</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>2</mn><mn>100</mn></mfrac><mo>&#xD7;</mo><mn>14</mn></mtd></mtr><mtr><mtd><mo>=</mo><mn>22</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>50</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>50</mn></mfrac><mo>&#xD7;</mo><mn>2</mn></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>44</mn><mn>2500</mn></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>11</mn><mn>625</mn></mfrac><msup><mi>m</mi><mn>3</mn></msup></mtd></mtr></mtable><mspace linebreak="newline"/><mtext>&#xA0;Here,&#xA0;</mtext><mi>r</mi><mo>=</mo><mfrac><mi>d</mi><mn>2</mn></mfrac><mtext>&#xA0;and is divided by&#xA0;</mtext><mn>100</mn><mtext>&#xA0;to convert in to metre</mtext></math>

Volume of iron rod = 0.88 m3

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, number of rods&#xA0;</mtext><mo>=</mo><mn>0.88</mn><mo>&#xD7;</mo><mfrac><mn>625</mn><mn>11</mn></mfrac><mo>=</mo><mn>50</mn></math>

 

  2. What will be the curved surface area of a right circular cylinder having length 160 cm and radius of the base is 7 cm?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Curved surface area of the cylinder&#xA0;</mtext><mo>=</mo><mn>2</mn><mi>&#x3C0;</mi><mi>r</mi><mi>h</mi><mo>=</mo><mn>2</mn><mo>&#xD7;</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>160</mn><mspace linebreak="newline"/><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>7040</mn><mo>&#xA0;</mo><mi>s</mi><mi>q</mi><mo>&#xA0;</mo><mi>c</mi><mi>m</mi></math>

 

  3. A rod of 1 cm diameter and 30 cm length is converted into a wire of 3 m length of uniform thickness. The diameter of the wire is

Solution: Given, r1 = 1 cm, h1 =30 cm, h2 = 300 cm

 

Volume of rod = volume of wire

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="1em"><mtr><mtd><mi>&#x3C0;</mi><msubsup><mi>r</mi><mn>1</mn><mn>2</mn></msubsup><msub><mi>h</mi><mn>1</mn></msub><mo>=</mo><mi>&#x3C0;</mi><msubsup><mi>r</mi><mn>2</mn><mn>2</mn></msubsup><msub><mi>h</mi><mn>2</mn></msub></mtd></mtr><mtr><mtd><mi>&#x3C0;</mi><mo>&#xD7;</mo><mo>(</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup><mo>&#xD7;</mo><mn>30</mn><mo>=</mo><mi>&#x3C0;</mi><mo>&#xD7;</mo><mo>(</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup><mo>&#xD7;</mo><mn>300</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msubsup><mi>r</mi><mn>2</mn><mn>2</mn></msubsup><mo>=</mo><mfrac><mn>30</mn><mn>300</mn></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msub><mi>r</mi><mn>2</mn></msub><mo>=</mo><mfrac><mn>1</mn><msqrt><mn>10</mn></msqrt></mfrac><mi>c</mi><mi>m</mi></mtd></mtr><mtr><mtd><mtext>diameter&#xA0;</mtext><mo>=</mo><mn>2</mn><msub><mi>r</mi><mn>2</mn></msub><mo>=</mo><mn>2</mn><mo>&#xD7;</mo><mfrac><mn>1</mn><msqrt><mn>10</mn></msqrt></mfrac><mo>=</mo><mfrac><mn>2</mn><msqrt><mn>10</mn></msqrt></mfrac><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

 

  4. A hollow cylinder made of wood has thickness 1 cm while its external radius is 3 cm. If the height of the cylinder is 8 cm, then find the volume, curved surface area and total surface area of the cylinder.

 

Solution: Radius r = Inner radius = External radius – Thickness = 3 – 1 = 2 cm

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, require volume&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>h</mi><mfenced separators="|"><mrow><msup><mi>R</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mi>r</mi><mn>2</mn></msup></mrow></mfenced><mo>=</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>8</mn><mfenced separators="|"><mrow><msup><mn>3</mn><mn>2</mn></msup><mo>&#x2212;</mo><msup><mn>2</mn><mn>2</mn></msup></mrow></mfenced><mo>=</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>8</mn><mo>&#xD7;</mo><mn>5</mn><mo>=</mo><mfrac><mn>880</mn><mn>7</mn></mfrac><mi>c</mi><msup><mi>m</mi><mn>3</mn></msup><mspace linebreak="newline"/><mtext>&#xA0;Curved surface area&#xA0;</mtext><mo>=</mo><mn>2</mn><mi>&#x3C0;</mi><mi>h</mi><mo>(</mo><mi>R</mi><mo>+</mo><mi>r</mi><mo>)</mo><mo>=</mo><mn>2</mn><mo>&#xD7;</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>8</mn><mo>&#xD7;</mo><mn>5</mn><mo>=</mo><mfrac><mn>1760</mn><mn>7</mn></mfrac><mi>s</mi><mi>q</mi><mi>c</mi><mi>m</mi></math>

 

  5. The ratio of the radii of two cylinders is 2: 3 and the ratio of their heights is 5: 3. The ratio of their volumes will be

 

Solution: Let the radii be 2r and 3r and heights be 5h and 3h.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, ratio of volumes&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><mo>&#xD7;</mo><mo>(</mo><mn>2</mn><mi>r</mi><msup><mo>)</mo><mn>2</mn></msup><mo>&#xD7;</mo><mn>5</mn><mi>h</mi></mrow><mrow><mi>&#x3C0;</mi><mo>&#xD7;</mo><mo>(</mo><mn>3</mn><mi>r</mi><msup><mo>)</mo><mn>2</mn></msup><mo>&#xD7;</mo><mn>3</mn><mi>h</mi></mrow></mfrac><mo>=</mo><mfrac><mn>20</mn><mn>27</mn></mfrac><mo>=</mo><mn>20</mn><mo>:</mo><mn>27</mn></math>

 

 

 


2 Volume and Surface Area of Cone and Sphere
N/A

 

Cone:

Cone is a solid or hollow body with a round base and pointed top. It is formed by the rotation of a triangle along any of the side.

 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;Volume&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Base area&#xA0;</mtext><mo>&#xD7;</mo><mtext>&#xA0;Height&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mi>h</mi><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Slant height&#xA0;</mtext><mo>(</mo><mi>I</mi><mo>)</mo><mo>=</mo><msqrt><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><msup><mi>h</mi><mn>2</mn></msup></msqrt><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Curved surface area&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mi>l</mi><mo>+</mo><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mo>(</mo><mo>&#x2223;</mo><mo>+</mo><mi>r</mi><mo>)</mo><mspace linebreak="newline"/><mi>W</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>,</mo><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>b</mi><mi>a</mi><mi>s</mi><mi>e</mi><mo>,</mo><mo>&#xA0;</mo><mi>h</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>H</mi><mi>e</mi><mi>i</mi><mi>g</mi><mi>h</mi><mi>t</mi><mo>&#xA0;</mo><mi>a</mi><mi>n</mi><mi>d</mi><mo>&#xA0;</mo><mi>l</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>s</mi><mi>l</mi><mi>a</mi><mi>n</mi><mi>t</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>i</mi><mi>g</mi><mi>h</mi><mi>t</mi><mo>.</mo></math>

 

Frustum of Cone:

  • If a cone is cut by a plane parallel to the base so as to divide the cone into two parts upper part and lower part, then the lower part is called frustum.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Volume&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><mi>h</mi></mrow><mn>3</mn></mfrac><mfenced separators="|"><mrow><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><mi>r</mi><mi>R</mi></mrow></mfenced><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Slant height&#xA0;</mtext><mo>(</mo><mi>I</mi><mo>)</mo><mo>=</mo><msqrt><msup><mi>h</mi><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>R</mi><mo>&#x2212;</mo><mi>r</mi><msup><mo>)</mo><mn>2</mn></msup></msqrt><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Total surface area&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mfenced separators="|"><mrow><msup><mi>R</mi><mn>2</mn></msup><mo>+</mo><msup><mi>r</mi><mn>2</mn></msup><mo>+</mo><mi>R</mi><mi>l</mi><mo>+</mo><mi>r</mi><mi>l</mi></mrow></mfenced><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Curved surface area&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mo>(</mo><mi>R</mi><mo>+</mo><mi>r</mi><mo>)</mo><mo>&#x2223;</mo><mspace linebreak="newline"/></math>

 

Sphere:

  • A sphere is a three-dimensional solid figure, which is made up of all points in the space, which lie at a constant distance from a fixed point.
  • That constant distance and fixed point are respectively called the radius and centre of the sphere.
  • In fact, a sphere is like a solid ball.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Volume of the sphere&#xA0;</mtext><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><mo>(</mo><mi>r</mi><msup><mo>)</mo><mn>3</mn></msup><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Total surface area&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mspace linebreak="newline"/><mi>W</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>R</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi></math>

 

Hollow Sphere or Spherical Shell:

  • It’s both external and internal surfaces are spherical and both the surfaces have a common central point.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Volume of hollow sphere&#xA0;</mtext><mo>=</mo><mfrac><mn>4</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><mfenced separators="|"><mrow><msup><mi>R</mi><mn>3</mn></msup><mo>&#x2212;</mo><msup><mi>r</mi><mn>3</mn></msup></mrow></mfenced><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Internal surface area&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;External surface area&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><msup><mi>R</mi><mn>2</mn></msup><mspace linebreak="newline"/><mi>W</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>,</mo><mo>&#xA0;</mo><mi>R</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>E</mi><mi>x</mi><mi>t</mi><mi>e</mi><mi>r</mi><mi>n</mi><mi>a</mi><mi>l</mi><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi><mo>&#xA0;</mo><mi>a</mi><mi>n</mi><mi>d</mi><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>I</mi><mi>n</mi><mi>t</mi><mi>e</mi><mi>r</mi><mi>n</mi><mi>a</mi><mi>l</mi><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi></math>

Hemisphere:

 

 

  • It is the half part of a sphere.

 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Volume of the hemisphere&#xA0;</mtext><mo>=</mo><mfrac><mn>2</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><mo>(</mo><mi>r</mi><msup><mo>)</mo><mn>3</mn></msup><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Total surface area&#xA0;</mtext><mo>=</mo><mn>3</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Curved surface area&#xA0;</mtext><mo>=</mo><mn>2</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mspace linebreak="newline"/><mi>W</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>,</mo><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>R</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi></math>

Examples:

 

 

  1. The diameter of a right circular cone is 14 m and its slant height is 10 m. Find its curved surface area, total surface area.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Here,&#xA0;</mtext><mi>r</mi><mo>=</mo><mfrac><mn>14</mn><mn>2</mn></mfrac><mo>=</mo><mn>7</mn><mi>m</mi><mtext>&#xA0;and&#xA0;</mtext><mi>l</mi><mo>=</mo><mn>10</mn><mi>m</mi><mspace linebreak="newline"/><mtext>&#xA0;Curved surface area&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mi>l</mi><mo>=</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>10</mn><mo>=</mo><mn>22</mn><mo>&#xD7;</mo><mn>10</mn><mo>=</mo><mn>220</mn><mi>s</mi><mi>q</mi><mi>m</mi><mspace linebreak="newline"/><mtext>&#xA0;Total surface area&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mo>(</mo><mi>r</mi><mo>+</mo><mi>l</mi><mo>)</mo><mo>=</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mn>7</mn><mo>(</mo><mn>7</mn><mo>+</mo><mn>10</mn><mo>)</mo><mo>=</mo><mn>22</mn><mo>&#xD7;</mo><mn>17</mn><mo>=</mo><mn>374</mn><mi>s</mi><mi>q</mi><mi>m</mi></math>

 

  2. The frustum of a right circular cone has the diameters of base 10 cm, of top 6 cm and a height of 5 cm. Find its slant height.

 

Solution: Lets draw a figure from given data

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Slant height&#xA0;</mtext><mo>=</mo><msqrt><msup><mi>h</mi><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>R</mi><mo>&#x2212;</mo><mi>r</mi><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>25</mn><mo>+</mo><mo>(</mo><mn>5</mn><mo>&#x2212;</mo><mn>3</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>25</mn><mo>+</mo><mn>4</mn></msqrt><mo>=</mo><msqrt><mn>29</mn></msqrt><mi>c</mi><mi>m</mi></math>

 

  3. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes?

Solution:

 

Given that the diameter of the Moon is approximately one-fourth of the diameter of the Earth.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>L</mi><mi>e</mi><mi>t</mi><mo>&#xA0;</mo><mi>t</mi><mi>h</mi><mi>e</mi><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi><mo>&#xA0;</mo><mi>o</mi><mi>n</mi><mo>&#xA0;</mo><mi>M</mi><mi>o</mi><mi>o</mi><mi>n</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>r</mi><mspace linebreak="newline"/><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>T</mi><mi>h</mi><mi>e</mi><mi>n</mi><mo>,</mo><mo>&#xA0;</mo><mi>w</mi><mi>e</mi><mo>&#xA0;</mo><mi>g</mi><mi>e</mi><mi>t</mi><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>E</mi><mi>a</mi><mi>r</mi><mi>t</mi><mi>h</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>4</mn><mi>r</mi><mspace linebreak="newline"/><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mfrac><mtext>&#xA0;Volume of Moon&#xA0;</mtext><mtext>&#xA0;Volume of Earth&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mrow><mfrac><mn>4</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><msup><mi>r</mi><mn>3</mn></msup></mrow><mrow><mfrac><mn>4</mn><mn>3</mn></mfrac><mi>&#x3C0;</mi><mo>(</mo><mn>4</mn><mi>r</mi><msup><mo>)</mo><mn>3</mn></msup></mrow></mfrac><mo>=</mo><mfrac><msup><mi>r</mi><mn>3</mn></msup><mrow><mn>64</mn><msup><mi>r</mi><mn>3</mn></msup></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>64</mn></mfrac><mo>=</mo><mn>1</mn><mo>:</mo><mn>64</mn></math>

 

  4. What will be the difference between total surface area and curved surface area of a hemisphere having 2 cm diameter?

 

Solution: Given diameter = 2 cm, so, radius = 1 cm.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Total surface area of hemisphere&#xA0;</mtext><mo>=</mo><mn>3</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mtext>&#xA0;and curved surface area&#xA0;</mtext><mo>=</mo><mn>2</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Difference between total surface area and curved surface area&#xA0;</mtext><mo>=</mo><mn>3</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>&#x2212;</mo><mn>2</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mi>&#x3C0;</mi><mo>(</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mi>&#x3C0;</mi><mtext>&#xA0;sq&#xA0;cm</mtext></math>

 

 

  5. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: When radius&#xA0;</mtext><mo>=</mo><mn>7</mn><mi>c</mi><mi>m</mi><mtext>&#xA0;Then, surface area&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><mo>(</mo><mn>7</mn><msup><mo>)</mo><mn>2</mn></msup><mspace linebreak="newline"/><mtext>&#xA0;When radius&#xA0;</mtext><mo>=</mo><mn>14</mn><mi>c</mi><mi>m</mi><mtext>&#xA0;Then, surface area&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><mo>(</mo><mn>14</mn><msup><mo>)</mo><mn>2</mn></msup><mspace linebreak="newline"/><mtext>&#xA0;Ratio of surface areas&#xA0;</mtext><mo>=</mo><mn>4</mn><mi>&#x3C0;</mi><mo>(</mo><mn>7</mn><msup><mo>)</mo><mn>2</mn></msup><mo>:</mo><mn>4</mn><mi>&#x3C0;</mi><mo>(</mo><mn>14</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><msup><mn>7</mn><mn>2</mn></msup><mo>:</mo><msup><mn>14</mn><mn>2</mn></msup><mo>=</mo><msup><mn>1</mn><mn>2</mn></msup><mo>:</mo><msup><mn>2</mn><mn>2</mn></msup><mo>=</mo><mn>1</mn><mo>:</mo><mn>4</mn></math>

 


3 Volume and Surface area of Prism and Pyramid
N/A

 

Prism:

  • A right prism is a solid whose top and bottom face are parallel to each other and are identical polygons that are parallel.
  • The faces joining the top and bottom faces are rectangles and are called lateral faces.
  • The distance between the base and the top is called height or length of the right prism.

 

  • Volume of prism = Area of base x Height of the prism
  • Lateral (vertical) surface area = Perimeter of base x Height of the prism
  • Total surface area = Lateral surface area + 2 x Area of base 

Pyramid:

  • A solid whose base is a polygon and whose faces are triangles, is called a pyramid. The triangular faces meet at a common point called vertex,
  • "A pyramid whose base is regular polygon and the foot of the perpendicular from the vertex of the base, coincides with the centre of the base, is called a right pyramid."

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;Volume of a pyramid&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Area of base&#xA0;</mtext><mi>x</mi><mtext>&#xA0;Height&#xA0;</mtext><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;Lateral surface area&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Perimeter of the base&#xA0;</mtext><mi>x</mi><mtext>&#xA0;Slant height</mtext></math>

  • Total surface area = Lateral surface area + Area of the base

Examples:

  1. The perimeter of the triangular base of a right prism is 60 cm and the sides of the base are in the ratio 5: 12: 13. Then, its volume will be

Solution: Let the sides of the base are 5s, 12s and 13s respectively.

Given, perimeter of base = 60 cm

So, 5s + 12s + 13s = 60 cm

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>30</mn><mi>s</mi><mo>=</mo><mn>60</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mspace width="1em"/><mi>s</mi><mo>=</mo><mfrac><mn>60</mn><mn>30</mn></mfrac><mo>=</mo><mn>2</mn></mtd></mtr></mtable></math>

The sides of base are 5s = 5 x 2 = 10 cm, 12s = 12 x 2 = 24 cm, 13s = 13 x 2 = 26 cm.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, volume of prism&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>10</mn><mo>&#xD7;</mo><mn>24</mn><mo>&#xD7;</mo><mn>50</mn><mo>=</mo><mn>6000</mn><mi>c</mi><msup><mi>m</mi><mn>3</mn></msup></math>

2. A prism and a pyramid have the same base and the same height. Find the ratio of the volumes of the prism and the pyramid.

Solution:

We know that,

Volume of the prism = (Area of the base) x (Height)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Volume of the pyramid&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mtext>&#xA0;(Area of the base)&#xA0;</mtext><mi>x</mi><mtext>&#xA0;(Height)</mtext><mspace linebreak="newline"/><mtext>&#xA0;Required ratio&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>A</mi><mo>&#xD7;</mo><mi>H</mi></mrow><mrow><mfrac><mn>1</mn><mn>3</mn></mfrac><mo>&#xD7;</mo><mi>A</mi><mo>&#xD7;</mo><mi>H</mi></mrow></mfrac><mo>=</mo><mn>3</mn><mo>:</mo><mn>1</mn></math>

Therefore, Ratio of the volumes of the prism and the pyramid = 3: 1

3. The base of a right prism is a square having side of 20 cm. If its height is 8 cm, then find the total surface area and volume of the prism.

Solution:

Given, side = 20 cm and height = 8 cm

  • Now, Total surface area = Lateral surface area + 2 x Area of base
  • Now, lateral surface area = Perimeter of the base x Height = [20 + 20 + 20 + 20] x 8 = 80 x 8 = 640 cm2
  • Area of base = Area of square = 20 x 20 = 400 cm2

On putting, these values in formula,

  • we get Total surface area = 640 + 2 x 400 = 640 + 800 = 1440 cm2
  • Volume of the prism = Area of base x Height = 20 x 20 x 8 = 3200 cm3

4. A prism has the base a right-angled triangle whose sides adjacent to the right angle are 12 cm and 15 cm long. The height of the prism is 20 cm. The density of the material of the prism is 4 g/cu cm. The weight of the prism is

Solution: Volume of prism = Area of base x height

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>12</mn><mo>&#xD7;</mo><mn>15</mn><mo>&#xD7;</mo><mn>20</mn></mtd></mtr><mtr><mtd><mo>=</mo><mn>1800</mn><mi>c</mi><msup><mi>m</mi><mn>3</mn></msup></mtd></mtr></mtable></math>

Therefore, Weight of prism = 1800 x 4

                                                   = 7200 g

                                                   = 72 kg

  5. Find the total surface area of a pyramid having a slant height of 10 cm and a base which is a square of side 2 cm (in cm)?

Solution: Total surface area of pyramid

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfenced open="[" close="]" separators="|"><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;(perimeter of the base) (slant height)&#xA0;</mtext></mrow></mfenced><mo>+</mo><mtext>&#xA0;Area of the base&#xA0;</mtext></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>10</mn><mo>+</mo><msup><mn>2</mn><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>=</mo><mn>44</mn><mi>c</mi><msup><mi>m</mi><mn>2</mn></msup></mtd></mtr></mtable></math>


4 Geometry (Lines and Angles)
N/A

  • Geometry is a branch of Mathematics which deals with the questions and concepts of shape, size, relative position of figures, their angles etc.

It can be broadly divided into two parts

Plane Geometry

  • Plane geometry is about flat shapes like line, circle and triangle.
  • These are two-dimensional figure which can easily be drawn on paper.

Solid Geometry

  • Solid geometry is about three-dimensional objects like cube, prism and sphere.
  • These are three-dimensional figure

Point:

A figure of which length, breadth and height cannot be measured is called a point. It is infinitesimal.

Line:

  • A line is defined by its length but has no breadth.
  • A line contains infinite points.
  • Three or more points are said to be collinear, if there is a line which contains all of them.

Plane:

 It is a flat surface having length and breadth both but no thickness. It is a 2-dimensional figure.

Parallel Lines:

  • Two lines in the same plane are said to be parallel, if they don't have any intersection point, however how far they are extended in either direction.
  • They remain at same distance for the whole length. The sign of parallel is '| |'.
  • Here, 1 and m are called parallel lines.

Transversal Lines:

  • A straight line that cuts two or more straight lines at distinct points is called a transversal.
  • In the figure given below 1 and m are parallel lines and p is a transversal.

Angle:

  • An angle is formed by two rays with a common initial point.
  • Let O is the initial point, then O is called the vertex. e.g., AOB = θ

Types of Angles:

1. Right angle: The angle whose value is 90° is called a right angle. θ = 90°)

2. Acute angle: The angle whose value lies between 0° and 90° is called an acute angle. (0°< θ < 90°)

 

 3. Obtuse angle: The angle whose value lies between 90° and 180° is called an obtuse angle. (90°<θ<180°)

 

 4. Straight angle: The angle whose value is 180° is called a straight angle. (θ =180°)

 

 5. Reflex angle: The angle whose value lies between 180° and 360° is called a reflex angle. (180°< θ <360°)

 

 6. Complete angle: The angle whose value is 360° is called a complete angle. (θ =360°)

 

 7. Supplementary angle: If the sum of two angles is 180°, then they are called supplementary angles. Let θ1 and θ2 be two angles, then θ1 + θ2 =180°

 

 8. Complementary angle: If the sum of two angles is 180°, then they are called supplementary angles. Let θ1 and θ2 be two angles, then θ1 + θ2 = 90°

Angle Bisector:

  • A line which cuts an angle into two equal angles is called an angle bisector
  • An angle bisector can be internal or external.

Internal angle bisector:

Here, two angles are formed AOC and BOC. Both angles are equal (θ) because OB is the internal angle bisector.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Here,&#xA0;</mtext><mi>&#x3B8;</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi mathvariant="normal">&#x2220;</mi><mi>A</mi><mi>O</mi><mi>C</mi><mo>=</mo><mi mathvariant="normal">&#x2220;</mi><mi>A</mi><mi>O</mi><mi>B</mi><mo>=</mo><mi mathvariant="normal">&#x2220;</mi><mi>B</mi><mi>O</mi><mi>C</mi></math>

External angle bisector:

Here, ∠A'OB and ∠BOC are equal and external bisector is OB.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Here,&#xA0;</mtext><mi>&#x3B8;</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi mathvariant="normal">&#x2220;</mi><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mi>O</mi><mi>C</mi><mo>=</mo><mi mathvariant="normal">&#x2220;</mi><msup><mi>A</mi><mi mathvariant="normal">&#x2032;</mi></msup><mi>O</mi><mi>B</mi><mo>=</mo><mi mathvariant="normal">&#x2220;</mi><mi>B</mi><mi>O</mi><mi>C</mi></math>

Examples:

  1. In the given figure, find x.

Solution: From figure

∠ABC + ∠DBC = 1800

(3x + 15)0 + (x + 5)0 = 1800

4x = 1600

x = 400

 2. In the given figure, straight lines AB and CD intersect at O. If <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">&#x2220;</mi><mi>&#x3B2;</mi></math> = 3∠p, then ∠p is equal to

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: From question,&#xA0;</mtext><mi mathvariant="normal">&#x2220;</mi><mi>&#x3B2;</mi><mo>=</mo><mn>3</mn><mi mathvariant="normal">&#x2220;</mi><mi>p</mi><mspace linebreak="newline"/><mtext>&#xA0;We know that,&#xA0;</mtext><mi mathvariant="normal">&#x2220;</mi><mi>&#x3B2;</mi><mo>+</mo><mi mathvariant="normal">&#x2220;</mi><mi>p</mi><mo>=</mo><msup><mn>180</mn><mo>&#x2218;</mo></msup></math>

∠p + 3∠p = 1800

4∠p = 1800

∠p = 450

 3. In the given figure, AB||CD. If ∠CAB = 800 and ∠EFC = 25°, then ∠CEF is equal to

Solution: Given, ∠CAB = 800 and ∠EFC = 25° and AB||CD

Let ∠CEF = xo

Here AB||CD and AF is transversal

So, ∠DCF = ∠CAB = 800 [since, corresponding angles]

In triangle CEF, side EC has been produced to D.

x + 25 = 80o

x = 55o

 4. In the given figure, AOB is straight line If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then ∠COD is equal to

Solution: Given AOB is a straight line.

So, ∠AOC + ∠EOB + ∠BOD = 180o

40 + 4x + 3x = 180

7x = 180 – 40

7x = 140

x = 20

Therefore, ∠COD = 4x° = 80o

 5. If every interior angle of regular octagon is 135°, then find the external angle of it.

Solution: Every external angle of octagon = 180o – Interior angle

                                                                             = 180o – 135o

                                                                             = 45o


5 Geometry (Triangle)
N/A

Triangle:

A triangle is a three - sided closed plane figure which is formed by joining 3 non-collinear points.

  • In figure A, B, C are three non-collinear points which forms ?ABC.
  •  Now, in a ?ABC,

There are three vertices A, B and C,

Three sides AB, BC and AC and

three angles ∠A, ∠B and ∠C and sum of these three angles is 180°. i.e., ∠A+ ∠B + ∠C =180°

Types of Triangles:

  1.  Equilateral triangle: A triangle having all sides equal is called an equilateral triangle. In this triangle each angle is congruent and is equal to 60°

 2. Scalene triangle: A triangle having all sides of different length is called a scalene triangle.

 3. Isosceles triangle: A triangle having two sides equal is called an isosceles triangle. In this triangle angles opposite to congruent sides are also equal.

 4.  Right angled triangle: A triangle one of whose angles measures 90° is called a right-angled triangle. In ?ABC, ∠B = 90°

 5.  Obtuse angled triangle: A triangle one of whose angles lies between 90° and 180° is called an obtuse angled triangle. In ?ABC, ∠A >90°

 6.  Acute angled triangle: A triangle whose each angle is less than 90° is called an acute angled triangle. In ?ABC, (∠A, ∠B, ∠C) < 90o

Properties of Triangles:

1. Sum of two sides is always more than third side

2. Difference of two sides is always less than third side

3. Greater angle has greater side opposite to it and smaller angle has smaller side opposite to it

4. The exterior angle is equal to the sum of two interior angles not adjacent to it, ∠ACD = ∠BCE = ∠A+ ∠B

Congruency of Triangles

Two triangles are congruent if they satisfy the following conditions.

  1. SSS congruency (Side - Side - Side): If the three sides of one triangle are equal to the corresponding three sides of another triangle, then two triangles are congruent.
  • In ?ABC and ?PQR, AB = PQ, BC = QR and AC = PR, then ?ABC = PQR

 2.  SAS congruency (Side - Angle - Side): If two sides and the angle included between them are equal to the corresponding side and angle included of other triangle are equal, then the two triangles are congruent.

  • In ?ABC and ?PQR, AB = PQ, BC = QR and ∠B = ∠Q, then ?ABC = ?PQR

 3.   ASA congruency (Angle - Side - Angle): If two angles and the side included between them are equal to the corresponding angles and side included of other triangle are equal, then two triangles are congruent.

  • In ?ABC and ?PQR, ∠B = ∠Q, ∠C = ∠R and BC = QR then, ?ABC = ?PQR

 4.  AAS congruency (Angle - Angle - Side): If two angles and the side other than the included side of one triangle are equal to the corresponding angles and the side other than included side of other triangle are equal, then the two triangles are congruent.

  • In ?ABC and ?PQR, ∠B = ∠Q, ∠C = ∠R and AC = PR then, ?ABC = ?PQR

 5.  RHS congruency (Right Angle-Hypotenuse-Side):

 If hypotenuse and one side of right-angled triangle are equal to hypotenuse and corresponding side of other triangle, then the two triangles are congruent.

  • In ?ABC and ?PQR, BC =QR AC=PR and ∠B = ∠Q= 90°, then ?ABC = ?PQR

Similarity of Triangles

Two triangles are said to be similar if they satisfy the following conditions.

  1.  AA similarity (Angle-Angle Similarity): If two angles of one triangle are equal to the two angles of the other triangle, then the two triangles are similar.
  2.  SAS similarity (Side-Side-Side Similarity): If two sides of one triangle are proportional to the corresponding side of other triangle and the angle included by them are equal, then the two triangles are similar.
  3. SSS similarity (Side-Angle-Side Similarity): If three sides of one triangle are proportional to the corresponding three sides of other triangle, then the two triangles are similar.

Properties of Similar Triangle:

  1. Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides
  2.  Ratio of the areas of two similar triangles is equal to the ratio of the squares of corresponding altitudes and medians
  3.  In two similar triangles ABC and PQR

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>P</mi><mi>Q</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mi>C</mi></mrow><mrow><mi>P</mi><mi>R</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>B</mi><mi>C</mi></mrow><mrow><mi>Q</mi><mi>R</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mtext>&#xA0;Perimeter of&#xA0;</mtext><mi mathvariant="normal">&#x25B3;</mi><mi>A</mi><mi>B</mi><mi>C</mi></mrow><mrow><mtext>&#xA0;Perimeter of&#xA0;</mtext><mi mathvariant="normal">&#x25B3;</mi><mi>P</mi><mi>Q</mi><mi>R</mi></mrow></mfrac></math>

Some Important Terms Related to Triangles:

Angle Bisector:

  • It is the bisector of an angle contained in the vertex of a triangle.
  • All the three angle bisectors of a triangle meet at a point called the incentre of the triangle.
  • The incentre is the centre of a circle which can be perfectly inscribed in the triangle.

  • In radius = JD = JE = JF
  • <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">&#x2220;</mi><mi>B</mi><mi>J</mi><mi>C</mi><mo>=</mo><mn>90</mn><mo>+</mo><mfrac><mrow><mi mathvariant="normal">&#x2220;</mi><mi>A</mi></mrow><mn>2</mn></mfrac></math>

In ,AD, BE and CF are angle bisectors and meet at incentre J.

 

 

Perpendicular Bisector:

  • It is the line passing through the mid-point of the side of a triangle and perpendicular to it.
  • All the three perpendicular bisectors of a triangle meet at a point called the circumcentre of the triangle.
  • The circumcentre is the centre of a circle which can be perfectly circumscribed about the triangle.
  • The circumradius = PC = QC = RC and QCR = 2P.
  • In PQR, the angle bisectors meet at circumcentre C.
  • Also, D, E and F are the mid-points of OR, RP and PQ, respectively

Median:

  • It is the line joining the mid-point of a side of a triangle with the vertex opposite to that side.
  • All the three medians of a triangle meet at a point called the centroid of the triangle.
  • They also intersect each other such that each median is split in the ratio of 1: 2 from the base side.
  • In ,AD, BE and CF are the medians and meet at the centroid G. D, E, F are mid-points of BC, CA and AB, respectively.

 

Examples:

  1. The angles of triangle are in the ratio 3:5:7. The triangle is

Solution: Let the angle measure (3x) °, (5x) ° and (7x) °

  • Then, 3x + 5x + 7x = 180 => 15x = 180 => x = 12
  • These angles are 36° ,60° and 84°
  • Therefore, the triangle is acute angled

 2. The side AC of a ABC is extended to ‘D’ such that BC = CD. If ACB is 70°, then what is ADB equal to?

 

Solution: Let’s draw a figure from given data

  • ACB + BCD = 180o
  • BCD = 180o – 70o = 110o

 

 

  • In  BC = CD

 

CBD = CDB…(i)

 

  • Also, BCD + ∠CBD + CDB = 180o

 

2∠CBD = 180o - BCD

 

CBD = 180o – 110o ð = 70o

 

Therefore, CDB = ADB =  = 35o

 

 

 3. ABC is a triangle right angled at A and a perpendicular AD is drawn on the hypotenuse BC. What is BC.AD equal to?

 

Solution:

 

 

  • In case of a right-angled triangle, if we draw a perpendicular from the vertex containing right angle to the hypotenuse, we get three triangles, two smaller and one original and these three triangles are similar triangles. 

here ?ABC ~ ?ABD ~ ?ADC

Therefore, BC. AD = AB. AC

 4. In a ?ABC, A = 90o, C = 55o and  What is the value of BAD?

 

 

Solution: Let’s draw a figure from the given data

In ?BAC, B = 180o – (90o + 55o) = 35o

 

Now, in ?ADB, ADB = 90o

ADB + DBA + BAD = 180o

BAD = 180o – 90o – 35o = 55o

 5. In a ?ABC, if A = 115°, C = 20° and D is a point on BC such that AD  BC and BD = 7 cm, then AD is of length

 

Solution:

Let us draw a figure from the data given

Given, In ?ABC, , A = 115o, C = 20o

 

B = 180o – (115o + 20o) = 45o

Now in  ?ABC,  = tan 45o

 

AD = BD = 7 cm


6 Geometry (Quadrilateral)
N/A

 

  • It is a plane figure bounded by four straight lines. It has four sides and four internal angles.
  • The sum of the internal angles of a quadrilateral is equal to 360°.

 

Parallelogram

 

A quadrilateral in which the opposite sides are equal and parallel, is called a parallelogram. In a parallelogram,

(i)   The sum of any two adjacent interior angles is equal to 180°.

 

 A + B = B + C = C + D = D + A =180°

(ii)   The opposite angles are equal in magnitudes A = C and B = D.

(iii)   line joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram.

 

(i)   line joining the mid-points of the adjacent sides of a parallelogram is a parallelogram.

()   The parallelogram inscribed in a circle is a rectangle and circumscribed about a circle is a rhombus.

 

(iAC2 + BD2 =2 (AB2 + BC2)

 

Rhombus

A parallelogram in which all the sides are equal, is called a rhombus.

(i)   The opposite sides are parallel and all the sides are of equal lengths. AB = BC = CD = DA.

(ii)  The sum of any two adjacent interior angles is equal to 180°. 

 

A + B = B + C = C + D = D + A = 180° 

(iii)  The opposite angles are equal in magnitudes, i.e., A = C and B = D.

(iThe diagonals bisect each other at right angles and form four right angled triangles.

(  Area of the four right triangles ?AOB = ?BOC = ?COD = ?DOA and each equal  the area of the rhombus.

(i)   Figure formed by joining the mid-points of the adjacent sides of a rhombus is a rectangle. 

Rectangle

A parallelogram in which the adjacent sides are perpendicular to each other and opposite side are equal is called a rectangle.

(i)  The diagonals of a rectangle are of equal magnitudes and bisect each other i.e., AC = BD and OA = OB = OC = OD.

 

(i)   The figure formed by joining the mid-points of adjacent sides of a rectangle is a rhombus.

(ii)   The quadrilateral formed by joining the mid-points of intersection of the angle bisectors of a parallelogram is a rectangle.

Square

 

A parallelogram in which all the sides are equal and perpendicular to each other, is called a square.

(i)   The diagonals bisect each other at right angles and form four isosceles right-angled triangles.

 

(ii)   The diagonals of a square are of equal magnitudes i.e., AC = BD.

 

(iii)   The figure formed by joining the mid-points of adjacent sides of a square is a square.

 

Trapezium

It is a quadrilateral where only one pair of opposite sides are parallel.

 

ABCD is a trapezium as AB || DC.

(i)   If the non-parallel sides i.e., (AD and BC) are equal, then diagonals will also be equal to each other,

(ii)   Diagonals intersect each other in the ratio of lengths of parallel sides.

(iii)   line joining the mid-points of non-parallel sides is half the sum of parallel sides and is called the median.

Cyclic Quadrilateral:

 

A quadrilateral whose vertices are on the circumference of a circle, is called a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral are supplementary, i.e.,  = 180°.

 

If the side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. i.e., ADC = CBE

 

Polygons:

A polygon is a closed plane figure bounded by straight lines.

 Convex polygon

A polygon in which none of its interior angles is more than 180° is called convex polygon.

Concave polygon

A polygon in which at least one angle is more than 180° is called concave polygon.

 Irregular polygon

 A polygon in which all the sides or angles are not of the same measure.

 Regular polygon

 A regular polygon has all its sides and angles equal.

(iii)   Each exterior angle of a regular polygon = 

(ii)   Each interior angle = 180 — Exterior angle.

 

(iii)   Sum of all interior angles = (2n -4) x 90°

 

(i)   Sum of all exterior angles =360°

 

()   Number of diagonals of polygon on n sides = 

Examples:

 

 1. The angles of a quadrilateral are in the ratios 3: 4: 5: 6. The smallest of these angles is

 

Solution:

 

Let the angles of the quadrilateral be (3x) °, (4x) °, (5x) ° and (6x) °.

 

Then, 3x + 4x + 5x + 6x = 360 => 18x = 360 => x = 20

 

Smallest angle = (3 x 20) ° = 60°

 

 2. In the give figure, AD || BC. Find the value of x.

Solution: Here, AD|| BC,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mi>O</mi><mi>A</mi></mrow><mrow><mi>O</mi><mi>C</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mi>D</mi></mrow><mrow><mi>O</mi><mi>B</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>3</mn><mrow><mi>x</mi><mo>&#x2212;</mo><mn>3</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>x</mi><mo>&#x2212;</mo><mn>5</mn></mrow><mrow><mn>3</mn><mi>x</mi><mo>&#x2212;</mo><mn>19</mn></mrow></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>9</mn><mi>x</mi><mo>&#x2212;</mo><mn>57</mn><mo>=</mo><msup><mi>x</mi><mn>2</mn></msup><mo>&#x2212;</mo><mn>8</mn><mi>x</mi><mo>+</mo><mn>15</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mi>x</mi><mn>2</mn></msup><mo>&#x2212;</mo><mn>17</mn><mi>x</mi><mo>+</mo><mn>72</mn><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mo>(</mo><mi>x</mi><mo>&#x2212;</mo><mn>8</mn><mo>)</mo><mo>(</mo><mi>x</mi><mo>&#x2212;</mo><mn>9</mn><mo>)</mo><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mn>8</mn><mo>,</mo><mn>9</mn></mtd></mtr></mtable></math>

 3. If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angle of the parallelogram is

Solution: Let the smallest angle be xo

Then, its adjacent angle = (2x – 24) o

Therefore, x + 2x – 24 = 180

ð 3x = 204

ð x = 68

Therefore, Largest angle = (2 x 68 – 24) o = (136 – 24) o = 112o

 4. A quadrilateral ABCD is inscribed in a circle. If AB is parallel to CD and AC = BD, then the quadrilateral must be a

Solution:

The quadrilateral must be trapezium because a quadrilateral where only one pair of opposite sides are parallel (in this case AB || CD) is trapezium.

 

 5. In the given figure, ABCD is a parallelogram in which BAD = 75° and CBD = 60°. Then, BDC is equal to

Solution: Given C = A = 75o

Since opposite angles of parallelogram are equal

In ?ABCD, CBD + BDC + BDC = 180o

60o + 75o + BDC = 180o

135o + BDC = 180o

BDC = 45o 


7 Geometry (Circle)
N/A

 

Circle

A circle is a set of points which are equidistant from a given point.

The given point is known as the centre of that circle.

In the given figure O is the centre of circle and r is the radius of circle.

 

Chord

A line segment whose end points lie on the circle.

  • AB is the chord in the given figure.
  • The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
  • AD = DB and OD  AB.

 

(i)   Equal chords of a circle are equidistant from the centre and vice-versa.

 

 

 

(ii)   Equal chords subtend equal angles at the centre and vice-versa

 

(iii)   In a circle or in congruent circles equal chords are made by equal arcs

 

(i)   If two chords AB and CD of a circle, intersect inside a circle (outside the circle when produced at point E) then, AE x BE = CE x DE

Secant

 

A line segment which intersects the circle at two distinct points, is called as secant.

 

 

In the given diagram PQ intersects circle at two points at A and B

 

Sector of Circle

The region enclosed by an arc of a circle and its two bounding radii is called a sector of the circle. Thus, in the adjoining figure, OABO is the sector of the circle C (o, r)

 

Segment of a Circle

A chord divides the circle into two regions. These two regions are called the segments of a circle.

 

In the figure, PSR is the major segment and PQR is minor segment. Angles made in the same segment are equal.

Angles and Central Angles

  • The angle subtended at the centre by any two points on the circumference of the circle is called central angle.

(i)   Angle in a semi-circle is a right angle.

 

(ii)   The angle subtended by an arc at the centre of the circle is twice the angle subtended by the arc at any point on the remaining part of the circle.

 

Tangent:

A line segment which has one common point with circumference of a circle i.e., it touches only at one point is called as tangent of circle.

 

In the given figure PQ is tangent which touches the circle at point R.

(i)   Radius is always perpendicular to tangent.

 

(ii)   The length of two tangents drawn from the external point to the circle are equal.

 

(iii)   The angle which a chord makes with a tangent at its point of contact is equal to any angle in the alternate segment. PTA = ABT, where AT is the chord and PT is the tangent to the circle.

(i)   If PT is a tangent (with P being an external point and T being the point of contact) and PAB is a secant to circle (with A and B as the points, where the secant cuts the circle), then PT2 =PA x PB.

Pair of Circle

 

(i)   (a) When two circles touch externally, then the distance between their centres is equal to the sum of their radii then, AB = AC + BC

(b) When two circles touch internally the distance between their centres is equal to the difference between their radii AB = AC - BC

(ii)   In a given pair of circles, there are two types of tangents. The direct tangents and the cross (or transverse) tangents. In the figure, AB and CD are the direct tangents and EH and GF are the transverse tangents.

 

When two circle of radii r1 and r2 have their centres at a distance d.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;The length of the direct common tangent&#xA0;</mtext><mo>=</mo><msqrt><msup><mi>d</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mfenced separators="|"><mrow><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2212;</mo><msub><mi>r</mi><mn>2</mn></msub></mrow></mfenced><mn>2</mn></msup></msqrt><mspace linebreak="newline"/><mo>&#x2219;&#xA0;&#xA0;&#xA0;&#xA0;</mo><mtext>&#xA0;The length of transverse tangent&#xA0;</mtext><mo>=</mo><msqrt><msup><mi>d</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mfenced separators="|"><mrow><msub><mi>r</mi><mn>1</mn></msub><mo>+</mo><msub><mi>r</mi><mn>2</mn></msub></mrow></mfenced><mn>2</mn></msup></msqrt></math>

Examples:

 

 1. Two circles of same radius 5 cm, intersect each other at A and B. If AB = 8 cm, then the distance between the centres is

Solution:

Let’s draw a figure from given data.

 

Let O and P be the radius of two circles.

 

From the figure, on joining AO and AP.

In ?AOX, PX = 3 cm

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>A</mi><msup><mi>O</mi><mn>2</mn></msup><mo>=</mo><mi>A</mi><msup><mi>X</mi><mn>2</mn></msup><mo>+</mo><mi>O</mi><msup><mi>X</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>(</mo><mn>5</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><msup><mfenced separators="|"><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mn>2</mn></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>O</mi><mi>X</mi><msup><mo>)</mo><mn>2</mn></msup></mtd></mtr><mtr><mtd><mn>25</mn><mo>=</mo><msup><mfenced separators="|"><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mn>2</mn></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>O</mi><mi>X</mi><msup><mo>)</mo><mn>2</mn></msup></mtd></mtr><mtr><mtd><mn>25</mn><mo>=</mo><msup><mfenced separators="|"><mfrac><mn>8</mn><mn>2</mn></mfrac></mfenced><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>O</mi><mi>X</mi><msup><mo>)</mo><mn>2</mn></msup></mtd></mtr><mtr><mtd><mn>25</mn><mo>=</mo><msup><mn>4</mn><mn>2</mn></msup><mo>+</mo><mo>(</mo><mi>O</mi><mi>X</mi><msup><mo>)</mo><mn>2</mn></msup></mtd></mtr><mtr><mtd><mo>(</mo><mi>O</mi><mi>X</mi><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mn>25</mn><mo>&#x2212;</mo><mn>16</mn><mo>=</mo><mn>9</mn></mtd></mtr><mtr><mtd><mi>O</mi><mi>X</mi><mo>=</mo><mn>3</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable><mspace linebreak="newline"/><mtext>&#xA0;Similarly, in&#xA0;</mtext><mi mathvariant="normal">&#x25B3;</mi><mi>A</mi><mi>P</mi><mi>X</mi><mo>,</mo><mi>P</mi><mi>X</mi><mo>=</mo><mn>3</mn><mi>c</mi><mi>m</mi></math>

 

Therefore, Distance between the centre = OX + PX = 3 + 3 = 6 cm

 

 2. In a ABC, O is its circumcentre and BAC = 50°. The measure of OBC is

Solution: The angle subtended by an arc at the centre of the circle is twice the angle subtended by the arc at any point on the remaining part of the circle.

BOC = 2 BAC = 2 X 50° = 100°

Now, in BOC, OB = OC

 

Let ∠OBC = ∠OCB= x

Therefore, x + x + 100o = 180o

 2x = 80o

x = 40o

 

 3. Find x in the given figure

 

Solution: If two chords of a circle, intersect inside a circle (outside a circle) at any point.

Then, PA x PB = PC x PD

ð 6 x 15 = 5 (x + 5)

ð x + 5 = 18

ð x = 13 cm

 

 1. If a regular hexagon is inscribed in a circle of radius r, then find the perimeter of the hexagon?

Solution: Here, OA = OB = AB = r

Therefore, perimeter of hexagon = 6 x AB = 6r

 

 2. R and r are the radii of two circles (R > 1). If the distance between the centres of the two circles be ‘d’, then length of common tangent of two circles is

Solution: Let us draw a figure from given data

 

Let the common tangent of both circles is PQ.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;From figure,&#xA0;</mtext><mi>P</mi><mi>Q</mi><mo>=</mo><mi>O</mi><mi>B</mi><mo>=</mo><msqrt><mo>(</mo><mi>A</mi><mi>B</mi><msup><mo>)</mo><mn>2</mn></msup><mo>&#x2212;</mo><mo>(</mo><mi>O</mi><mi>A</mi><msup><mo>)</mo><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mo>(</mo><mi>d</mi><msup><mo>)</mo><mn>2</mn></msup><mo>&#x2212;</mo><mo>(</mo><mi>R</mi><mo>&#x2212;</mo><mi>r</mi><msup><mo>)</mo><mn>2</mn></msup></msqrt></math>

 

 

 

 

 


8 Heights and Distances
N/A

  1. In a right-angled triangle OAB, where ∠BOA = θ

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>sin</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mtext>&#xA0;Perpendicular&#xA0;</mtext><mtext>&#xA0;Hypotenuse&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>O</mi><mi>B</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>cos</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mtext>&#xA0;Base&#xA0;</mtext><mtext>&#xA0;Hypotenuse&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mi>A</mi></mrow><mrow><mi>O</mi><mi>B</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>tan</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mtext>&#xA0;Perpendicular&#xA0;</mtext><mtext>&#xA0;Base&#xA0;</mtext></mfrac><mo>=</mo><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>O</mi><mi>A</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>cosec</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mi>sin</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mi>B</mi></mrow><mrow><mi>A</mi><mi>B</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>sec</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mi>cos</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mi>B</mi></mrow><mrow><mi>O</mi><mi>A</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>&#x2219;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mi>cot</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mi>tan</mi><mo>&#xA0;</mo><mi>&#x3B8;</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>O</mi><mi>A</mi></mrow><mrow><mi>A</mi><mi>B</mi></mrow></mfrac></mtd></mtr></mtable></math>

 2. Trigonometrical Identities:

  • sin2 θ + cos2 θ = 1
  • 1 + tan2 θ = sec2 θ
  • 1 + cot2 θ = cosec2θ

 3. Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

Therefore, Angle of elevation of P from O = ∠AOP

 4. Angle of Depression:

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Examples:

  1. The angle of elevation of the sun, when the length of the shadow of a tree is  times the height of the tree, is
  1. Solution:

  • Let AB be the tree and AC be its shadow.
  • Let ∠ACB = 0
  • θ = 30o

 2. The top of a 15-metre-high tower makes an angle of elevation of 60o with the bottom of an electric pole and angle of elevation of 30o with the top of the pole. What is the height of the electric pole?

Solution:

Let AB be the tower and CD be the electric pole.

Then, ∠ACB= 60o, ∠EDB= 60o and AB = 15m

Let CD = h. Then, BE = (AB – AE) = (AB – CD) = (15 – h)

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>A</mi><mi>C</mi></mrow></mfrac><mo>=</mo><mi>tan</mi><mo>&#xA0;</mo><msup><mn>60</mn><mn>0</mn></msup><mo>=</mo><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>A</mi><mi>C</mi><mo>=</mo><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><msqrt><mn>3</mn></msqrt></mfrac><mo>=</mo><mfrac><mn>15</mn><msqrt><mn>3</mn></msqrt></mfrac></mtd></mtr><mtr><mtd><mfrac><mrow><mi>B</mi><mi>E</mi></mrow><mrow><mi>D</mi><mi>E</mi></mrow></mfrac><mo>=</mo><mi>tan</mi><mo>&#xA0;</mo><msup><mn>30</mn><mn>0</mn></msup><mo>=</mo><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>D</mi><mi>E</mi><mo>=</mo><mo>(</mo><mi>B</mi><mi>E</mi><mo>&#xD7;</mo><msqrt><mn>3</mn></msqrt><mo>)</mo><mo>=</mo><msqrt><mn>3</mn></msqrt><mo>(</mo><mn>15</mn><mo>&#x2212;</mo><mi>h</mi><mo>)</mo></mtd></mtr><mtr><mtd><mi>A</mi><mi>C</mi><mo>=</mo><mi>D</mi><mi>E</mi><mo>=&gt;</mo><mfrac><mn>15</mn><msqrt><mn>3</mn></msqrt></mfrac><mo>=</mo><msqrt><mn>3</mn></msqrt><mo>(</mo><mn>15</mn><mo>&#x2212;</mo><mi>h</mi><mo>)</mo></mtd></mtr><mtr><mtd><mn>3</mn><mi>h</mi><mo>=</mo><mo>(</mo><mn>45</mn><mo>&#x2212;</mo><mn>15</mn><mo>)</mo></mtd></mtr><mtr><mtd><mi>h</mi><mo>=</mo><mn>10</mn><mi>m</mi></mtd></mtr></mtable></math>

 3. A ladder leaning against a wall makes an angle of 600 with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall.

Solution: Let AB be the wall and BC be the ladder.

Then, ∠ACB = 600 and BC = 19 m

Let AC = x metres

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi>A</mi><mi>C</mi></mrow><mrow><mi>B</mi><mi>C</mi></mrow></mfrac><mo>=</mo><mi>cos</mi><mo>&#xA0;</mo><msup><mn>60</mn><mn>0</mn></msup></mtd></mtr><mtr><mtd><mfrac><mi>x</mi><mn>19</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></mtd></mtr><mtr><mtd><mi>x</mi><mo>=</mo><mfrac><mn>19</mn><mn>2</mn></mfrac><mo>=</mo><mn>9.5</mn></mtd></mtr></mtable></math>

Therefore, distance of the foot of the ladder from the wall = 9.5 m

 4. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 300 with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60o. What is the distance between the base of the tower and the point P?

Solution: Let us draw a figure from given data.

  • Here one of the AB, AD and CD must have given.
  • So, the data is inadequate.

 5. Two ships are sailing in the sea on he to sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed rom the ships are 30o and 450 respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Solution:

Let us draw a figure from given data

Let AB be the light house and C and D be the positions of the ships.

Then, AB = 100 m, ∠ACB = 300 and ∠ADB = 450

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>A</mi><mi>C</mi></mrow></mfrac><mo>=</mo><mi>tan</mi><mo>&#xA0;</mo><msup><mn>30</mn><mn>0</mn></msup><mo>=</mo><mfrac><mn>1</mn><msqrt><mn>3</mn></msqrt></mfrac><mo>=&gt;</mo><mi>A</mi><mi>C</mi><mo>=</mo><mi>A</mi><mi>B</mi><mo>&#xD7;</mo><msqrt><mn>3</mn></msqrt><mo>=</mo><mn>100</mn><msqrt><mn>3</mn></msqrt><mi>m</mi></mtd></mtr><mtr><mtd><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>A</mi><mi>D</mi></mrow></mfrac><mo>=</mo><mi>tan</mi><mo>&#xA0;</mo><msup><mn>45</mn><mn>0</mn></msup><mo>=</mo><mn>1</mn><mo>=&gt;</mo><mi>A</mi><mi>D</mi><mo>=</mo><mi>A</mi><mi>B</mi><mo>=</mo><mn>100</mn><mi>m</mi></mtd></mtr><mtr><mtd><mtext>&#xA0;Therefore,&#xA0;</mtext><mi>C</mi><mi>D</mi><mo>=</mo><mo>(</mo><mi>A</mi><mi>C</mi><mo>+</mo><mi>A</mi><mi>D</mi><mo>)</mo><mo>=</mo><mo>(</mo><mn>100</mn><msqrt><mn>3</mn></msqrt><mo>+</mo><mn>100</mn><mo>)</mo><mi>m</mi><mo>=</mo><mn>100</mn><mo>(</mo><msqrt><mn>3</mn></msqrt><mo>+</mo><mn>1</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>100</mn><mo>&#xD7;</mo><mn>2.73</mn><mo>)</mo><mi>m</mi><mo>=</mo><mn>273</mn><mi>m</mi></mtd></mtr></mtable></math>


9 Similar Triangles
N/A

Two triangles are said to be similar if they have equal pair of corresponding angles and same ratio of corresponding sides.

They have same shape but their sizes are different.

Similar triangles are denoted by ~

In the above figure, these two triangles are said to be similar only if

  1. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Properties of similar triangles:

  1. Each pair of corresponding angles are equal
  2. The ratio of corresponding sides is same
  3. Both have same shape but sizes may be different

Similar triangles theorems:

Angle-Angle Similarity

Two triangles are said to be similar if any two angles of a triangle are equal to any two angles of another triangle.

In the above figure, if ∠A = ∠D and ∠B = ∠E then ΔABC ~ ΔDEF

Side-Angle-Side Similarity

If two sides of a triangle are proportional to the corresponding sides of another triangle, and the angle included by them in both the triangle are equal, then two triangles are said to be similar.

Thus, if ∠A = ∠D and AB/DE = AC/DF then ΔABC ~ΔDEF.

 From the congruency, AB/DE = BC/EF = AC/DF and ∠B = ∠E and ∠C = ∠F

Side-Side-Side Similarity

If all the three sides of a triangle are in proportion to the three sides of another triangle, then the two triangles are similar.

Thus, if AB/DE = BC/EF = AC/DF then ΔABC ~ΔDEF.

From this result, we can infer that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Properties of Similar Triangle

  1. Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides
  2.  Ratio of the areas of two similar triangles is equal to the ratio of the squares of corresponding altitudes and medians
  3. In two similar triangles ABC and PQR

Examples:

  1. Let ΔPQR, ΔDEF are similar triangles. If PQ = 6 and DE = 7, find A(ΔPQR)/A(ΔDEF)

Solution: A(ΔPQR)/A(ΔDEF) = PQ2/DE2 = 36/49

  1. If three sides of a triangle ABC and XYZ having coordinates A(-4, 0), B(0, 8), C(4, 0), X(-2,2), Y(0,6) and Z(2, 2). Prove that these triangles are similar.

Solution: Given Coordinates of the vertices

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mi>A</mi><mi>B</mi><mo>=</mo><msqrt><mfenced separators="|"><mrow><msup><mn>4</mn><mn>2</mn></msup><mo>+</mo><msup><mn>8</mn><mn>2</mn></msup></mrow></mfenced></msqrt><mo>=</mo><msqrt><mn>80</mn></msqrt></mtd></mtr><mtr><mtd><mi>B</mi><mi>C</mi><mo>=</mo><msqrt><mfenced separators="|"><mrow><msup><mn>4</mn><mn>2</mn></msup><mo>+</mo><msup><mn>8</mn><mn>2</mn></msup></mrow></mfenced></msqrt><mo>=</mo><msqrt><mn>80</mn></msqrt></mtd></mtr><mtr><mtd><mi>C</mi><mi>A</mi><mo>=</mo><msqrt><mfenced separators="|"><msup><mn>8</mn><mn>2</mn></msup></mfenced></msqrt><mo>=</mo><mn>8</mn></mtd></mtr><mtr><mtd><mi>X</mi><mi>Y</mi><mo>=</mo><msqrt><mfenced separators="|"><mrow><msup><mn>2</mn><mn>2</mn></msup><mo>+</mo><msup><mn>4</mn><mn>2</mn></msup></mrow></mfenced></msqrt><mo>=</mo><msqrt><mn>20</mn></msqrt></mtd></mtr><mtr><mtd><mi>Y</mi><mi>Z</mi><mo>=</mo><msqrt><mfenced separators="|"><mrow><msup><mn>2</mn><mn>2</mn></msup><mo>+</mo><msup><mn>4</mn><mn>2</mn></msup></mrow></mfenced></msqrt><mo>=</mo><msqrt><mn>20</mn></msqrt></mtd></mtr><mtr><mtd><mi>Z</mi><mi>X</mi><mo>=</mo><msqrt><mfenced separators="|"><msup><mn>4</mn><mn>2</mn></msup></mfenced></msqrt><mo>=</mo><mn>4</mn></mtd></mtr></mtable></math>

Calculate the ratios of the lengths of corresponding sides

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mfrac><mrow><mi>A</mi><mi>B</mi></mrow><mrow><mi>X</mi><mi>Y</mi></mrow></mfrac><mo>=</mo><mfrac><msqrt><mn>80</mn></msqrt><msqrt><mn>20</mn></msqrt></mfrac><mo>=</mo><mn>2</mn></mtd></mtr><mtr><mtd><mfrac><mrow><mi>B</mi><mi>C</mi></mrow><mrow><mi>Y</mi><mi>Z</mi></mrow></mfrac><mo>=</mo><mfrac><msqrt><mn>80</mn></msqrt><msqrt><mn>20</mn></msqrt></mfrac><mo>=</mo><mn>2</mn></mtd></mtr><mtr><mtd><mfrac><mrow><mi>C</mi><mi>A</mi></mrow><mrow><mi>Z</mi><mi>X</mi></mrow></mfrac><mo>=</mo><mfrac><mn>8</mn><mn>4</mn></mfrac><mo>=</mo><mn>2</mn></mtd></mtr></mtable></math>

From this we can tell that 

As, the lengths of the corresponding sides are proportional, we can say that triangles are similar.

  3. Find the length PQ in the triangle shown below

Solution:

∠PQR = ∠PST, ∠PRQ = ∠PTS and ∠P is the common angle => the two triangles ΔPQR and ΔPST are similar.

  2PQ = PQ + 6

  PQ = 6

  4. ABC and XYZ are two similar triangles with ZC = ZZ, whose areas are respectively 32 and 60.5 If XY = 7.7 cm, then what is AB equal to?

Solution: For similar triangles, ratio of areas is equal to the ratio of the squares of any two corresponding sides.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mi>A</mi><msup><mi>B</mi><mn>2</mn></msup><mo>=</mo><mn>31.36</mn></mtd></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mi>A</mi><mi>B</mi><mo>=</mo><mn>5.6</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

  5. ABC is a triangle right angled at A and a perpendicular AD is drawn on the hypotenuse BC. What is BCAD equal to?

Solution: In case of a right-angled triangle, if we draw a perpendicular from the vertex containing right angle to the hypotenuse,

we get three triangles, two smaller and one original and these three triangles are similar triangles.

So, ΔABC ~ ΔABD ~ ΔADC

Therefore, BC. AD = AB. AC


10 Area and Perimeter of Triangle
N/A

Area:

  • Total space enclosed by the boundary of a plane figure is called the area of that particular figure.
  • In another words, the area of a figure is a measure associated with the part of plane enclosed in the figure.
  • Area is measured in square unit like, square metre, square centimetre etc.

Perimeter:

  • Perimeter is the length of border around any enclosed plane.
  • Therefore, sum of the sides of a plane figure is the perimeter of that particular figure.
  • Unit of perimeter is same as the unit of sides like metre, centimetre etc.

Triangle:

  • A figure enclosed by three sides is known as a triangle.
  • A triangle has three angles with total sum of 180°.
  • Adjoining figure represents a triangle with sides, AB, BC, CA and ZA, ZB, ZC are the three angles of the triangle. Various types of triangles are discussed below.

Equilateral Triangle:

It has all three sides equal and each angle equal to 60o

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;(i)&#xA0;</mtext><mspace width="1em"/><mtext>&#xA0;Area&#xA0;</mtext><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>4</mn></mfrac><msup><mi>a</mi><mn>2</mn></msup><mo>=</mo><mn>0.433</mn><msup><mi>a</mi><mn>2</mn></msup><mspace linebreak="newline"/><mtext>&#xA0;(</mtext><mi>i</mi><mtext>i)&#xA0;&#xA0;Height&#xA0;</mtext><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>2</mn></mfrac><mi>a</mi><mo>=</mo><mn>0.866</mn><mi>a</mi><mspace linebreak="newline"/><mo>(</mo><mi>i</mi><mi>i</mi><mi>i</mi><mo>)</mo><mo>	</mo><mi>P</mi><mi>e</mi><mi>r</mi><mi>i</mi><mi>m</mi><mi>e</mi><mi>t</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>3</mn><mi>a</mi><mo>,</mo><mo>&#xA0;</mo><mi>w</mi><mi>h</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>&#xA0;</mo><mi>a</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>s</mi><mi>i</mi><mi>d</mi><mi>e</mi></math>

Isosceles Triangle:

It has any two sides and two angles equal and altitude drawn on non-equal side bisects it.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;(i)&#xA0;</mtext><mspace width="1em"/><mtext>&#xA0;Area&#xA0;</mtext><mo>=</mo><mfrac><mi>b</mi><mn>4</mn></mfrac><msqrt><mn>4</mn><msup><mi>a</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mi>b</mi><mn>2</mn></msup></msqrt><mspace linebreak="newline"/><mtext>&#xA0;(ii) Height&#xA0;</mtext><mo>=</mo><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mfenced separators="|"><mfrac><mi>b</mi><mn>2</mn></mfrac></mfenced><mn>2</mn></msup></msqrt><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msqrt><mn>4</mn><msup><mi>a</mi><mn>2</mn></msup><mo>&#x2212;</mo><msup><mi>b</mi><mn>2</mn></msup></msqrt><mspace linebreak="newline"/><mtext>&#xA0;(iii) Perimeter&#xA0;</mtext><mo>=</mo><mi>a</mi><mo>+</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>=</mo><mn>2</mn><mi>a</mi><mo>+</mo><mi>b</mi></math>

where, a = Each of two equal sides b = Third side

Scalene Triangle:

It has three unequal sides.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>(i)&#xA0;</mtext><mspace width="1em"/><mtext>&#xA0;Area&#xA0;</mtext><mo>=</mo><msqrt><mi>s</mi><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>a</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>b</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>c</mi><mo>)</mo></msqrt><mspace linebreak="newline"/><mtext>&#xA0;Where,&#xA0;</mtext><mi>s</mi><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi></mrow><mn>2</mn></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>a</mi><mo>,</mo><mi>b</mi><mtext>&#xA0;and&#xA0;</mtext><mi>c</mi><mtext>&#xA0;are sides of the triangle and&#xA0;</mtext><mi>s</mi><mtext>&#xA0;is semi-perimeter.</mtext><mspace linebreak="newline"/><mtext>&#xA0;(ii) Perimeter&#xA0;</mtext><mo>=</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi></math>

Right Angled Triangle:

 It is a triangle with one angle equal to 90°

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;(i)&#xA0;</mtext><mspace width="1em"/><mtext>&#xA0;Area&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Base&#xA0;</mtext><mo>&#xD7;</mo><mtext>&#xA0;Height&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mi>b</mi><mo>&#xD7;</mo><mi>p</mi><mspace linebreak="newline"/><mo>(</mo><mi>i</mi><mi>i</mi><mo>)</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>&#xA0;</mo><mo>	</mo><mi>P</mi><mi>e</mi><mi>r</mi><mi>i</mi><mi>m</mi><mi>e</mi><mi>t</mi><mi>e</mi><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>p</mi><mo>&#xA0;</mo><mo>+</mo><mo>&#xA0;</mo><mi>b</mi><mo>&#xA0;</mo><mo>+</mo><mo>&#xA0;</mo><mi>h</mi><mo>&#xA0;</mo><mspace linebreak="newline"/><mtext>&#xA0;(iii)&#xA0;</mtext><mspace width="1em"/><msup><mi>h</mi><mn>2</mn></msup><mo>=</mo><msup><mi>p</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>=&gt;</mo><mi>h</mi><mo>=</mo><msqrt><msup><mi>p</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup></msqrt></math>

where, p = Perpendicular, b =Base and h = Hypotenuse

h2 = p2 + b2

Isosceles Right-Angled Triangle:

It is a triangle with one angle equal to 90° and two sides containing the right angle are equal.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;(i)&#xA0;</mtext><mspace width="1em"/><mtext>&#xA0;Area&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><msup><mi>a</mi><mn>2</mn></msup><mspace linebreak="newline"/><mtext>&#xA0;(</mtext><mi>i</mi><mtext>i) Perimeter&#xA0;</mtext><mo>=</mo><mi>a</mi><mo>+</mo><mi>a</mi><mo>+</mo><mi>d</mi><mo>=</mo><mn>2</mn><mi>a</mi><mo>+</mo><mi>d</mi></math>

Properties of Triangle:

 1. Sum of any two sides of a triangle is greater than the third side

 2. Side opposite to the greatest angle will be the greatest and side opposite to the smallest angle will be the smallest

3. Among all the triangles that can be formed with a given perimeter, the equilateral triangle will have the maximum area

4. The lines joining the mid-points of sides of a triangle to the opposite vertex are called medians. In the given figure, AF, BE and CD are medians

5. The point where the three medians of a triangle meet are called centroid. In the given figure, O is the centroid. The centroid divides each of the median in the ratio of 2: 1

6. The median of a triangle divides it into two triangles of equal areas

7. The incentre and circumcentre lies at a point that divides the height in the ratio 2:1. i.e., the circumradius is always twice the median

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;8. Radius of an incircle of an equilateral triangle of side a is&#xA0;</mtext><mfrac><mi>a</mi><mrow><mn>2</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac><mtext>&#xA0;and area is&#xA0;</mtext><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>a</mi><mn>2</mn></msup></mrow><mn>12</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;9. Radius of circumcircle of an equilateral triangle of side a will be&#xA0;</mtext><mfrac><mi>a</mi><msqrt><mn>3</mn></msqrt></mfrac><mtext>&#xA0;and area is&#xA0;</mtext><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>a</mi><mn>2</mn></msup></mrow><mn>3</mn></mfrac></math>

10. The area of the triangle formed by joining the mid-points of the sides of a given triangle is <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>4</mn></mfrac><mtext>&#xA0;th</mtext></math> of the area of the given triangle

Examples:

  1. The perimeter of an equilateral triangle is 51 cm. Find its area.

 Sol. Given, perimeter of an equilateral triangle is 51 cm.

Let each side of triangle be a cm, then sum of sides = 51 cm

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>3</mn><mi>a</mi><mo>=</mo><mn>51</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>a</mi><mo>=</mo><mn>17</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Area of equilateral triangle&#xA0;</mtext><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>4</mn></mfrac><mo>(</mo><mn>17</mn><msup><mo>)</mo><mn>2</mn></msup><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>4</mn></mfrac><mo>(</mo><mn>17</mn><mo>&#xD7;</mo><mn>17</mn><mo>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>4</mn></mfrac><mo>(</mo><mn>289</mn><mo>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mn>289</mn><msqrt><mn>3</mn></msqrt></mrow><mn>4</mn></mfrac></mtd></mtr></mtable></math>

  2. The area of a right-angled triangle is 10 sq cm. If its perpendicular is equal to 20 cm, find its base.

Solution: Given, area = 10 sq cm

Perpendicular = 20 cm

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;We know that Area&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Base&#xA0;</mtext><mo>&#xD7;</mo><mtext>&#xA0;Perpendicular</mtext><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mspace width="1em"/><mn>10</mn><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mtext>&#xA0;Base&#xA0;</mtext><mo>&#xD7;</mo><mn>20</mn><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mspace width="1em"/><mtext>&#xA0;Base&#xA0;</mtext><mo>=</mo><mn>1</mn><mi>c</mi><mi>m</mi></math>

  3. Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of $ 5 per sqm.

Solution: Let’s check if it is right angled triangle

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><msup><mi>h</mi><mn>2</mn></msup><mo>=</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><msup><mi>p</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mrow><mo>(</mo><mn>25</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mo>(</mo><mn>15</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>20</mn><msup><mo>)</mo><mn>2</mn></msup></mrow></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mn>625</mn><mo>=</mo><mn>225</mn><mo>+</mo><mn>400</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>&#x21D2;</mo><mo>&#xA0;</mo><mn>625</mn><mo>=</mo><mn>625</mn></mtd></mtr></mtable></math>

Hence it is a right-angled triangle

So, the triangular filed is right angled at B.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Area of the field&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mi>A</mi><mi>B</mi><mo>&#xD7;</mo><mi>B</mi><mi>C</mi><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>15</mn><mo>&#xD7;</mo><mn>20</mn></mtd></mtr><mtr><mtd><mo>=</mo><mn>150</mn><msup><mi>m</mi><mn>2</mn></msup></mtd></mtr></mtable></math>

From this the cost of sowing seed is $ 5 per sq m.

Therefore, cost of sowing seed for 150 m2 = 150 x 5 = $ 750

  4. A ?DEF is formed by joining the mid-points of the sides of ?ABC. Similarly, a ?PQR is formed by joining the mid-points of the sides of the ?DEF. If the sides of the ?PQR are of lengths 1, 2 and 3 units, what is the perimeter of the ?ABC?

Solution: Given lengths are 1, 2 and 3 units

  • Perimeter of ?PQR = 1 + 2 + 3 = 6 units

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;Now, in&#xA0;</mtext><mi mathvariant="normal">&#x25B3;</mi><mi>D</mi><mi>E</mi><mi>F</mi><mo>,</mo><mfrac><mrow><mi>D</mi><mi>Q</mi></mrow><mrow><mi>D</mi><mi>F</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mi>P</mi><mi>Q</mi></mrow><mrow><mi>F</mi><mi>E</mi></mrow></mfrac></math>

  • So, 2PQ = FE
  • Similarly, DF = 2PR and DE = 2QR
  • Therefore, perimeter of ?DEF = 2 x 6 = 12 units
  • Similarly, Perimeter of ?ABC = 2 x perimeter of ?DEF

= 2 x 12 = 24 units

  5. Two isosceles triangles have equal vertical angles and their corresponding sides are in the ratio of 3: 7. What is the ratio of their areas?

Solution: Here, given that triangles are equiangular and hence they are similar.

Ratio of their areas = ratio of squares of corresponding sides

                                    = (3)2: (7)2

                                    = 9: 49


11 Area and perimeter of Quadrilateral
N/A

Quadrilateral:

  • A figure enclosed by four sides is called a quadrilateral.
  • A quadrilateral has four angles and sum of these angles is equal to 360°.
  • Various types of quadrilaterals are discussed below

Square:

It is a parallelogram with all 4 sides equal and each angle is equal to 90°.

  1. Area = (side)2 = a2
  2. Perimeter = 4 x side = 4a
  3. Diagonal(d) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mi>a</mi><msqrt><mn>2</mn></msqrt></math>

Where, a = side and d = diagonal

Properties of Square:

 1. Diagonal of a square are equal and bisect each other at right angles (90°)

2. All square are rhombus but converse is not true

3. Diagonal is the diameter of the circumscribing circle that circumscribes the square and circumradius <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>a</mi><msqrt><mn>2</mn></msqrt></mfrac></math>

4. If area of two squares is in the ratio of A1: A2 then ratio of their perimeter is given <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msub><mi>A</mi><mn>1</mn></msub></msqrt><mo>:</mo><msqrt><msub><mi>A</mi><mn>2</mn></msub></msqrt></math>

Rectangle:

  • It is a parallelogram with opposite sides equal and each angle is equal to 90°.

  1. Area = Length x Breadth = L x B
  2. Perimeter = 2 (L + B)
  3. <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Diagonal&#xA0;</mtext><mo>(</mo><mi>d</mi><mo>)</mo><mo>=</mo><msqrt><msup><mi>L</mi><mn>2</mn></msup><mo>+</mo><msup><mi>B</mi><mn>2</mn></msup></msqrt></math>

Properties of Rectangle:

 1. The diagonals of a rectangle are of equal lengths and they bisect each other

 2. All rectangles are parallelograms but reverse is not true.

Parallelogram:

 A quadrilateral, in which opposite sides are parallel is called a parallelogram.

  1. Area = Base x Height = b x h
  2. Perimeter = 2 (a + b)

Note: Opposite angles are equal in a parallelogram but they are not right angle.

Properties of Parallelogram:

  1. Diagonals of a parallelogram bisect each other.
  2. A parallelogram inscribed in a circle is a rectangle.
  3. A parallelogram circumscribed about a circle is a rhombus.
  4. Each diagonal of a parallelogram divides it into two triangles of equal area
  5. A parallelogram and a rectangle have equal areas if they are on the same base and between the same parallel lines
  6. The opposite angles of parallelogram are equal
  7. The sum of the squares of the four sides is equal to the sum of squares of diagonal

Trapezium

 It is a quadrilateral with any one pair of opposite sides parallel

  1. <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Area&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>[</mo><mtext>&#xA0;sum of the parallel sides&#xA0;</mtext><mi>x</mi><mtext>&#xA0;Height&#xA0;</mtext><mo>]</mo><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>(</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>)</mo><mi>h</mi></math> 

    Where a and b are parallel sides and h is the height or perpendicular distance between a and b.

  2. Perimeter =AB + BC + CD + AD 
  3.  Area of trapezium, when the lengths of parallel and non-parallel sides are given

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mrow><mi>a</mi><mo>+</mo><mi>b</mi></mrow><mi>k</mi></mfrac><msqrt><mi>s</mi><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>a</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>b</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>c</mi><mo>)</mo></msqrt><mfenced open="[" close="" separators="|"><mrow><mtext>&#xA0;where,&#xA0;</mtext><mi>k</mi><mo>=</mo><mi>a</mi><mo>&#x2212;</mo><mi>b</mi><mtext>&#xA0;and&#xA0;</mtext><mi>s</mi><mo>=</mo><mfrac><mrow><mi>k</mi><mo>+</mo><mi>c</mi><mo>+</mo><mi>d</mi></mrow><mn>2</mn></mfrac></mrow></mfenced></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;(iv)&#xA0;&#xA0;&#xA0;Perpendicular distance '&#xA0;</mtext><mi>h</mi><mtext>&#xA0;' between the two parallel sides&#xA0;</mtext><mo>=</mo><mfrac><mn>2</mn><mi>k</mi></mfrac><msqrt><mi>s</mi><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>a</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>b</mi><mo>)</mo><mo>(</mo><mi>s</mi><mo>&#x2212;</mo><mi>c</mi><mo>)</mo></msqrt></math>

Rhombus:

  • It is a parallelogram with all 4 sides equal.
  • The opposite angles in a rhombus are equal but they are not right angle.

Where, a = side, d1 and d2 are diagonals.

Properties of Rhombus:

  1. A rhombus has unequal diagonal and they bisect each other at right angles (90°)
  2. A rhombus may or may not be a square but all squares are rhombus
  3. All rhombus is parallelogram but reverse is not true

Examples:

  1. A square field has its area equal to 289 sq m. Find its side and perimeter.

Solution: Given area of square = 289 sq m.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">&#x21D2;</mo><mtext>&#xA0;Side&#xA0;</mtext><mi>x</mi><mtext>&#xA0;side&#xA0;</mtext><mo>=</mo><msup><mi>s</mi><mn>2</mn></msup><mo>=</mo><mn>289</mn><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mi>s</mi><mo>=</mo><msqrt><mn>289</mn></msqrt><mo>=</mo><mn>17</mn><mi>c</mi><mi>m</mi></math>

Perimeter = 4 x s = 4 x 17 = 68 cm

  2. The length and breadth of a rectangle are 6 cm and 4 cm, respectively. What will be its diagonal?

Solution: Given that, Length(L) = 6 cm and Breadth(B) = 4 cm and diagonal =?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>According to the formula,&#xA0;</mtext><mi>d</mi><mo>=</mo><msqrt><msup><mi>L</mi><mn>2</mn></msup><mo>+</mo><msup><mi>B</mi><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><msup><mn>6</mn><mn>2</mn></msup><mo>+</mo><msup><mn>4</mn><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mn>36</mn><mo>+</mo><mn>16</mn></msqrt><mo>=</mo><msqrt><mn>52</mn></msqrt><mo>=</mo><msqrt><mn>13</mn><mo>&#xD7;</mo><mn>4</mn></msqrt><mo>=</mo><mn>2</mn><msqrt><mn>13</mn></msqrt><mi>c</mi><mi>m</mi></math>

  3. The base of a parallelogram is thrice of its height. If the area of the parallelogram is 2187 sq cm, find its height.

Solution: Area of parallelogram = base x height

Let height = p and base = 3p

According to the question, 3p x p = 2187

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>3</mn><msup><mi>p</mi><mn>2</mn></msup><mo>=</mo><mn>2187</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mi>p</mi><mn>2</mn></msup><mo>=</mo><mfrac><mn>2187</mn><mn>3</mn></mfrac><mo>=</mo><mn>729</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>a</mi><mo>=</mo><mn>27</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

  4. The area of a trapezium is 384 cm. If its parallel sides are in the ratio 3: 5 and the perpendicular distance between them is 12 cm, the smaller of the parallel sides is

Solution: Given parallel sides are in the ratio 3: 5

Let the sides of trapezium be 5x and 3x respectively.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#x2217;</mo><mo>(</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>x</mi><mo>)</mo><mo>&#x2217;</mo><mn>12</mn><mo>=</mo><mn>384</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>8</mn><mi>x</mi><mo>=</mo><mfrac><mrow><mn>384</mn><mi>x</mi><mn>2</mn></mrow><mn>12</mn></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mn>64</mn><mn>8</mn></mfrac><mo>=</mo><mn>8</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

Length of smaller of the parallel sides = 8 x 3 = 24 cm

  5. If the diagonals of a rhombus are 4.8 cm and 1.4 cm, then what is the perimeter of the rhombus?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: perimeter of rhombus&#xA0;</mtext><mo>=</mo><mn>2</mn><msqrt><msubsup><mi>d</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>d</mi><mn>2</mn><mn>2</mn></msubsup></msqrt><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mn>2</mn><msqrt><mo>(</mo><mn>4.8</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>1.4</mn><msup><mo>)</mo><mn>2</mn></msup></msqrt></mtd></mtr><mtr><mtd><mo>=</mo><mn>2</mn><msqrt><mn>23.04</mn><mo>+</mo><mn>1.96</mn></msqrt></mtd></mtr><mtr><mtd><mo>=</mo><mn>2</mn><msqrt><mn>25</mn></msqrt></mtd></mtr><mtr><mtd><mo>=</mo><mn>2</mn><mo>&#xD7;</mo><mn>5</mn></mtd></mtr></mtable></math>

perimeter of rhombus = 10 cm


12 Area and Circumference of Circle
N/A

It is a plane figure enclosed by a line on which every point is equidistant from a fixed point (centre) inside the curve

Sector: Sector is a part of area of circle between two radii and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3B8;</mi></math> is the angle enclosed between two radii

Semi-circle:

A circle when separated into two parts along its diameter, then each half part is known as semicircle.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>(i)&#xA0;&#xA0;Area of semicircle&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac><mspace linebreak="newline"/><mo>(</mo><mi>i</mi><mi>i</mi><mo>)</mo><mo>	</mo><mtext>&#xA0;Perimeter&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mo>+</mo><mn>2</mn><mi>r</mi><mspace linebreak="newline"/><mi>H</mi><mi>e</mi><mi>r</mi><mi>e</mi><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>r</mi><mi>a</mi><mi>d</mi><mi>i</mi><mi>u</mi><mi>s</mi><mo>&#xA0;</mo><mi>o</mi><mi>f</mi><mo>&#xA0;</mo><mi>s</mi><mi>e</mi><mi>m</mi><mi>i</mi><mo>-</mo><mi>c</mi><mi>i</mi><mi>r</mi><mi>c</mi><mi>l</mi><mi>e</mi></math>

Circular Ring:

(i)  Area = R2 – r2

(ii)  Difference in circumference of both rings = <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>(</mo><mn>2</mn><mi>&#x3C0;</mi><mi>R</mi><mo>&#x2212;</mo><mn>2</mn><mi>&#x3C0;</mi><mi>r</mi><mo>)</mo><mo>=</mo><mn>2</mn><mi>&#x3C0;</mi><mo>(</mo><mi>R</mi><mo>&#x2212;</mo><mi>r</mi><mo>)</mo></math>

Examples:

  1. If the area of a semi-circle be 77 sq m, find its perimeter?

Solution: According to the question, Area of semi-circle = 77 m

\<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac><mo>=</mo><mn>77</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mfrac><mrow><mn>77</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>7</mn></mrow><mn>22</mn></mfrac></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>r</mi><mo>=</mo><mn>7</mn><mi>m</mi></mtd></mtr></mtable></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Circumference of semi-circle&#xA0;</mtext><mo>=</mo><mi>&#x3C0;</mi><mi>r</mi><mo>+</mo><mn>2</mn><mi>r</mi><mo>=</mo><mo>(</mo><mi>&#x3C0;</mi><mo>+</mo><mn>2</mn><mo>)</mo><mi>r</mi><mo>=</mo><mfenced separators="|"><mrow><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>+</mo><mn>2</mn></mrow></mfenced><mo>&#xD7;</mo><mn>7</mn><mo>=</mo><mn>36</mn><mi>m</mi></math>

  1. The area of a circle is increased by 22 sq cm when its radius is increased by 1 cm. Find the original radius of the circle.

 

Solution: Let original radius be r.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Then, according to the question,&#xA0;</mtext><mi>&#x3C0;</mi><mo>(</mo><mi>r</mi><mo>+</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup><mo>&#x2212;</mo><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup><mo>=</mo><mn>22</mn></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>&#x3C0;</mi><mo>&#xD7;</mo><mfenced open="[" close="]" separators="|"><mrow><mo>(</mo><mi>r</mi><mo>+</mo><mn>1</mn><msup><mo>)</mo><mn>2</mn></msup><mo>&#x2212;</mo><msup><mi>r</mi><mn>2</mn></msup></mrow></mfenced><mo>=</mo><mn>22</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mn>22</mn><mn>7</mn></mfrac><mo>&#xD7;</mo><mo>(</mo><mi>r</mi><mo>+</mo><mn>1</mn><mo>+</mo><mi>r</mi><mo>)</mo><mo>(</mo><mi>r</mi><mo>+</mo><mn>1</mn><mo>&#x2212;</mo><mi>r</mi><mo>)</mo><mo>=</mo><mn>22</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>2</mn><mi>r</mi><mo>+</mo><mn>1</mn><mo>=</mo><mn>7</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>2</mn><mi>r</mi><mo>=</mo><mn>6</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>r</mi><mo>=</mo><mn>3</mn><mi>c</mi><mi>m</mi></mtd></mtr></mtable></math>

Therefore, original radius of the circle = 3 cm

 

  3. The ratio of the areas of the circumcircle and the incircle of a square is

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution: Ratio of the area of the circumcircle and incircle of a square&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><mo>(</mo><mtext>&#xA0;Diagonal&#xA0;</mtext><msup><mo>)</mo><mn>2</mn></msup></mrow><mrow><mi>&#x3C0;</mi><mo>(</mo><mtext>&#xA0;side&#xA0;</mtext><msup><mo>)</mo><mn>2</mn></msup></mrow></mfrac><mspace linebreak="newline"/><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mrow><mo>(</mo><mi>sid</mi><mi>e</mi><mo>&#xA0;</mo><mi>x</mi><msqrt><mn>2</mn></msqrt><msup><mo>)</mo><mn>2</mn></msup></mrow><mrow><mo>(</mo><mtext>&#xA0;side&#xA0;</mtext><msup><mo>)</mo><mn>2</mn></msup></mrow></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>2</mn><mn>1</mn></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mn>2</mn><mo>:</mo><mn>1</mn></mtd></mtr></mtable></math>

 

  4. AB and CD are two diameters of a circle of radius r and they are mutually perpendicular. What is the ratio of the area of the circle to the area of the triangle ACD?

Solution:

 

Let’s, draw a figure from the given data

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mtext>&#xA0;Required ratio&#xA0;</mtext><mo>=</mo><mfrac><mtext>&#xA0;Area of circle&#xA0;</mtext><mrow><mtext>&#xA0;Area of triangle&#xA0;</mtext><mi>A</mi><mi>C</mi><mi>D</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup></mrow><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>2</mn><mi>r</mi><mo>&#xD7;</mo><mi>r</mi></mrow></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mi>&#x3C0;</mi><mo>:</mo><mn>1</mn></mtd></mtr></mtable></math>

 

  5. The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle

 

Solution: Let’s draw a figure from given data

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Area of the triangle&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mn>8</mn><mo>&#xD7;</mo><mn>4</mn><mo>=</mo><mn>16</mn><mi>s</mi><mi>q</mi><mi>c</mi><mi>m</mi><mspace linebreak="newline"/><mtext>&#xA0;Area of semi-circle&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>&#x3C0;</mi><msup><mi>r</mi><mn>2</mn></msup></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mn>16</mn><mi>&#x3C0;</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mn>8</mn><mi>&#x3C0;</mi><mspace linebreak="newline"/><mtext>&#xA0;Required area&#xA0;</mtext><mo>=</mo><mn>8</mn><mi>&#x3C0;</mi><mo>&#x2212;</mo><mn>16</mn><mo>=</mo><mn>8</mn><mo>(</mo><mi>&#x3C0;</mi><mo>&#x2212;</mo><mn>2</mn><mo>)</mo><mi>s</mi><mi>q</mi><mi>c</mi><mi>m</mi></math>

 


13 Volume and Surface Area of Cube and Cuboid
N/A

 

  • Volume and surface area are related to solids or hollow bodies.
  • These bodies occupy space and have usually three dimensions length, breadth and height.

Volume:

 

  • Space occupied by an object is called the 'volume' of that particular object.

 

  • It is always measured in cube unit like cubic meter, cubic centimetre etc

 

Surface Area:

  • Surface area of a solid body is the area of all of its surfaces together.
  • Surface area is measured in square unit like square centimetre, square metre etc.

 

Cube:

 

  • A solid body having 6 equal faces with equal length, breadth and height is called a cube.
  • In fact, each face of a cube is a square

  • Volume of the cube = a3
  • Lateral surface area of the cube = 4a2
  • Total surface area of the cube = 6a2
  • Diagonal of the cube <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><mn>3</mn></msqrt><mi>a</mi></math>

where, a is Side (edge) of the cube

Cuboid:

  • A rectangular solid body having 6 rectangular faces is called a cuboid.

  • Volume of the cuboid = bh
  • Lateral surface area of a cuboid = 2 (l + b) h
  • Total surface area of the cuboid = 2 (lb + bh + lh)
  • Diagonal of cuboid <math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>l</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><msup><mi>h</mi><mn>2</mn></msup></msqrt></math>

where, = Length, b - Breadth and h – Height

Some other cube or cuboidal shaped object are as follows.

Room

 

  • A rectangular room has 4 walls (surfaces) and opposite walls have equal areas.

  • Total area of walls = 2 (l + b) x h
  • Total volume of the room = lbh

 

where, = Length, b = Breadth and h = Height

Box

  • A box has its shape like cube or cuboid.
  • The amount that a box can hold or contain, is called the capacity of the box.
  • In fact, capacity is the internal volume

  • Surface area of an open box = 2 (length + breadth) x Height + Length x Breadth

        = 2 x (l+ b) x h + l x b

  • Capacity of box = (l- 2t) (b – 2t) (h – 2t) where, t = Thickness of the box
  • Volume of the material of the box = External volume - Internal volume (capacity) = lbh-{l -2t)(b - 2t) {h - 2t) where, = Length, b = Breadth and h = Height

 

Note For calculation of any of the parameter, length, breadth and height should be in same unit.

Examples:

  1. Find the volume and surface area of a cuboid 18 m long, 14 m broad and 7 m high.

Solution:

  • Volume of the cuboid = Length x Breadth x Height
  • Here, length = 18 m, breadth = 14 m and height = 7 m = 18 X 14 X 7= 1764m3
  • Surface area = 2(lb + bh + lh) = 2(18 x 14 + 14 x 7 + 18 x 7) = 2(252+ 98+ 126) = (2 x 476) = 952 sq m

 

  2. A wooden box measures 10 cm x 6 cm x 5 cm. Thickness of wood is 2 cm. Find the volume of the wood required to make the box.

Solution:

External volume = 10 x 6 x 5 = 300 cm3

Internal volume (l - 2t) (b - 2t) (h - 2t) = (10 - 4) X (6 - 4) X (5 - 4) = 6 x 2 x 1= 12 cm3

Volume of the wood = External volume - Internal volume = 300 - 12 = 288 cm3

 

  3. The surface area of a cube is 486 sq cm. Find its volume.

Sol. Let the edge of cube = a

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>According to the question,&#xA0;</mtext><mn>6</mn><msup><mi>a</mi><mn>2</mn></msup><mo>=</mo><mn>486</mn><mo>=&gt;</mo><msup><mi>a</mi><mn>2</mn></msup><mo>=</mo><mfrac><mn>480</mn><mn>6</mn></mfrac><mo>=</mo><mn>81</mn><mspace linebreak="newline"/><mo stretchy="false">&#x21D2;</mo><mtext>&#xA0;Therefore&#xA0;</mtext><mi>a</mi><mo>=</mo><msqrt><mn>81</mn></msqrt><mo>=</mo><mn>9</mn><mi>c</mi><mi>m</mi></math>

Volume = a3 = 93 = 9 x 9 x 9 = 729 cm3

 

  4. Three cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube. Find half of the surface area of the new cube.

Solution: Volume (new cube) = (13 + 63 + 83) = 729 cm3

ð a3 = 729

ð a = 9 cm

 

  5. A metal box measures 20 cm x 12 cm x 5 cm. Thickness of the metal is 1 cm. Find the volume of the metal required to make the box.

Solution: External volume = 20 x 12 x 5 = 1200 cm3

Internal volume = (20 – 2) x (12 – 2) x (5 – 2) = 18 x 10 x 3 = 540 cm3

 

Therefore, volume of the metal = External volume – internal volume = 1200 – 540 = 660 cm3

 

 

 


1 Word Problems Based on Numbers
N/A

  • Numbers play an important role in our day-to-day life.
  • Puzzles based on these numbers are known as word problems.
  •  To solve such problems, you have to extract the information correctly and form equations based on given information.
  • The equations formed can be single variable, multi variables, linear, quadratic etc., depending on the type of problem asked.

Types of Word Problems Based on Numbers:

There are basically following types of questions that are asked on word problem on numbers.

 Type 1: Based on Operation with Numbers:

These types of questions include the operations like subtraction, addition, multiplication, division of number with other number, calculation of average of consecutive numbers, calculation of parts of a number, operation on even or odd numbers, calculation of sum or difference of reciprocal of numbers etc

  • Consecutive natural numbers can be assumed as x - 2, x -1, x, x +1, x + 2......
  • Consecutive even/odd natural numbers can be assumed x-3, x-1, x + 1, x + 3.......
  • If sum of two numbers is given as S, then take one number as x and other as (S - x)
  • If difference of two numbers is d, then take one number as x and other as (x + d) or (x - d)

Type 2: Based on Formation of Number with Digits:

These types of questions include formation of a number with digits and its difference with reciprocal of the same number, calculation of a number, if a number is added or subtracted to it. The digits get reversed etc.

  • A two-digit number with x as unit digit and y as ten's digit is formed as (10y + x) and if the digits are reversed, then number is represented as (10x + y).
  • A three-digit number with x as unit digit, y as ten's digit and z as hundred's digit is formed as (100z +10y + x).

Type 3: Question Regarding Calculation of Heads and Feet of Animals:

If a group of animals having either two feet (like ducks, hens etc) or four feet (like horses, cows etc) is there and total number of heads in the group are Hand number of feet of these animals are L, then Number of animals with four feet =

Number of animals with two feet = total number of heads – total number of four feeted animals

Examples:

  1. A number is 25 more than its two-fifth. Find the number.

Solution. Let the number be y.

Then, according to the question,  

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mfrac><mrow><mn>3</mn><mi>y</mi></mrow><mn>5</mn></mfrac><mo>=</mo><mn>25</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>y</mi><mo>=</mo><mfrac><mn>125</mn><mn>3</mn></mfrac></mtd></mtr></mtable></math>

 2. The sum of the digits of a two-digit number is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number.

Solution: Suppose the two-digit number = 10x + y and

 sum of the digits = x + y = 10…(i)

After reversing the digits, the new number = 10y + x

According to the question, (10x + y) - (10y + x) = 18

9x - 9y = 18

 x - y = 2 ...(ii)

On adding Eqs. (i) and (ii), we get

x + y = 10

x - y = 2

2x = 12

x = 6

On placing the value of x in Eq(i), we get y = 4

Therefore, original number = 10x + y = 10 x 6 + 4 = 64

 3. In a park, there are some cows and some ducks. If total number of heads in the park are 68 and number of their legs together is 198, then find the number of ducks in the park.

Solution: Given L = 198 and H = 68

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#x2219;&#xA0;&#xA0;&#xA0;So, number of cows&#xA0;</mtext><mo>=</mo><mfrac><mrow><mi>L</mi><mo>&#x2212;</mo><mn>2</mn><mi>H</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mn>198</mn><mo>&#x2212;</mo><mo>(</mo><mn>2</mn><mo>&#xD7;</mo><mn>68</mn><mo>)</mo></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mn>198</mn><mo>&#x2212;</mo><mn>136</mn></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mn>62</mn><mn>2</mn></mfrac><mo>=</mo><mn>31</mn></math>

  • Therefore, number of ducks = number of heads – number of cows = 68 – 31 = 37

 4. The sum of five consecutive odd numbers is equal to 175. What is the sum of the second largest number and the square of the smallest number amongst them together?

Solution: Sum of five consecutive odd numbers

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>+</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo>+</mo><mi>x</mi><mo>+</mo><mn>4</mn><mo>+</mo><mi>x</mi><mo>+</mo><mn>6</mn><mo>+</mo><mi>x</mi><mo>+</mo><mn>8</mn><mo>=</mo><mn>175</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>20</mn><mo>=</mo><mn>175</mn></mtd></mtr><mtr><mtd><mo stretchy="false">&#x21D2;</mo><mi>x</mi><mo>=</mo><mfrac><mrow><mn>175</mn><mo>&#x2212;</mo><mn>20</mn></mrow><mn>5</mn></mfrac><mo>=</mo><mn>31</mn></mtd></mtr></mtable></math>

Sum of the second largest and square of smallest one = (31 + 6) + (31)2 = 37 + 961 = 998

 5. A chocolate has 12 equal pieces. Peter gave 1/4th of it to Jack, 1/3rd of it to John and 1/6th of it to Fiza. The number of pieces of chocolate left with Peter is

Solution: The number of pieces of chocolate left with Peter = 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mfenced separators="|"><mfrac><mrow><mn>3</mn><mo>+</mo><mn>4</mn><mo>+</mo><mn>2</mn></mrow><mn>12</mn></mfrac></mfenced></mtd></mtr><mtr><mtd><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>9</mn><mn>12</mn></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mn>12</mn><mo>&#x2212;</mo><mn>9</mn></mrow><mn>6</mn></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mn>3</mn><mn>12</mn></mfrac></mtd></mtr></mtable></math>

Hence, number of pieces of chocolate left with Peter is 3


2 Simplification
N/A

A complex arithmetical expression can be converted into a simple expression by simplification. VBODMAS' Rule

To simplify arithmetic expressions, which involve various operations like brackets, multiplication, addition, etc; a particular sequence of the operations has to be followed.

The operations have to be carried out in the order, in which they appear in the word VBODMAS, where different letters of the word stand for following operations.

Order of removing brackets

First Small brackets (Circular brackets) ‘()'

 Second Middle brackets (Curly brackets) ’{}'

Third Square brackets (Big brackets) '[]'

V = Vinculum or Bar

B = Bracket

O = Of

D = Division

M = Multiplication

A = Addition

S = Subtraction

Order of above-mentioned operations is same as the order of letters in the 'VBODMAS' from left to right as

The order will be as follows:

First -> Vinculum bracket is solved

Second -> Brackets are to be solved in order given above, [first, then second, the third]

Third -> Operation of ‘Of’ is done,

Fourth -> Operation of division is performed,

Fifth -> Operation of multiplication is performed,

Sixth -> Operation of addition is performed,

Seventh -> Operation of subtraction is performed.

Note:

Absolute value of a real number

If m is a real number, then its absolute value is defined as

Example: |3| = 3 and |-3| = -(-3) = 3

Basic Formulae:

  1. (a+ b) (a -b) = (a2-b2)
  2. (a + b)2 = a2+ b2+ 2ab
  3. (a - b)2 = a2+ b2- 2ab
  4. (a + b+ c)2 = a2+ b2+ c2+ 2ab + 2bc +2ca
  5. (a + b)3 = a3 + b3 + 3ab (a + b)
  6. (a - b)3 = a3 - b3 - 3ab (a - b)
  7. (a3 + b3) = (a + b) (a2- ab+ b2)
  8. (a3 - b3) = (a - b) (a2 + ab+ b2)
  9. a3 + b3 + c3 - 3abc = (a + b+ c) (a2+ b2+ c2- ab – bc - ca)
  10. when a + b +c = 0, then a3 + b3 + c3 = 3abc
  11. (a + b)2-(a - b)2 = 4ab
  12. (a + b)2 + (a - b)2 = 2(a2+ b2)

Examples:

  1. Find the value of a3 + b3 + c3 -3abc, when a = 225, b = 226, c = 227

Solution: We know that,

a3 + b3 + c3 - 3abc = (a + b + c)  [(a - b)2 + (b – c)2 + (c – a)2]

                                               = (225 + 226 + 227)  [ 1 + 1 + 4]

                                              = 678 x 3

                                             = 2034

 2. Simplify <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2212;</mo><mo>[</mo><mn>9</mn><mo>&#x2212;</mo><mo>{</mo><mn>18</mn><mo>&#x2212;</mo><mo>(</mo><mn>15</mn><mo>&#x2212;</mo><mover accent="true"><mrow><mn>12</mn><mo>&#x2212;</mo><mn>9</mn></mrow><mo accent="false">&#xAF;</mo></mover><mo>)</mo><mo>}</mo><mo>]</mo></math>

Solution: Given expression <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>6</mn><mo>&#x2212;</mo><mo>[</mo><mn>9</mn><mo>&#x2212;</mo><mo>{</mo><mn>18</mn><mo>&#x2212;</mo><mo>(</mo><mn>15</mn><mo>&#x2212;</mo><mover accent="true"><mrow><mn>12</mn><mo>&#x2212;</mo><mn>9</mn></mrow><mo accent="false">&#xAF;</mo></mover><mo>)</mo><mo>}</mo><mo>]</mo></math>

                                                = 6 – [9 – {18 – (15 – 12 + 9)}]

                                                = 6 – [9 – {18 – 12}]

                                                = 6 – [9 – 6]

                                                = 6 – 3

                                                = 3

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;3. If&#xA0;</mtext><mi>x</mi><mo>+</mo><mfrac><mn>1</mn><mi>x</mi></mfrac><mo>=</mo><mn>2</mn><mtext>, then&#xA0;</mtext><mfrac><mrow><mn>2</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>2</mn></mrow><mrow><mn>3</mn><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>3</mn></mrow></mfrac><mo>=</mo><mtext>&#xA0;?&#xA0;</mtext></math>

Solution: Given expression 

On dividing numerator and denominator by x,

we get

 4. A man divides $8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew and each son receives five times as much as each nephew, how much does each daughter receive?

Solution: Let the share of each nephew be $ p.

Then, share of each daughter = $ 4p

Share of each son = $ 5p

So, 5 x 5p + 4 x 4p + 2 x p = 8600

  25p + 16p + 2p = 8600

  43p = 8600

  p = 200

So, share of each daughter = $ (4 x 200) = $ 800

 5. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets $ 2.40 per hour for regular work and $ 3.20 per hours for overtime. If he earns $ 432 in 4 weeks, then how many hours does he work for?

Solution: Suppose the man works overtime for t hours.

Now, working hours in 4 weeks = 5 x 8 x 4 = 160

 160 x 2.40 + t x 3.20 = 432

  3.20t = 432 – 384

  t = 15

Therefore, total hours of work = (160 + 15) = 175.

 


3 Co-Prime Numbers
N/A

Co-prime numbers are the numbers having common factor as only 1. A pair of numbers whose highest common factor is 1 are said to be co-prime.

Let p and q are two positive integers, if they have 1 as their only common factor and thus HCF (x, y) = 1 they are called as Co-prime numbers.

  • Co-prime numbers need not be prime numbers always.
  • For example, {4 and 7}, {21 and 22} are co-prime numbers.

Finding co-prime numbers:

Let us take an example of {5, 12}

5 is a prime number -> 5 x 1

12 is not a prime number-> 2 x 2 x 3 x 1

Here 5 and 12 have only 1 as highest common factor

So, they are co-prime numbers

Important Points to remember:

  • 1 is co-prime with every number
  • Two prime numbers are always co-prime to each other because prime number has only 2 factors (number itself and 1)
  • Any two successive numbers are always co-prime
  • The sum of any two co-prime numbers is always coprime with their product
  • The numbers whose unit digit is 0 and 5 can never be co-prime to each other
  • Two even numbers cannot be co-prime to each other because they have a common factor two

Examples:

  1. Check if 5, 7 and 9 are co-prime numbers?

Solution: Here 5 and 7 are prime numbers -> 5 x 1 = 5 and 7 x 1 = 7

                                    9 can be written as 3 x 3 x 1 = 9

The common factor of all the three is 1

So, they are co-prime numbers

  1. Show that 850 and 1000 are not co-prime numbers

Solution: here 850 -> 2 x 425

                               -> 2 x 5 x 85

                               -> 2 x 2 x 5 x 5 x 17

                        1000 -> 2 x 500

                                  -> 2 x 2 x 250

                                  -> 2 x 2 x 2 x 5 x 5 x 5

Here we have common factors as 2 and 5

So, they are not co-prime numbers

  1. Write any one co-prime number of 123 other than 122 and 124

Solution: here 123 -> 3 x 41 x 1

Given any one co-prime number of 123

So, 125 -> 5 x 5 x 5 x 1

Here 123 and 125 does not highest common factor other than 1.

So, 123 and 125 are co-prime numbers

  1. 143 and 146 are co-prime numbers. True or False?

Solution:

143 -> 11 x 13 x 1

146 -> 2 x 73 x 1

No, highest common factor other than 1

True, they are co-prime numbers

  1. 11, 12 are co-prime numbers, check if sum of the co-prime numbers is co-prime with the product of the co-prime numbers.

Solution: sum = 11 + 12 = 23, product = 11 x 12 = 132

Here sum 23 is a prime number -> 23 x 1

Product 132 -> 2 x 2 x 3 x 11 x 1

23 and 132 does not have any common factor other than 1.

So, sum of the co-prime numbers is co-prime with the product of the co-prime numbers

 


4 Numbers
N/A

Theory:

 i.            Types of Numbers:

                                    I.            Natural numbers: Natural numbers are counting numbers. For example N = {1,2, 3...}.  All-natural numbers are positive. Zero is not a natural number.

                                  II.            Whole Numbers: All-natural numbers and zero form the set of whole numbers. For example W = {0,1,2,3,...}

                                III.            Integers: Whole numbers and negative numbers form the set of integers. For example / = {...,-4,-3,-2,-1,0,1,2,3,4,...}. Integers are of two types.

·         Positive Integers: Natural numbers are called as positive integers. For example I + = {1,2,3,4,...}

·         Negative Integers: Negative of natural numbers are called as negative integers. For example I~ ={-1,-2,-3,-4,...}. 

                               IV.            Even Numbers: A counting number which is divisible by 2, is called an even number. For example: 2, 4, 6, 8, 10, 12, ... etc.

                                 V.            Odd Numbers: A counting number which is not divisible by 2, is known as an odd number. For example: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ... etc.

                               VI.            Prime Numbers: A counting number is called a prime number when it is exactly divisible by, 1 and itself. For example: 2, 3, 5, 7, 11, 13, ... etc.

                             VII.            Composite Numbers: Composite numbers are non-prime natural numbers. They must have atleast one factor apart from 1 and itself. For example: 4, 6, 8, 9, etc.

                           VIII.            Coprimes: Two natural numbers are said to be coprimes, if their HCF is 1. For example (7, 9), (15, 16). Coprime numbers may or may not be prime

                                IX.            Rational numbers: A number that can be expressed as p/q is called a rational number, where p and q are inteqers and q is not equal to zero

                                  X.            Irrational numbers:The numbers that cannot be expressed in the form of p/q are called irrational numbers, where p and q are integers and q is not equal to 0.

                                XI.            Real numbers: Real numbers include rational and irrational numbers both. For example: 

ii.            Divisibility Tests:

                                    I.            Divisibility by 2: When the last digit of a number is either 0 or even, then the number is divisible by 2. For example 12, 86, 472, 520, 1000 etc., are divisible by 2.

                                  II.            Divisibility by 3: When the sum of the digits of a number is divisible by 3, then the number is divisible by 3. For example (i) 1233: 1 + 2 + 3 + 3 = 9, which is divisible by 3, so 1233 must be divisible by 3. (ii) 156:1 + 5 + 6 = 12, which is divisible by 3, so 156 must be divisible by 3.

                                III.            Divisibility by 4: When the number made by last two-digits of a number is divisible by 4, then that particular number is divisible by 4. Apart from this, the number having two or more zeroes at the end, is also divisible by 4. For example (i) 6428 is divisible by 4 as the number made by its last two digits i.e., 28 is divisible by 4. (ii) The numbers 4300, 153000, 9530000 etc., are divisible by 4 as they have two or more zeroes at the end.

                               IV.            Divisibility by 5: Numbers having 0 or 5 at the end are divisible by 5. For example 45, 4350, 135, 14850 etc., are divisible by 5 as they have 0 or 5 at the end.

                                 V.            Divisibility by 6: When a number is divisible by both 3 and 2, then that particular number is divisible by 6 also. For example 18, 36, 720, 1440 etc., are divisible by 6 as they are divisible by both 3 and 2.

                               VI.            Divisibility by 7: A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either zero or a multiple of 7. For example 658 is divisible by 7 because 65 - 2 X 8 = 65 - 16 = 49. As 49 is divisible by 7, the number 658 is also divisible by 7

                             VII.            Divisibility by 8: When the number made by last three digits of a number is divisible by 8, then the number is also divisible by 8. Apart from this, if the last three or more digits of a number are zeroes, then the number is divisible by 8. For example (i) 2256 As 256 (the last three digits of 2256) is divisible by 8, therefore 2256 is also divisible by 8. (ii) 4362000 As 4362000 has three zeroes at the end. Therefore it will definitely divisible by 8.

                           VIII.            Divisibility by 9:When the sum of all the digits of a number is divisible by 9, then the number is also divisible by 9. For example (i) 936819 9+3 + 6 + 8 + 1 + 9= 36 which is divisible by 9. Therefore, 936819 is also divisible by 9. (ii) 4356 4 + 3 + 5 + 6 = 18 which is divisible by 9. Therefore, 4356 is also divisible by 9.

                                IX.            Divisibility by 10: When a number ends with zero, then it is divisible by 10. For example 20, 40, 150, 123450, 478970 etc., are divisible by 10 as these all end with zero.

                                  X.            Divisibility by 11:When the sums of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11. For example (i) 2865423: Let us see Sum of digits at odd places (A) = 2 + 6+4 + 3 = 15 Sum of digits at even places (B) = 8 + 5 + 2 = 15 =>A = B Hence, 2865423 is divisible by 11. (ii) 217382 Let us see Sum of digits at odd places (A) = 2 + 7 + 8 = 17 Sum of digits at even places (B) = 1 + 3 + 2 = 6 A- B = 17-6 = 11 Clearly, 217382 is divisible by 11.

                                XI.            Divisibility by 12: A number which is divisible by both 4 and 3 is also divisible by 12. For example 2244 is divisible by both 3 and 4. Therefore, it is divisible by 12 also.

                              XII.            Divisibility by 14: A number which is divisible by both 7 and 2 is also divisible by 14. For example 1232 is divisible by both 7 and 2. Therefore, it is divisible by 14 also.

                            XIII.            Divisibility by 15: A number which is divisible by both 5 and 3 is divisible by 15 also. For example 1275 is divisible by both 5 and 3. Therefore, it is divisible by 15 also.

                           XIV.            Divisibility by 16: A number is divisible by 16 when the number made by its last 4-digits is divisible by 16. For example 126304 is divisible by 16 as the number made by its last 4-digits i.e., 6304 is divisible by 16.

                             XV.            Divisibility by 18: A number is divisible by 18 when it is even and divisible by 9. For example 936198 is divisible by 18 as it is even and divisible by 9.

                           XVI.            Divisibility by 25: A number is divisible by 25 when its last 2-digits are either zero or divisible by 25. For example 500, 1275, 13550 are divisible by 25 as last 2-digits of these numbers are either zero or divisible by 25.

                         XVII.            Divisibility by 125: A number is divisible by 125 when the number made by its last 3-digits is divisible by 125. For example 630125 is divisible by 125 as the number made by its last 3-digits are divisible by 125.

 

Basic Formulae:

  1. (a+ b) (a -b) = (a2-b2)
  2. (a + b)2 = a2+ b2+ 2ab
  3. (a - b)2 = a2+ b2- 2ab
  4. (a + b+ c)2 = a2+ b2+ c2+ 2ab + 2bc +2ca
  5. (a3 + b3) = (a + b) (a2- ab+ b2)
  6. (a3 - b3) = (a - b) (a2 + ab+ b2)
  7. a3 + b3 + c3 - 3abc = (a + b+ c) (a2+ b2+ c2- ab – bc - ca)
  8. when a + b +c = 0, then a3 + b3 + c3 = 3abc
  9. (a + b)2-(a - b)2 = 4ab
  10. (a + b)2 + (a - b)2 = 2(a2+ b2)
  11. Division algorithm: Dividend = (Divisor * Quotient) + Remainder
  12. (xm – am) is divisible by (x - a) for all values of m.
  13. (xm - am) is divisible by (x + a) for even values of m.
  14. (xm + am) is divisible by (x + a) for odd values of m.
  15. Number of prime factors of ap + bq + cr+ ds is p + q + r + s, where a, b, c and d are prime number.
  16. Arithmetic Progression:

An A.P with first term a and common difference d is given by a, (a+ d), (a+ 2d), (a + 3d).

The nth term of the A.P is given by Tn = a (n – 1) d.

Sum of n terms in the A.P is given by Sn= (n/2) (2a+ (n-1) d) which is (n/2) (first term + last term).

  1. Sum of first n natural numbers = (n(n+1))/2
  2. Sum of first n odd numbers = n2
  3. Sum of first n even numbers = n(n+1)
  4. Sum of square of first n natural numbers = (n(n+1) (2n+1))/6
  5. Sum of cube of first n natural numbers = (n(n+1))2/4
  1. Geometrical Progression:

A G.P with first term a and common ratio r is a, ar, ar2, ar3,…..

The nth term of the G.P is given by Tn = arn-1.

Sum of n terms in the G.P is given by Sn= (a(1-rn)/(1-r)).

Example Problems

  1. A computer program was tested 300 times before its release. The testing was done in three stages of 100 tests each. The software failed 15 times in Stage I, 12 times in Stage II, 8 times in Stage III, 6 times in both Stage I and Stage II, 7 times in both Stage II and Stage III, 4 times in both Stage I and Stage III, and 4 times in all the three stages. How many times the software failed in a single stage only?

Solution:

 Assume that the software fails a, b, and c times in a single stage, in two stages, and in all stages respectively.

 Therefore b + 3c = 6+ 7 + 4 = 17 but c = 4, hence b = 5

Similarly, we have a + 2b + 3c = 15 + 12 + 8 = 35

a = 35 – 12 – 10 = 35 – 22 = 13

 

  1. What is the sum of all positive integers lying between 200 and 400 that are multiples of 7? 

Solution: Least number divisible by 7 and above 200 is 203.

Greatest number divisible of 7 and below 400 is 399.

Total numbers divisible by 7 between 200 to 400 are 29

Now, sum of n terms of AP = (n/2) (first term + last term) where, first term = 203, last term = 399 and n = 29

sum of n terms of AP = (29/2) (203+399) = 8729

  1. A number when divided by the sum of 555 and 445 gives two times their differences as quotient and 30 as the remainder. The number is:

Solution:

Required number = ((555+ 445) *2*110) + 30 = 220030

Hence option (d) is the answer

Dividend = (divisor * quotient) + Remainder.

  1. What is the unit digit in (795- 358)?

Solution: Unit digit in 795 = unit digit in [(74)23 x 73]

                                            = unit digit in [(unit digit in (2401))23 x (343)]

                                            = unit digit in (123 x 343)

                                            = unit digit in (343)

                                            = 3

              Unit digit in 358 = unit digit in [(44)14 x 32]

                                            = unit digit in [(unit digit in (81))14 x (9)]

                                            = unit digit in (114 x 9)

                                            = unit digit in (1 x 9)

                                            = 9

unit digit in (795- 358) = unit digit in (343-9) = unit digit in (334) = 4.

  1. What will be the remainder when 17200 is divided by 18?

(a) 5

(b) 3

(c) 2

(d) 1

Solution: When m is even, (xm - am) is completely divisible by (x + a).

                 (17200-1200) is completely divisible by (17+1) i.e. 18

                 (17200 – 1) is divisible by 18.

                 On dividing 17200 by 18  be get 1 as remainder

                Hence, option (d) is the correct answer.


5 HCF and LCM
N/A

Theory:

  1. Factors and Multiples: If a number x divides another number y exactly (without leaving any remainder), then x is a factor of y and y is a multiple of x.

Factors Set of numbers which exactly divides the given number.

Multiples Set of numbers which are exactly divisible by the given number.

Common Multiple: A common multiple of two or more numbers is a number which is completely divisible (without leaving remainder) by each of them.

For example,we can obtain common multiples of 4, 6 and 12 as follows

 Multiples of 4 = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, ...}

 Multiples of 6 = {6, 12, 18, 24, 30, 36, 42, 48, ...}

 Multiples of 12 = {12, 24, 36, 48, 60, ...}

∴ Common multiples of 4, 6 and 12 = {12, 24, 36, 48, ...}

  1. Least Common Multiple (L.C.M): The LCM of two or more given numbers is the least number to be exactly divisible by each of them.

 For example, we can obtain LCM of 4 and 12 as follows:

Multiples of 4 = 4,8,12,16,20, 24,28,32,36, .........

Multiples of 12 =12, 24,36,48,60,72..........

 Common multiples of 4 and 12 =12,24,36, .........

∴ LCM of 4 and 12 =12

There are two methods of finding the L.C.M of a given set of numbers:

  1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M is the product of the highest powers of all the factors.

Ex. 1:

Find the LCM of 8,12 and 15.

Factors of 8 = 2x2x2 = 23, Factors of 12 = 2 x 2 x 3 = 22 x 31, Factors of 15 = 3 x 5 = 31 x 51. Here, the prime factors that occur in the given numbers are 2, 3 and 5 and their highest powers are 3, 1 and 1 .∴ Required LCM = 23 X 31 X 51 = 8 X 3 X 5 = 120.

  1. Division Method: Arrange the given numbers in a row in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M of the given numbers.

Ex. 1: What will be the LCM of 15, 24, 32 and 45?

LCM of 15, 24,32 and 45 is calculated as

 

∴ Required LCM = 2x2x2x3x5x4x3= 1440 Note Start division with the least prime number

  1. Highest Common Factor (HCF):

HCF of two or more numbers is the greatest number which divides each of them exactly. For example, 6 is the HCF of 12 and 18 as there is no number greater than 6 that divides both 12 and 18. Similarly, 3 is the highest common factor of 6 and 9.

There are two methods to calculate the HCF of two or more numbers which are explained below:

  1. Prime Factorisation Method: Express the each one of he given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

Ex. 1: Find the HCF of 24, 30 and 42.

 Resolving 24,30 and 42 into their prime factors,

∴ Factors of 24 = 2x2x2x3 = (23 x 31)

 Factors of 30 = 2 X 3 X 5 = (21 X 31 X 51)

Factors of 42 = 2 x 3 x 7 = (21 x 31 x 71).

The product of common prime factors with the least powers = 2 X 3 =6. So, HCF of 24, 30 and 42 = 6.

  1. Division Method:Suppose we have to find the H.C.F of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F of more than two numbers :Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers.

  1. Method to Calculate LCM and HCF of Fractions:

 The LCM and HCF can be obtained by the following formulae:

              LCM of fractions = (LCM of numerators)/ (HCF of denominators)

              HCF of fractions = (HCF of numerators)/ (LCM of denominators)

Note:

 1. All the fractions must be in their lowest terms. If they are not in their lowest terms, then conversion in the lowest form is required before finding the HCF or LCM

 2. The required HCF of two or more fractions is the highest fraction which exactly divides each of the fractions

 3. The required LCM of two or more fractions is the least fraction/integer which is exactly divisible by each of them

 4. The HCF of numbers of fractions is always a fraction but this is not true in case of LCM

  1. Product of two numbers =Product of their H.C.F and L.C.M

Examples:

  1. In a fire range, 4 shooters are firing at their respective targets. The first, the second, the third and the fourth shooter hit the target once in every 5 s, 6 s, 7 s and 8 s, respectively. If all of them hit their targets at 9:00 am, when will they hit their targets together again?

Solution: Time after which they will hit the target again together = LCM (5,6,7 and 8)

= 5X3X7X2X4 =840s.

 They will hit together again at 9: 14 am

  1. What is the HCF of 8 (x5 - x3 + x) and 28(x6 +l)?

Solution: Let p(x) = 8(x5 – x3 + x) = 4 * 2 * x (x4 – x2 + 1) and

q(x) = 28(x6 + 1) = 7 X 4 [(x2)3 + (l)3] = 4 * 7 * (x2 + 1) (x4 – x2 + 1)

∴HCF of p(x) and q(x) = 4 (x4 – x2 + 1)

  1. The H.C.F of (9/10), (12/25), (18/35) and (21/40) is:

Solution: Required H.C.F = (H.C.F OF 9, 12, 18, 21)/ (L.C.M OF 10, 25, 35, 40)

                                             = 3/2800

  1. Two numbers, both greater than 29, have H.C.F. 29 and L.C.M. 4147. The sum of the numbers is:

Solution: Product of numbers = 29 x 4147

Let the numbers be 29a and 29b. Then, 29a X 29b = (24 x 4147) -> ab = 143.

Now, co-primes with product 143 are (1, 143) and (11,13).

So, the numbers are (29 x 1, 29 x 143) and (29 x 11, 29 x13).

Since both numbers are greater than 29, the suitable pair is (29 x 11, 29 x 13) i.e., (319, 377).

The required sum = (319 + 377) = 696.

  1. How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60?

Solution: LCM of 32, 40, 48 and 60 = 480.

The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.

Hence, required number of numbers are 4.


6 Simple Fractions
N/A

Fraction:

A digit which can be represented in p/q form, where q * 0, is called a fraction. Here, p is called the numerator and q is called the denominator.

For example, 3/5 is a fraction, where 3 is called numerator and 5 is called denominator.

Simple Fraction:

The fraction which has a denominator other than the power of 10 is called a simple fraction.

For example: 3/7, 5/11, etc.

Simple fraction is also known as vulgar fraction

   i.  Types of Simple Fractions

There are following types of fractions:

  1. Proper Fraction When the numerator of a fraction is less than its denominator, then fraction is called proper fraction

Example: ½, 21/43.

  1. Improper Fraction When the numerator of a fraction is greater than its denominator, then fraction is called improper fraction

Example: 17/13, 18/14,

  1. Compound Fraction A fraction, in which numerator or denominator or both are in fraction, then it is called compound fraction.

Example: 1/ (7/9), (13/11)/17,

  1. Inverse Fraction If we inverse the numerator and the denominator of a fraction, then the resultant fraction will be the inverse fraction of the original fraction.

Example: let given function is 3/7, then inverse function is 7/3.

  1. Mixed Fraction The fraction, which is the combination of integer and fraction, is called mixed fraction.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mn>3</mn><mfrac><mn>2</mn><mn>5</mn></mfrac><mo>,</mo><mn>7</mn><mfrac><mn>1</mn><mn>9</mn></mfrac></math>

  1. Continuous Fraction It has no certain definition but only say that a fraction contains additional fractions in its denominators, is called continuous fraction.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mn>2</mn><mo>+</mo><mfrac><mn>1</mn><mrow><mn>2</mn><mo>+</mo><mfrac><mn>2</mn><mrow><mn>5</mn><mo>+</mo><mfrac><mn>2</mn><mn>3</mn></mfrac></mrow></mfrac></mrow></mfrac><mo>,</mo><mn>6</mn><mo>+</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><mfrac><mn>1</mn><mrow><mn>1</mn><mo>+</mo><mfrac><mn>2</mn><mn>3</mn></mfrac></mrow></mfrac></mrow></mfrac></math>

   ii.  Operations on Simple Fractions

 Addition of Simple Fractions

  1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are added and their addition is divided by denominator.

Example: (1/4) + (2/4) = (1+2) (1/4) = (3/4)

  1.  When Denominators are Different If denominators of fractions are not same, then make their denominators equal (by taking their LCM) and then add their numerators.

Example: (1/2) + (1/3) + (1/4) = ((1 x 6) + (1 x 4) + (1 x 3))/12 = (6 + 4 + 3)/12 = 13/12

Subtraction of Simple Fractions

1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are subtracted and their subtraction is divided by the denominator.

Example:(3/4) -(1/4) = (3-1) (1/4) = (2/4) = 1/2

2. When Denominators are Different If denominators of fractions are not same, then make their denominators equal and then subtract their numerators.

Example: (2/3) - (1/2) = ((2 x 2) - (3 x 1))/6 = (4 - 3)/6 = 1/6

Multiplication of Simple Fractions

  1. To multiply two or more simple fractions, multiply their numerators and denominators.

Example:(1/2) x (3/4) = (1x3) / (2 x 4) = (3/8)

  2. If fractions are given in mixed form, first convert them into improper fraction and then multiply.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mfenced separators="|"><mrow><mn>2</mn><mfrac><mn>4</mn><mn>5</mn></mfrac></mrow></mfenced><mo>&#xD7;</mo><mfenced separators="|"><mrow><mn>1</mn><mfrac><mn>8</mn><mn>3</mn></mfrac></mrow></mfenced><mo>=</mo><mo>(</mo><mn>14</mn><mo>/</mo><mn>5</mn><mo>)</mo><mo>&#xD7;</mo><mo>(</mo><mn>11</mn><mo>/</mo><mn>3</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>154</mn><mo>/</mo><mn>15</mn><mo>)</mo></math>

Division of Simple Fractions

To divide two fractions, first fraction is multiplied by the inverse of second fraction

Example: (2/3) + (3/5) = (2/3) x (5/3) = 10/9

   iii.  Comparison of Simple Fractions

 Following are some techniques to compare fractions.

  1. Cross Multiplication Method:

Example: If (a/b) and (c/d) are two fractions, then i) if ad>bc, then (a/b) > (c/d)

ii) if ad<bc, then (a/b) < (c/d), iii) if ad=bc, then (a/b) =(c/d)

  1. By Changing Fractions in Decimal Form: To compare two or more fractions, first convert fractions into decimal form and then compare.

Example: Between (1/7) and (2/9), which fraction is bigger?

Solution: (1/7) = 0.14 and (2/9) = 0.22. It is clear that 0.22 > 0.14. Therefore (2/9) > (1/7).

  1. By Equating Denominators of Given Fractions: For comparison of fractions, take LCM of the denominators of all fractions, so that the denominators of all fractions are same. Now, the fraction having largest numerator is the largest fraction.

Example: Arrange the following fractions in decreasing order (3/5), (7/9), (11/13).

LCM of 5,9 and 13 = 5 x 9 x 13 = 585

(3/5) = (3 x 117)/ (5 x 117) = 351/585;

(7/9) = (7 x 65)/ (9 x 65) = 455/585;

(11/13) = (11 x 45)/ (13 x 45) = 495/585

Now, the fraction having largest numerator will be largest.

Therefore, decreasing order will be (495/585), (455/585), (351/585)

Hence, order will be (495/585), (455/585), (351/585).

Hence, order is (11/13), (7/9), (3/5).

  1. By Equating Numerators of Given Fractions: For comparison of fractions, take LCM of the numerator of all fractions, so that numerators of all the fractions are same. Now, the fraction having smallest numerator will be largest.

Example: Which fraction is largest among (3/13), (2/15), (4/17)?

Solution: LCM of 2, 3 and 4 = 2 x 2 x 3 = 12

(3/13) = (3 x 4)/ (13 x 4) = 12/52 = (2 x 6)/ (15 x 6) = (12/90) and (4/17) = (3 x 4)/(3 x 17) = (12/51)

Now, the fraction having smallest denominator will be largest.

Hence, (4/17) is the biggest number

Important Facts Related to Simple Fractions:

  1. If in a fraction, numerator is equal to denominator, then the value of fraction is equal to 1
  2.  If the numerator of a fraction is always non-zero and denominator is zero, then the value of fraction is infinity (°°).
  3. If the numerator of a fraction is zero and denominator is not equal to zero, then the value of fraction is zero.
  4.  If the numerator or denominator of any fraction is either multiplied or divided by same number, then the value of fraction remains unchanged.
  5.  If the numerator and denominator have no common factor other than 1, then the fraction is said to be in its lowest form
  6.  A fraction is a rational number as it can be expressed in the form of p/q and q * 0.

Basic Formulae:

  • To represent any fraction in simplified form, divide its numerator and denominator by their HCF.
  • If in the given fractions, the difference between numerator and denominator are same, then fraction having larger numerator is the largest and fraction having smaller numerator is the smallest.
  •  Formula 3 if in the given fractions, the numerators are increasing by a definite value and the denominator is also increasing by a definite value but the value of denominator is greater than numerator, then the fraction having smaller numerator will be the smallest fraction and the fraction having larger numerator will be the largest fraction.
  • If any number is divided by (a/b) instead of multiplying by (a/2), then the obtained value will be x greater than original value and the given number will be (abx)/ (b2 – a2).

Examples:

  1. 13 Jack was to find 6/7 of a fraction. Instead of multiplying, he divided the fraction by 6/7 and the result obtained was 13/70 greater than original value. Find the fraction given to Jack?

Solution: Given, a = 6, b = 7 and x = (13/70)

Therefore, required fraction = (abx)/ (b2 – a2) = (6 x 7 x (13/70))/ (72 – 62) = (6 x 13)/ (10 x 13) = (3/5)

Hence, fraction given to Jack is (3/5).

  1. Arrange the given fractions in increasing order, (4/5), (5/6), (6/7)

Solution: Since, all the fractions have difference in numerator and denominator are same.

Therefore, increasing order (4/5), (5/6), (6/7).

  3. Out of the fractions (5/7), (4/9), (6/11), (2/5) and (3/4), what is the difference between the largest and the smallest fractions?

Solution: (5/7) = 0.71, (4/9) = 0.44, (6/11) = 0.54, (2/5) = 0.40, (3/4) = 0.75

Here, the largest fraction = (3/4) and the smallest fraction = (2/5)

So required difference = (3/4) – (2/5) = (15 – 8)/20 = 7/20

  4. The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator, then find the fraction.

Solution: Let denominator of fraction = x

Then, numerator = x – 4.

Therefore Fraction = (x-4)/x. Now, according to the question,

8[(x -4) – 2] = (x + 1)

 -> (x -4) – 2 = ((x +1)/8)

-> x – 6 = (x+1)/8

 -> 8(x – 6) = x + 1

-> 8x – 48 = x + 1

-> 7x = 49,

x = 49/7 -> x = 7, Therefore fraction = (7 – 4)/ 7 = (3/7).

  5. 4/7 of a pole is in the mud. When 1/3 of it is pulled out, 250 cm of the pole is still in the mud. Find the full length of the pole.

Solution: Total length of pole = (Length of pole in mud)/ (Remaining part of pole in mud)

                                                     = 250/ ((4/7) – (1/3))

                                                     = 1050

                Therefore, length of pole = 1050.


7 Decimal Fractions
N/A

Fraction:

A digit which can be represented in p/q form, where q * 0, is called a fraction. Here, p is called the numerator and q is called the denominator.

For example, 3/5 is a fraction, where 3 is called numerator and 5 is called denominator.

Decimal Fraction:

If the fraction has denominator in the powers of 10, then fraction is called decimal fraction.

Example: 10th part of unit = (1/10) = 0.1, 10th part of 6 = (6/10) = 0.6

To convert a decimal fraction into a vulgar fraction, place 1 in the denominator under the decimal point. Then, after removing the decimal point, place as many zeroes after it as the number of digits after the decimal point. Finally, reduce the fraction to its lowest terms.

Example: 0.23 = (23/100), 0.0035 = 35/10000 = 7/2000

Note:  Placing zeroes to the right of a decimal fraction, it does not make any change in value Hence, 0.5, 0.50, 0.500 and 0.5000 are equal.

If numerator and denominator of a fraction have same number of decimal places, then each of the decimal points be removed

Thus, (1.84/2.99) = (184/299) = (8/13)

Types of Decimal Fractions

1.     Recurring Decimal Fraction: The decimal fraction, in which one or more decimal digits are repeated again and again, is called recurring decimal fraction. To represent these fractions, a line is drawn on the digits which are repeated.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mo>(</mo><mn>2</mn><mo>/</mo><mn>3</mn><mo>)</mo><mo>=</mo><mn>0.6666</mn><mo>&#x2026;</mo><mo>=</mo><mn>0</mn><mo>.</mo><mover accent="true"><mn>6</mn><mo accent="false">&#xAF;</mo></mover><mo>;</mo><mo>(</mo><mn>22</mn><mo>/</mo><mn>7</mn><mo>)</mo><mo>=</mo><mn>3.142857142857</mn><mo>=</mo><mn>3</mn><mo>.</mo><mover accent="true"><mn>142857</mn><mo accent="false">&#xAF;</mo></mover></math>

2.    Pure Recurring Decimal Fraction: When all the digits in a decimal fraction are repeated after the decimal point, then the decimal fraction is called as pure recurring decimal fraction.

To convert pure recurring decimal fractions into simple fractions (vulgar form), write down the repeated digits only once in numerator and place as many nines in the denominator as the number of digits repeated.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mn>0</mn><mo>.</mo><mover accent="true"><mn>5</mn><mo accent="false">&#xAF;</mo></mover><mo>=</mo><mo>(</mo><mn>5</mn><mo>/</mo><mn>9</mn><mo>)</mo></math>

Since, there is only 1 repeated digit. Therefore, only single 9 is placed in denominator.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Example:&#xA0;</mtext><mn>0</mn><mo>.</mo><mover accent="true"><mn>53</mn><mo accent="false">&#xAF;</mo></mover><mo>=</mo><mo>(</mo><mn>53</mn><mo>/</mo><mn>99</mn><mo>)</mo></math>

Since, there are only 2 repeated digits. Therefore, two 9's are placed in denominator.

3.     Mixed Recurring Decimal Fraction: A decimal fraction in which some digits are repeated and some are not repeated after decimal is called as mixed recurring decimal fraction.

Example: 0.1733333… = <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0.17</mn><mover accent="true"><mn>3</mn><mo accent="false">&#xAF;</mo></mover></math>

To convert mixed recurring decimal fractions into simple fractions, in the numerator, take the difference between the number formed by all the digits after decimal point (repeated digits will be taken only once) and the number formed by non-repeating digits. In the denominator, place as many nines as there are repeating digits and after nine put as many zeroes as the number of non-repeating digits.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Thus, 0.&#xA0;</mtext><mover accent="true"><mn>16</mn><mo accent="false">&#xAF;</mo></mover><mo>=</mo><mo>(</mo><mo>(</mo><mn>16</mn><mo>&#x2212;</mo><mn>1</mn><mo>)</mo><mo>/</mo><mn>90</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>15</mn><mo>/</mo><mn>90</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>6</mn><mo>)</mo><mo>;</mo><mn>0.22</mn><mover accent="true"><mn>73</mn><mo accent="false">&#xAF;</mo></mover><mo>=</mo><mo>(</mo><mo>(</mo><mn>2273</mn><mo>&#x2212;</mo><mn>22</mn><mo>)</mo><mo>/</mo><mn>9900</mn><mo>)</mo><mo>=</mo><mo>(</mo><mn>2251</mn><mo>/</mo><mn>9900</mn><mo>)</mo></math>

Operations on Decimal Fractions:

Addition and Subtraction of Decimal Fractions:

To add or subtract decimal fractions, the given numbers are written under each other such that the decimal points lie in one column and the numbers so arranged can now be added or subtracted as per the conventional method of addition and subtraction.

Example: (i) 353.5 + 2.32 + 43.23 =? (ii) 1000 - 132.23 =?

Solution: 

Multiplication of Two or More Decimal Fractions: Given fractions are multiplied without considering the decimal points and then in the product, decimal point is marked from the right-hand side to as many places of decimal as the sum of the numbers of decimal places in the multiplier and the multiplicand together.

Example: (i) 4.3 x 0.13 =? (ii) 1.12 x 2.3 x 4.325 =?

Sol. (i) 43 x 13 = 559

Sum of the decimal places = (1 + 2) = 3; Therefore, required product = 0.559 (ii) 112 x 23 x 4325 = 11141200 Sum of the decimal places = (2 + 1 + 3) = 6 ∴ Required product = 11.141200

Multiplication of Decimal Fraction by an Integer:

Given integer is multiplied by the fraction without considering the decimal point and then in the product, decimal is marked as many places before as that in the given decimal fraction.

Example: Find the value of the following. (i) 19.72x4 (ii) 0.0745x10 (iii) 3.52x14

Sol. (i) 19.72 x 4

Multiplying without taking decimal point into consideration 1972 x 4 = 7888

So, 19.72 x 4= 78.88

Since, in the given decimal fraction, decimal point is two places before. So, in the product, decimal point will also be put two places before.

Similarly, (ii) 0.0745 x 10 = 0.745 (iii) 3.52 x 14 = 49.28

Dividing a Decimal Fraction by an Integer:

Do simple division i.e., divide the given decimal number without considering the decimal point and place the decimal point as many places of decimal as in the dividend.

Example: Suppose we have to find the quotient (0.0204 <math xmlns="http://www.w3.org/1998/Math/MathML"><mover accent="true"><mi>A</mi><mo>~</mo></mover></math>. 17). Now, 204 <math xmlns="http://www.w3.org/1998/Math/MathML"><mover accent="true"><mi>A</mi><mo>~</mo></mover></math>. 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 <math xmlns="http://www.w3.org/1998/Math/MathML"><mover accent="true"><mi>A</mi><mo>~</mo></mover></math>. 17 = 0.0012

Division of Decimal Fractions:

In such divisions, dividend and divisor both are multiplied first by a suitable multiple of 10 to convert divisor into a whole number and then above-mentioned rule of division is followed.

Example: Thus, (0.00066/0.11) = ((0.00066 x 100)/ (0.11 x 100)) = (0.066/11) = 0.006

Basic Examples

1.      when 0.252525…... is converted into a fraction, then find the result.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Solution:&#xA0;</mtext><mn>0.252525</mn><mo>&#x2026;</mo><mo>&#x2026;</mo><mo>=</mo><mn>0</mn><mo>.</mo><mover accent="true"><mn>25</mn><mo accent="false">&#xAF;</mo></mover><mo>=</mo><mn>25</mn><mo>/</mo><mn>99</mn><mo>.</mo></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;2.&#xA0;&#xA0;&#xA0;Find the value&#xA0;</mtext><mn>27</mn><mo>&#xD7;</mo><mn>1</mn><mo>.</mo><mover accent="true"><mn>2</mn><mo accent="false">&#xAF;</mo></mover><mo>&#xD7;</mo><mn>5.526</mn><mover accent="true"><mn>2</mn><mo accent="false">&#xAF;</mo></mover><mo>&#xD7;</mo><mn>0</mn><mo>.</mo><mover accent="true"><mn>6</mn><mo accent="false">&#xAF;</mo></mover></math>

Solution: 27 x ((12-1)/9) x ((5.5262-5526)/9000) x (6/9)

                 = 27 x (11/9) x (49736/9000) x (6/9)

                 = (1094192/9000)

 

                 = 121.576888….

                 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>121.576</mn><mover accent="true"><mn>8</mn><mo accent="false">&#xAF;</mo></mover></math>

3.     In the year 2020, Henry gets $3832.5 as his pocket allowance. Find his pocket allowance per day?

Solution: Henry’s pocket allowance = $3832.50

Total days in 2020 (general year) = 365 days

Allowance per day = (3832.5/365) = $10.5

4.     When 52416 is divided by 312, the quotient is 168. What will be the quotient when 52.416 is divided by 0.0168?

Solution: Given, (52416/312) = 168 -> (52416/168) = 312.

Now, (5.376/16.8) = (53.76/168) = ((53.76/168) x (1/100)) = (32/100) = 0.32

5.     ((36.54)2 – (3.46)2)/? = 40

Solution: ((36.54)2 – (3.46)2)/x = 40. Then, x = ((36.54)2 – (3.46)2)/40 = ((36.54)2 – (3.46)2)/ (36.54 + 3.46) = (36.54 – 3.46) = 33.08

 

Since ((a)2 – (b)2/ (a + b)) = (a – b).


1 Statistics
N/A

Statistics is a branch of mathematics that deals with numbers and analysis of the data. Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data.

In short, Statistics deals with collecting, classifying, arranging, and presenting collected numerical data in simple comprehensible ways.

With the help of statistics, we are able to find various measures of central tendencies and the deviation of data values from the center.

Basic Formulae:

  • Mean is defined as the sum of all the elements of a set divided by the number of elements.

  • Median is the middle value of a dataset.
  • If a set consists of an odd number of values, then the middle value will be the median of the set.
  •  If the set consists of an even number of sets, then the median will be the average of the two middle values.

Median(M) = If n is odd, then 

If n is even, then 

  • The mode in a dataset is the value that is most frequent in the dataset.

Mode = The value which occurs most frequently

  • The standard deviation is defined as the square rooting of the variance of the data.

Standard Deviation(S) = 

Where, x = observations given

 = Mean

 = Total number of observations

 

  1. In a group of 10, 4 students were selected at random and their total marks in the final assessments are recorded, which are 81, 83, 98, 76. Find the standard deviation of their marks.

Solution: Here N = 4

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x21D2;</mo><mo>&#xA0;</mo><mfrac><mrow><munderover><mo>&#x2211;</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>4</mn></munderover><mo>&#x200A;</mo><mo>(</mo><mi>x</mi><mo>&#x2212;</mo><mn>84.5</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>=</mo><mfrac><mrow><mo>(</mo><mn>81</mn><mo>&#x2212;</mo><mn>84.5</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>83</mn><mo>&#x2212;</mo><mn>84.5</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>98</mn><mo>&#x2212;</mo><mn>84.5</mn><msup><mo>)</mo><mn>2</mn></msup><mo>+</mo><mo>(</mo><mn>76</mn><mo>&#x2212;</mo><mn>84.5</mn><msup><mo>)</mo><mn>2</mn></msup></mrow><mn>4</mn></mfrac><mo>=</mo><mfrac><mn>269</mn><mn>4</mn></mfrac><mo>=</mo><mn>67.25</mn></math>

 2. Weight of girls = {40, 45, 50, 45, 55, 45, 60, 45}. Find the mode?

Solution: Here we have 45 as repeating value

Since only on value is repeating it is a unimodal list.

S0, mode = 45

 3. Find the median of the data: 24, 36, 45, 18, 20, 26, 38

Solution: Arrange them in ascending order 18, 20, 24, 26, 36, 38, 45

Median = middle most observation or  term when n is odd

So, median = 26

 4. All the students in a mathematics class took a 100-point test. Eight students scored 100, each student scored at least 55, and the mean score was 75.

What is the smallest possible number of students in the class?

Solution:

  • Let the number of students be 
  • Then the sum of their scores is at least 
  • we need to achieve the mean 75, which is equivalent to achieving the sum 75n.
  • we must have 
  • The smallest integer n for which this is true is n = 18.
  • To finish our solution, we now need to find one way how 18 students could have scored on the test.
  • We have 18 x 75 = 1350 points to divide among them.
  • The eight 100s make 800, hence we must divide the remaining 550 points among the other 10 students.
  • This can be done by giving 55 points to each of them.
  • Hence the smallest possible number of students is 18.

 5. A manager has given a test to his team in which 20% are women and 80% are men. The average score on the test was 70. Women all received the same score, and the average score of the men was 60. What score did each of the woman receive on the test?

  • Solution:

    • Let Total students = 100, Men = 80, Women = 20.
    • Sum of all the scores = 100 x 70 = 7000.
    • Sum of all the scores of men = 60 * 90 = 5400.
    • Difference in scores = 7000 - 5400= 1600.
    • 1600 is the scores obtained by women.
    • It's stated in the question that all Girls got the same score
    • so, score each of the woman received on the test 1600/20 = 80.


1 Permutations and Combinations in Depth
N/A

  1. It is a COMINATION if the order of selection DOES NOT matter

Combinations have the formula of nCr

Where:

               n = number of things to be chosen from (the population)

               r = number in group/team/category

 The numerator is just the factorial n! taken “r” times, and the denominator is just the factorial r! taken “r” times

Ex: 5C2   

  A larger factorial can be simplified down by simply subtracting (n-r) for r

      Ex: 19C17 = 19C2 where r = (19 - 17) = 2

 2. It is a PERMUTATION if the order of selection DOES matter

Permutations have the formula of nPr = nCr x r! 

  Permutations can be simplified to just the numerator portion of the combination 

EXAMPLE 1: If there are 4 tennis players and you need to make teams of 2 consisting of a captain and vice-captain, how many different teams can you make?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><msub><mi>C</mi><mn>2</mn></msub><mo>&#xD7;</mo><mn>2</mn><mo>!</mo><mo>=</mo><mfrac><mrow><mn>4</mn><mo>&#xD7;</mo><mn>3</mn></mrow><mrow><mn>1</mn><mo>&#xD7;</mo><mn>2</mn></mrow></mfrac><mo>&#xD7;</mo><mo>(</mo><mn>1</mn><mo>&#xD7;</mo><mn>2</mn><mo>)</mo><mo>=</mo><mn>4</mn><mo>&#xD7;</mo><mn>3</mn><mo>=</mo><mn>12</mn></math>

COMBINATION & PERMUTATION RULES EXPLAINED WITH EXAMPLES:

  1. If you have 10 children, how many ways can they stand in a row? 

Ans-> 10!

  1. You have 4 physics books, 3 math books, and 2 chemistry books. How many ways can they be arranged on a shelf?

Ans-> 9! -> There are 9 totally different books

  1. You have the same set of books as above, but all physics books must stay together, all math books must stay together, and all chemistry books must stay together on the shelf. How many ways can they be arranged?

Ans-> 3! x 4! x 3! x 2!  = 1,728

Think of each set of books going into a “box” or a “set” on the shelf. This creates 3 different “sets” – one for physics, one for chemistry, and one for math. Then within each “box” or “set”, the different books can be arranged as many times as there are books.

So, there are 3 different book types, giving 3!

Then there are 4 physics books, 3 math, and 2 chem books, giving 4!, 3!, and 2! Respectively, totalling 3! x (4! x 3! x 2!) = 1,728

  1. Same set of books as above, however all of the books are identical for each type (all physics books are the same, math is the same, chem are the same). How many ways can they be arranged on the shelf? 

Ans-> 3!  There is no differentiation between the books, except the type of book they are

  1. How many ways can the word “at” be arranged? (Or any 2-character combination) 

Ans-> 2!

  1. How many ways can the word “top” be arranged? (Or any 3-character combination) 

Ans -> 3! 

  1. How many ways can the word “too” be arranged? (Or any 3-character word with 2 identical characters)

Any X character combination with Y identical characters has  number of arrangements where X is the total number of characters and Y is the number of identical characters. If there are more than 1 set of identical characters, put them all in the denominator.

 8. How many ways can the word “mathematics” be arranged?

  because there are 2 instances of “m”, “a” and “t” each

 9. A person is standing downtown at point A and wishes to get to the other side of downtown to point B. There are 5 streets and 4 crossroads. How many ways can he get from A to B?

There are 9 total streets, and no matter what, he must go 4 ways right and 5 ways up.

 10. There are 6 people sitting at a circular table. How many ways can they be seated? 

Ans -> (6 -1)!

For a “circular problem” you have to subtract 1 from the possibilities – any arrangement will always start and end with the same person.

 11. A father has the following bills in his pocket - $5, $10, and $20. His son asks for some spending money for the weekend. How many ways can the father give his son some money? 

Ans -> 23

For each bill, the father has the option to either give it to his son or not. Then the father could also give a combination of any of the bills or not 

 12. The same scenario exists, but the father must give his son at least something to spend

Ans -> 23 – 1   you have to subtract the scenario of giving him neither of any of the bills

 13. At Walmart there is a box of 100 different $0.99 items in it at the register. How many ways can a shopper grab some of the items?

Ans: 2100            she has the option to grab or not grab each item

         2100 – 1     if she must choose at least 1 item

 15. A teacher has a pen, an eraser, and tape. She wants to distribute these among 2 students. How many ways can she distribute these items where all 3 must be distributed among the 2 students? 

Ans -> 23 = 8

Rn is the general equation used for these question types, where n is the number of objects to be distributed and R is the number of people (or groups) among which the n objects are to be distributed.

If you want to include the option of not distributing anything, add 1  Rn +1

 15. There are now 3 students in the class and the teacher has a pen, eraser, tape, and stapler. How many ways can she distribute the items if all must be distributed? If none can be distributed?

Ans -> 34; 34 + 1

 16. A teacher has 5 identical chocolates and wants to give them to her 3 students. How many ways can she give the chocolates to the students?

Ans-> (5+3-1) C3 – 1 = 7C2 

For N identical items being distributed among R people (or groups) the following formula should be used: N+(R-1) C (R-1) 

If all items must be distributed, then the use the following formula: (N-1) C (R-1)

 17. A teacher has 5 identical chocolates to give to 3 students and she must give each student something. How many ways can she distribute the chocolates?

(5 – 1)C3 – 1 = 4C2

 18. How many ways can you pick fruit from a “For Sale” basket at the grocery store that has 5 oranges, 6 apples, and 7 bananas in it?

(5 + 1) (6 + 1) (7 + 1) = 6 x 7 x 8 = 336

The “ + 1” is to account for the option of not picking that particular item

 19. If you must pick at least one fruit from the basket, how many ways can you pick fruit?

(5 + 1)(6 + 1)(7 + 1) – 1 = (6 x 7 x 8) – 1 = 336 – 1 = 335

 20. How many ways can you pick fruit from a basket with 10 apples, 15 oranges, and 25 other different types of fruit?

(10 + 1)(15 + 1)(225) = (11)(16)(225) = (11)(229)

The oranges and apples are identical; the 25 other types are all different, and for each one you can either chose one or not

 21. A man, his wife, and their child go to a movie theatre with 5 seats and all three must sit together. How many ways can they sit at the theatre?

[5 – (3+1)] x 3!

3! is for the different combinations the father, mother, and child can sit together (similar to the “boxes” or “sets” of the similar textbooks)

[n – (r+1)] gives you the number of seating arrangements where n is the number of seats and r is number of people

 22. A man, wife, daughter, and son go to a movie theatre with 10 seats and they all want to sit together. How many ways can they sit?

(10 – 4 + 1) x 4! = 7 x 4!

 23. In how many ways can 8 tennis players be divided into 2 distinct teams of 4 each?

(4 x 2)! / 4!2

Use the formula (m x n)! / n!m

Or (m x n)! / (n!m x m!) for non-distinct teams

 24. How many ways can 8 tennis players be divided into 2 non distinct teams of 4 each?

(4 x 2)! / (4!2 x 2!) 

 25. You have 10 pants, 20 shirts, and 5 shoes in your closet. You’re packing for a trip and want to take two of each. How many ways can you pack a suitcase? 

10C2 x 15C2 x 5C2 = 5 x 9 x 10 x 19 x 5 x 2

 26. There are 6 total horses in a race, two of which are ridden by Bob and John. How many ways can John be ahead of Bob?

 27.  How many ways can you rearrange the word “courage” with the vowels staying in alphabetical order?

7! Represents the total number of letters in “courage” and 4! Represents the vowels that must stay in order 

 28. You want to create a 5-character password, and the first character must be a consonant, the second must be a vowel, and the rest can be digits. How many 5 passwords can be created? 

21 x 5 x 10 x 10 x 10      21 consonants, 5 vowels, and 10 different digits

 29. There are 5 men and 3 women in a department.  They need to form a 3-member committee with at least one woman in the committee?

 30. 

3 member committee out of 8 staff = 8c3 = (8 * 7 * 6)/(1 * 2 * 3) = 56

All men committees = 5c3 = (5 * 4)/1 * 2 = 10

Required Committees = Total possible committees - All men committees) = 56 – 10 = 46

 31.  A coin is tossed 5 times (or 5 coins are tossed one time).  What are the different outcomes?

First let us do it for 3 tosses. 

T T T

 

H T T

T H T

T T H

 

H H T

H T H

T H H

 

H H H

Total Sample Space = = 8

Outcomes = 2 ( Head or Tail)

Experiment = Number of tosses = 3 

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mo>&#xA0;</mo><mi>S</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mo>&#xA0;</mo><mi>S</mi><mi>p</mi><mi>a</mi><mi>c</mi><mi>e</mi><mo>=</mo><mo>(</mo><mi>O</mi><mi>u</mi><mi>t</mi><mi>c</mi><mi>o</mi><mi>m</mi><mi>e</mi><mi>s</mi><msup><mo>)</mo><mrow><mi>E</mi><mi>x</mi><mi>p</mi><mi>e</mi><mi>r</mi><mi>i</mi><mi>m</mi><mi>e</mi><mi>n</mi><mi>t</mi><mi>s</mi></mrow></msup><mo>=</mo><msup><mn>2</mn><mn>3</mn></msup><mo>=</mo><mn>8</mn></math>

If it is 5 tosses = = 32 is the total sample space

 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Z</mi><mi>e</mi><mi>r</mi><mi>o</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><mo>&#xA0;</mo><msup><mo>&#x2192;</mo><mn>3</mn></msup><msub><mi>c</mi><mn>0</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>0</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>3</mn></msup><mo>=</mo><mn>1</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>0</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>3</mn></msup><mo>=</mo><mn>1</mn><mo>/</mo><mn>8</mn><mspace linebreak="newline"/><mi>O</mi><mi>n</mi><mi>e</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><msup><mo>&#x2192;</mo><mn>3</mn></msup><msub><mi>c</mi><mn>1</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>1</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mn>3</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>1</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mn>3</mn><mo>/</mo><mn>8</mn><mspace linebreak="newline"/><mi>T</mi><mi>w</mi><mi>o</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>3</mn></msup><msub><mi>c</mi><mn>2</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>2</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>1</mn></msup><mo>=</mo><mn>3</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>2</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>1</mn></msup><mo>=</mo><mn>3</mn><mo>/</mo><mn>8</mn><mspace linebreak="newline"/><mi>T</mi><mi>h</mi><mi>r</mi><mi>e</mi><mi>e</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>3</mn></msup><msub><mi>c</mi><mn>3</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>3</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>0</mn></msup><mo>=</mo><mn>1</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>3</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>0</mn></msup><mo>=</mo><mn>1</mn><mo>/</mo><mn>8</mn></math>  

Total = 8/8 = 1

For five tosses:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mo>&#xA0;</mo><mi>S</mi><mi>a</mi><mi>m</mi><mi>p</mi><mi>l</mi><mi>e</mi><mo>&#xA0;</mo><mi>S</mi><mi>p</mi><mi>a</mi><mi>c</mi><mi>e</mi><mo>=</mo><mo>(</mo><mi>O</mi><mi>u</mi><mi>t</mi><mi>c</mi><mi>o</mi><mi>m</mi><mi>e</mi><mi>s</mi><msup><mo>)</mo><mrow><mi>E</mi><mi>x</mi><mi>p</mi><mi>e</mi><mi>r</mi><mi>i</mi><mi>m</mi><mi>e</mi><mi>n</mi><mi>t</mi><mi>s</mi></mrow></msup><mo>=</mo><msup><mn>2</mn><mn>5</mn></msup><mo>=</mo><mn>32</mn></math>

If it is 5 tosses = = 32 is the total sample space

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Z</mi><mi>e</mi><mi>r</mi><mi>o</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>0</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>0</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>5</mn></msup><mo>=</mo><mn>1</mn><mo>*</mo><mo>(</mo><mo>&#xBD;</mo><msup><mo>)</mo><mn>0</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>5</mn></msup><mo>=</mo><mn>1</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>O</mi><mi>n</mi><mi>e</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>1</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>1</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>4</mn></msup><mo>=</mo><mn>5</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>1</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>4</mn></msup><mo>=</mo><mn>5</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>T</mi><mi>w</mi><mi>o</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>2</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>2</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>3</mn></msup><mo>=</mo><mn>10</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>2</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>3</mn></msup><mo>=</mo><mn>10</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>T</mi><mi>h</mi><mi>r</mi><mi>e</mi><mi>e</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>3</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>3</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mn>10</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>3</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>2</mn></msup><mo>=</mo><mn>10</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>F</mi><mi>o</mi><mi>u</mi><mi>r</mi><mo>&#xA0;</mo><mi>h</mi><mi>e</mi><mi>a</mi><mi>d</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>4</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>4</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>1</mn></msup><mo>=</mo><mn>5</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>4</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>1</mn></msup><mo>=</mo><mn>5</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>F</mi><mi>i</mi><mi>v</mi><mi>e</mi><mo>&#xA0;</mo><mi>t</mi><mi>o</mi><mi>s</mi><mi>s</mi><mi>e</mi><mi>s</mi><msup><mo>&#x2192;</mo><mn>5</mn></msup><msub><mi>c</mi><mn>5</mn></msub><mo>(</mo><mi>H</mi><msup><mo>)</mo><mn>5</mn></msup><mo>*</mo><mo>(</mo><mi>T</mi><msup><mo>)</mo><mn>0</mn></msup><mo>=</mo><mn>1</mn><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>5</mn></msup><mo>*</mo><mo>(</mo><mn>1</mn><mo>/</mo><mn>2</mn><msup><mo>)</mo><mn>0</mn></msup><mo>=</mo><mn>1</mn><mo>/</mo><mn>32</mn><mspace linebreak="newline"/><mi>P</mi><mo>(</mo><mi>a</mi><mi>t</mi><mo>&#xA0;</mo><mi>l</mi><mi>e</mi><mi>a</mi><mi>s</mi><mi>t</mi><mo>&#xA0;</mo><mi>o</mi><mi>n</mi><mi>e</mi><mo>&#xA0;</mo><mi>H</mi><mi>e</mi><mi>a</mi><mi>d</mi><mo>)</mo><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>1</mn><mo>&#xA0;</mo><mo>&#x2013;</mo><mo>&#xA0;</mo><mi>A</mi><mi>l</mi><mi>l</mi><mo>&#xA0;</mo><mi>T</mi><mi>a</mi><mi>i</mi><mi>l</mi><mi>s</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>1</mn><mo>&#xA0;</mo><mo>&#x2013;</mo><mo>&#xA0;</mo><mn>1</mn><mo>/</mo><mn>32</mn><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mn>31</mn><mo>/</mo><mn>32</mn></math>

 32. There are 5 Black cars and 3 White cars in a car dealership.  How many ways one can pick two cars in the following cases:

Black             White                   Pick

5                    3                            2

B                    B                            5c2/8c2 = 10/28

W                   W                         3c2/8c2 = 3/28

B                    W                          5c1 * 3c1/8c2 = 5 * 3/{(8 * 7)/(1 * 2)} = 15/28

 

32.

There are 5 Black cars, 4 Red cars and 3 White cars. 

Pick any two cars:

  1. Both Black 5c2/12c2
  2. Both Red  4c2/12c2
  3. Both White 3c2/12c2
  4. Different Color BR + BW + RW {(5c1 * 4c1) + (5c1 * 3c1) + (4c1 * 3c1)}/12c2
  5. Same Color (5c2 + 4c2 + 3c2)/12c2
  6. Any two Colors BR + BW + RW {(5c1 * 4c1) + (5c1 * 3c1) + (4c1 * 3c1)}/12c2   

Pick any 3 cars

  1. All three same color (5c3 + 4c3 + 3c3)/12c3
  2. All three different color  (5c1 * 4c1 * 3c1)/12c3

33.

7 married couples-Mixed double teams without married couples

The number of ways a lawn tennis mixed double can be made up from seven married couple if no husband and wife play in same set is.

Answer: 36

Total = 14

7 husbands * 7 wives = 49 ways

Out of above 49, 7 are married to each other

Mixed double teams without married couple = 49 – 7 = 36

34.

Arrangements

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

Answer: 80

First member can be chosen from one of the couples in 5c1 ways

Four couples left. 

The remaining two members can be chosen in 2^4 ways from 4 couples = 5 * 2^4 = 80 ways

or

10c3 = 10.9.8/3.2 = 120

1 couple can be selected in 5c1 ways = 5

This can couple can be with any other 8 people in 8 ways

1 couple in 3 = 5 * 8 = 40

Answer = 120-40 = 80


2 Permutations
N/A

Permutation

Each of the different arrangements which can be made by taking some or all of a given number of things or objects at a time, is called a permutation.

Permutation implies arrangement, where order of the things is important.

For example:

 The permutations of three items a, b and c taken two at a time are ab, ba, ac, ca, cb and be.

Since, the order in which the items are taken, is important, ab and ba are counted as two different permutations.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Number of all permutations of&#xA0;</mtext><mi>n</mi><mtext>&#xA0;things, taken&#xA0;</mtext><mi>r</mi><mtext>&#xA0;at a time, is given&#xA0;</mtext><mi>b</mi><mi>y</mi><msup><mo>:</mo><mi>n</mi></msup><msub><mi>P</mi><mi>r</mi></msub><mo>=</mo><mfrac><mrow><mi>n</mi><mo>!</mo></mrow><mrow><mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mi>r</mi><mo>)</mo><mo>!</mo></mrow></mfrac></math>

Cases of Permutation

There are several cases of permutation

  1. Formation of numbers with given digits

In these types of question, it is asked to form numbers with some different digit. These digits can be used with repetition or without repetitions.

  2. Formation of words with given letters

These questions are very much similar to previous case questions but here in place of numbers, word is formed from a set of English alphabets given in the form of a word.

Important point:

Number of permutations of n objects out o which p are alike and are of one type, q are alike and are of second type and r are alike and are of third type  <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>n</mi><mo>!</mo></mrow><mrow><mi>p</mi><mo>!</mo><mi>q</mi><mo>!</mo><mi>r</mi><mo>!</mo></mrow></mfrac></math>

  3. Arrangement of persons in a row or at a round table

 These types of question are based on arrangement of person (boy or girls etc) in a straight line facing some direction or around some circular object like table etc.

Note: Number of permutations of n objects taken all at a time is n! when repetition is not allowed.

  4Arrangement of books on a shelf, etc

 In such questions arrangement of books is done into a shelf in a row or one over the other.

Note:

Questions based on sending invitation to different persons are similar to questions based on arrangement of books.

Number of permutations of n different objects taken i at a time, when repetition is allowed = ni

Examples:

  1. How many numbers of four digits can be formed with the digits 1, 2, 3, 4 and 5? (Repetition of digits is not allowed.)

Solution: There are five numbers and number of places to be filled up = 4

<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mtext>&#xA0;So, the required number of numbers is&#xA0;</mtext><mn>5</mn></msup><msub><mi>P</mi><mn>4</mn></msub><mo>=</mo><mfrac><mrow><mn>5</mn><mo>!</mo></mrow><mrow><mo>(</mo><mn>5</mn><mo>&#x2212;</mo><mn>4</mn><mo>)</mo><mo>!</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>5</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>1</mn></mrow><mn>1</mn></mfrac><mo>=</mo><mn>120</mn></math>

  1. In how many ways, can the letters of the word 'DIRECTOR' be arranged, so that the three vowels are never together?

Solution; Total number of letters = 8 and total number of vowels = 3

Here, R occurs two times,

Therefore, total number of arrangements when there is no reaction = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>8</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac></math>= 20160, but when three vowels are together, regarding them as one letter, we have only 5 + 1 = 6

These 6 letters can be arranged in <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mn>6</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac></math> ways, since R occurs twice.                                                                             

Also, three vowels can be arranged among themselves in 3! ways.

Hence, number of arrangements when the three vowels are together = <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>3</mn><mo>!</mo><mo>&#xD7;</mo><mfrac><mrow><mn>6</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac><mo>=</mo><mn>2160</mn></math>

Number of arrangements, so that the three vowels are never together = 20160 - 2160= 18000

  1. In how many different ways 8 girls can be seated in a row?

Solution: Number of ways in which 8 girls can be seated in a row = 8! = 8x7x6x5x4x3x2x1= 40320

  4. Find the number of permutations that can be made from the letters of the word 'OMEGA' in such a way that Vowels occupying odd places.

Solution:

 Three vowels (O, E, A) can be arranged in the odd places in 3! ways

i.e., 1st position, 3rd position, 5th position

And two consonants (M, G) can be arranged in the even places in 2! ways

i.e., 2nd place and 4th place

...  Total number of ways = 3! X 2! = 12

  5. A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?

Solution: The first marble can be put into the pockets in 4 ways,

 In the same way second and third.

Thus, the number of ways in which the child can put the marbles = 4X4X4= 64 ways


3 Combination
N/A

Combination of things means selection of things. Here, order of things has no importance.

For example: The combination of two letters from the group of three letters A, B and C would be as follows AB, BC, AC.

Here, we make groups. So, AB or BA as a group is same.

Obviously, if order matters, then AB and BA are not same.

<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mtext>&#xA0;Formula for combination,&#xA0;</mtext><mi>n</mi></msup><msub><mi>C</mi><mi>r</mi></msub><mo>=</mo><mfrac><mrow><mi>n</mi><mo>!</mo></mrow><mrow><mi>r</mi><mo>!</mo><mo>(</mo><mi>n</mi><mo>&#x2212;</mo><mi>r</mi><mo>)</mo><mo>!</mo></mrow></mfrac></math>

It signifies number of groups formed from n different things, when r things are taken into consideration.

Important Points:

  • nCn = nC0 = 1
  • nCr -1 + nCr = n+1Cr
  • nC1 + nC2 + …. + nCn = 2n – 1
  • nCr = nCn-r or nCr = (nPr)/r!
  • nC0 + nC1 + nC2 + …. + nCn = 2n
  • if there are n points in a plane out of which m are collinear, then (i) Number of straight lines formed = nC2 - mC2 + 1 (ii) Number of triangles formed = nC3 - mC3

Cases of Combination

There are several cases of Combination,

  1. Formation of committee from a given set of persons

 These questions are based on formation of a committee consisting of some members (male and/ or female) from a group of persons following a certain condition.

  1. Selection of questions from question paper etc

In such question, a question paper is given with one or more parts and the different ways in which some specified number of questions can be attempted is asked.

Factorial

Factorial of a number can be defined as the product of all natural numbers up to that number i. e.,

n! = n x (n-l) x (n-2) x (n-3) x (n-4) x..... x 1= n x (n-1)!

4! = 4x3x2x1=4x3!

11! = 11x10x9x8x7x6x5x4x3x2x1

Note: Factorial of negative number and integers is not defined nPn = n, nP0 = 1

Fundamental Principles of Counting

 Multiplication Principle

  • If an operation can be performed in m different ways, following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m x n ways.
  • This can be extended to any finite number of mutually exclusive operations.

Addition Principle

If an operation can be performed in m different ways and another operation, which is independent of the first operation, can be performed in n different ways, then either of the two operations can be performed in (m + n) ways.

This can be extended to any finite number of mutually exclusive operations.

Examples:

  1. In how many ways can 5 members form a committee out of 10 be selected so that (i) two particular members must be included. (ii) two particular members must not be included.

Solution: (i) When two particular members are included then, we have to select 5—2 = 3 members out of 10 – 2 = 8

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Therefore, the required number of ways&#xA0;</mtext><msup><mo>=</mo><mn>8</mn></msup><msub><mi>C</mi><mn>3</mn></msub><mo>=</mo><mfrac><mrow><mn>8</mn><mo>!</mo></mrow><mrow><mn>5</mn><mo>!</mo><mn>3</mn><mo>!</mo></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>8</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>6</mn></mrow><mn>6</mn></mfrac><mo>=</mo><mn>56</mn></math>

  2. A question paper has two parts, part A and part B, each containing 10 questions. If the student has to choose 8 from part A and 5 from part B, in how many ways can he choose the question?

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Solution: The required number of ways&#xA0;</mtext><msup><mo>=</mo><mn>10</mn></msup><msub><mi>C</mi><mn>8</mn></msub><msup><mo>&#xD7;</mo><mn>10</mn></msup><msub><mi>C</mi><mn>5</mn></msub><mo>=</mo><mfrac><mrow><mn>10</mn><mo>&#xD7;</mo><mn>9</mn></mrow><mn>2</mn></mfrac><mo>&#xD7;</mo><mfrac><mrow><mn>10</mn><mo>&#xD7;</mo><mn>9</mn><mo>&#xD7;</mo><mn>8</mn><mo>&#xD7;</mo><mn>7</mn><mo>&#xD7;</mo><mn>6</mn></mrow><mrow><mn>5</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>3</mn><mo>&#xD7;</mo><mn>2</mn></mrow></mfrac></math>

= 5 x 9 x 3 x 2 x 7 x 6 = 11340

  3. A hall has 12 gates. In how many ways, can a man enter the hall through one gate and come out through a different gate?

Solution: Since, there are 12 ways of entering into the hall, the man come out through a different gate in 11 ways.

Hence, by the fundamental principle of multiplication, total number of ways is 12 x 11 = 132.

  4. In a plane, there are 11 points, out of which 5 are collinear. Find the number of tri angles made by these points.

Solution: Here n = 11, m = 5

Then, required number of triangles = nC3 - mC3

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><msup><mo>=</mo><mn>11</mn></msup><msub><mi>C</mi><mn>3</mn></msub><msup><mo>&#x2212;</mo><mn>5</mn></msup><msub><mi>C</mi><mn>3</mn></msub></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mn>11</mn><mo>&#xD7;</mo><mn>10</mn><mo>&#xD7;</mo><mn>9</mn></mrow><mrow><mn>3</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>1</mn></mrow></mfrac><mo>&#x2212;</mo><mfrac><mrow><mn>5</mn><mo>&#xD7;</mo><mn>4</mn><mo>&#xD7;</mo><mn>3</mn></mrow><mrow><mn>3</mn><mo>&#xD7;</mo><mn>2</mn><mo>&#xD7;</mo><mn>1</mn></mrow></mfrac></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mo>=</mo><mn>165</mn><mo>&#x2212;</mo><mn>10</mn><mo>=</mo><mn>155</mn></mtd></mtr></mtable></math>

  5. In how many ways, can 24 persons be seated around a circular table, if there are 13 seats?

Solution: First, we select 13 persons out of 24 persons in 24C13 ways.

 Now, these 13 persons can be seated in 12! Ways around a table.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>So, required number of ways&#xA0;</mtext><msup><mo>=</mo><mn>24</mn></msup><msub><mi>C</mi><mn>13</mn></msub><mo>&#xD7;</mo><mn>12</mn><mo>!</mo><mo>=</mo><mfrac><mrow><mn>24</mn><mo>!</mo></mrow><mrow><mn>13</mn><mo>!</mo><mo>(</mo><mn>24</mn><mo>&#x2212;</mo><mn>13</mn><mo>)</mo><mo>!</mo></mrow></mfrac><mo>&#xD7;</mo><mn>12</mn><mo>!</mo><mo>=</mo><mfrac><mrow><mn>24</mn><mo>!</mo></mrow><mrow><mn>13</mn><mo>!</mo><mo>(</mo><mn>11</mn><mo>)</mo><mo>!</mo></mrow></mfrac><mo>&#xD7;</mo><mn>12</mn><mo>!</mo><mo>=</mo><mfrac><mrow><mn>24</mn><mo>!</mo></mrow><mrow><mn>13</mn><mo>&#xD7;</mo><mn>11</mn><mo>!</mo></mrow></mfrac></math>


1 Probability – Cards
N/A

Probability means the chances of happening/occurring of an event.

  • There are total of 52 cards in a deck of playing cards
  •  It consists of 4 suits.
  • There are 13 cards of each suit Clubs, Diamonds, Hearts and Spades.
  • Spades, clubs are black in colour and hearts, diamonds are red in colour.
  • Each suit contains an ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2.
  • King, Queen, and Jack are called the face cards.
  • In the deck of 52 playing cards, there are 12 face cards and there are 4 Aces, 4 Jacks, 4 Queens and 4 Kings.
  • Whole together there are 26 red and 26 black cards.
  • The four different suits of cards are shown below in the picture.

  • In the above four suits we have 13 cards in each suit
  • Picture showing all the 52 cards in the pack of playing cards.

  • Here we have
  • Club - 13 cards
  • Diamond - 13 cards
  •  Heart - 13 cards
  • Spade - 13 cards
  • No. of black cards - 26
  • No. of red cards - 26
  • No. of Ace cards - 4
  • No. of Jack cards - 4
  •  No. of Queen cards - 4
  • No. of King cards - 4
  • No. of face cards – 12
  • Probability 

Examples:

  1. From a pack of 52 cards two are drawn with replacement. The probability, that the first is a spade and the second is a queen, is

Solution: Given two cards are drawn with replacement.

The probability that card is a spade P(s) 

The probability that card is a queen P(q) 

Required probability = P(s) x P(q) 

 2. Find the probability of drawing a king or an ace from a pack of playing cards.

Solution:

As there are four kings and four aces, the number of favourable cases = 8

The required probability 

 3. What is the probability of drawing a red card from a pack of cards?

Solution: The total number of outcomes = 52

The number of favourable outcomes = 26

Therefore, required probability 

 4. One card is drawn from a well-shuffled pack of 52 cards. What is the probability, that it is not the king of diamonds?

Solution: The king of diamonds can be drawn in only 1 way

 Since in a pack of cards there is only one king of diamonds

P(A) = Probability of drawing the king of diamonds 

Hence the probability of not drawing an ace of hearts 

 5. From a pack of 52 cards, two cards are drawn, what is the probability that both are hearts or both are jacks?

Solution:

Total number of ways = 52C2

Both are hearts = 13C2

Both are jacks = 4C2

So, required probability = (13C2 + 4C2)/ 52C2

 


2 Probability-Dice and Coins
N/A

Sample Space:

The set of all possible outcomes of an experiment is called the sample space, denoted by S. An element of S is called a sample point.

An Event:

Any subset of a sample space is an event.

Tossing a Coin

On tossing a coin certainty of occurrence of each of a head and a tail are the same.

Hence amount certainty of occurrence of each of a head and a tail is 50% i.e., 50/100 = 2

Therefore, is the amount of certainty of occurrence of a head (or a tail) on tossing a coin

So, we say that the probability of getting H is 1/2 or the probability of getting T is 1/2.

In the experiment of tossing a coin, the sample space has two points corresponding to head (H) and Tail (T) i.e., S {H, T}.

Here Let A be event of occurrence a head (or Tail) and S be the sample space

  • P(A) 

Throwing a Die

On throwing a dice certainty of occurrence of each of the numbers 1, 2, 3, 4, 5 and 6 on its top face are the same

Dice: Dice is a cuboid having one of the numbers 1, 2, 3, 4, 5 and 6 on each of its six faces

When a single die is thrown, there are six possible outcomes 1, 2, 3, 4, 5 and 6.

The probability of getting any one of these numbers is 

When we throw a dice then any one of the numbers 1, 2, 3, 4, 5 and 6 will come up.

So, the sample space, S = {1, 2, 3, 4, 5, 6}

Here Let A be event of occurrence any number from 1 to 6 and S be the sample space.

  • P(A) 

CONDITIONAL PROBABILITY:

Let A and B be two events associated with a random experiment.

Then, the probability of occurrence of A under the condition that B has already occurred and P (B) ≠ 0, is called the conditional probability of occurrence of A when B has already occurred and it is denoted by P (A/B).

Thus, P (A/B) = Probability of occurrence of A, if B has already occurred and P (B) ≠ 0

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mi>B</mi><mo>)</mo></mrow></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mfrac><mrow><mi>n</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>n</mi><mo>(</mo><mi>S</mi><mo>)</mo></mrow></mfrac><mfrac><mrow><mi>n</mi><mo>(</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>n</mi><mo>(</mo><mi>S</mi><mo>)</mo></mrow></mfrac></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mi>n</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>n</mi><mo>(</mo><mi>B</mi><mo>)</mo></mrow></mfrac></mtd></mtr></mtable></math>

Similarly, P(B/A) = Probability of occurrence of B, if A has already occurred and P(B) ≠ 0

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnspacing="0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em" columnalign="right left right left right left right left right left right left"><mtr><mtd><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mfrac></mtd></mtr><mtr><mtd><mo>=</mo><mfrac><mrow><mi>n</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>n</mi><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mfrac></mtd></mtr></mtable></math>

Examples:

  1. Two dice are thrown simultaneously. The probability of obtaining a total score of seven is

Solution:

Two dice are thrown then we have 6 × 6 exhaustive cases

So, n = 36.

Let A be the event of “total score of 7”

When 2 dice are thrown then A = [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)].

Total number of favourable cases = 6

Therefore, P(A) 

 2. The probability of getting head and tail alternately in three throws of a coin (or a throw of three coins), is

Solution:

Total probable ways = 8

Favourable number of ways = HTH, THT

 Required probability 

 3. Suppose six coins are tossed simultaneously. Then the probability of getting at least one head is:

Solution: Given six coins are tossed, then the total no. of outcomes = (2)6 = 64

Now, probability of getting no head 

Probability of getting at least one head 

 4. A dice is thrown twice. The probability of getting 4,5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw is

Solution:

Let P (A) be the probability of the event of getting 4, 5 or 6 in the first throw

Let P (B) be the probability of the event of getting 1, 2, 3 or 4 in the second throw,

Then P (A and B) = P(A). P(B) 

 5. A coin is tossed and a dice is rolled. The probability that the coin shows the tail and the dice shows 5 is

Solution:

 Probability of getting a tail on tossing a coin (P1

 Probability of getting a five on rolling a dice (P2

These two events are independent.

Required Probability 


3 Probability
N/A

Probability means the chances of happening/occurring of an event. So, in this chapter we discuss about the predictability of an event to happen/occur. We usually predict about many events based on certain parameters.

For example:

  • Getting a head or tail, when a coin is tossed
  • Getting a number from 1 to 6, when a die is rolled

The better we know about the parameters related to an event better will be the accuracy of the result predicted.

Mathematically, we can say that probability of happening an event is equal to the ratio of number of favourable outcomes to number of possible outcomes.

It is represented as shown below Probability happening of an event P

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mtext>&#xA0;Number of favourable outcomes&#xA0;</mtext><mtext>&#xA0;Total number of possible outcomes&#xA0;</mtext></mfrac></math>

Terms Related to Probability

 Various terms related to probability are as follows

Experiment:

An action where the result is uncertain even though the all-possible outcomes related to it is known in advance. This is also known as random experiment, e. g., Throwing a die, tossing a coin etc.

Sample Space

A sample space of an experiment is the set of all possible outcomes of that experiment. It is denoted by S.

 For example: If we throw a die, then sample space S = {1, 2, 3, 4, 5, 6} If we toss a coin, then sample space S = {Head, Tail}

Possible outcomes

All possibilities related to an event are known as possible outcomes.

Tossing a Coin When a coin is tossed, these are two possible outcomes.

 So, we say that the probability of getting H is 1/2 or the probability of getting T is 1/2,

Throwing a Die When a single die is thrown, there are six possible outcomes 1, 2, 3, 4, 5 and 6.

The probability of getting any one of these numbers is <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>6</mn></mfrac></math>

Event

Event is the single result of an experiment, e. g., Getting a head is an event related to tossing of a coin.

Types of Events

Various types of events are as follows

Certain and Impossible Events

A certain event is certain to occur, i.e., S (sample space) is a certain event.

Probability of certain event is 1, i.e., P(S) = 1.

An impossible event has no chance of occurring.

Probability of impossible event is 0, i. e., P (0) = 0.

Equally Likely Events

  • Events related to an experiment are said to be equally likely events, if probability of occurrence of each event is same.

 For example:

When a dice is rolled the possible outcome of getting an odd number = possible outcome of getting an even number = 3.

So, getting an even number or odd number are equally likely events.

Complement of an Event

  • The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A.
  • The complement of event A is represented by A (read as A bar).

The probability of complement of an event can be found by subtracting the given probability from <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><mi>P</mi><mo>(</mo><mover accent="true"><mi>A</mi><mo stretchy="false">&#xAF;</mo></mover><mo>)</mo><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mi>P</mi><mo>(</mo><mi>A</mi><mo>)</mo></math>

 

Mutually Exclusive and Exhaustive events:

 

  • Mutually exclusive: refers to two (or more) events that cannot both occur when the random experiment is formed. A ∩ B = Ø

 

Two events, A and B, are said to be mutually exclusive if the occurrence of A prohibits the occurrence of B (and vice versa)

 

  • Exhaustive Events: Two events, A and B, are said to be exhaustive if at least one of them will definitely occur.

 

Dependent Events:

 

  • Two events are called dependent, if the outcome or occurrence of the first affects the outcome or occurrence of the second, so that the probability is changed.

 

Independent Events

 

  • Two events A and B are called independent, if occurring or non-occurring of A does not affect the occurring or non-occurring of B.
  • If A and B are independent events, then P (A and B) = P (A B) = P(A) -P (B)

 

For example: getting head after tossing a coin and getting a 5 on a rolling single 6-sided die are independent events.

 

Rules/Theorems Related to Probability

 

The various theorems related to probability are discussed below

 

Addition Rule of Probability:

 

When two events A and B are mutually exclusive, the probability that A or B will occur, is the sum of the probability of each event.

 

 P (A or B) = P(A) + P(B) and P (A U B) = P (A) + P (B)

 

But when two events A and B are non-mutually exclusive, the probability that A or B will occur, is

 

 P (A or B) = P (A) + P(B) – P (A and B)

 

 P (A U B) = P(A) + P(B)- P (A B).

 

Multiplication Theorem of Probability

 

When two events A and B are mutually exclusive, the probability that A and B will occur simultaneously is given as P (A B) = P(A) * P (B/A) and P (A B) = P(A) * P(B) (A and B are independent event).

 

Law of Total Probability:

 

The rule states that if the probability of an event is unknown, it can be calculated using the known probabilities of several distinct events.

 

Mathematically, the total probability rule can be written in the following equation:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>(</mo><mi>A</mi><mo>)</mo><mo>=</mo><munder accentunder="false"><mo>&#x2211;</mo><mi>n</mi></munder><mo>&#x200A;</mo><mi>P</mi><mfenced separators="|"><mrow><mi>A</mi><mo>&#x2229;</mo><msub><mi>B</mi><mi>n</mi></msub></mrow></mfenced></math>

 

Where: n = number of events

 

Bn = the distinct event.

 

For Example: There are three events: A, B, and C. Events B and C are distinct from each other while event A intersects with both events.

 

We do not know the probability of event A.

 

 However, we know the probability of event A under condition B and the probability of event A under condition C.

 

 The total probability rule states that by using the two conditional probabilities, we can find the probability of event A.

 

Conditional Probability

 

The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already been occurred.

 

The notation for conditional probability is P (B/A), It is pronounced as the probability of happening of an event B given that A has already been happened.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mfenced separators="|"><mfrac><mi>A</mi><mi>B</mi></mfrac></mfenced><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mi>B</mi><mo>)</mo></mrow></mfrac><mtext>&#xA0;and&#xA0;</mtext><mi>P</mi><mfenced separators="|"><mfrac><mi>B</mi><mi>A</mi></mfrac></mfenced><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>&#x2229;</mo><mi>B</mi><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mfrac></math>

Points to be noted while we find probability of cards

 

i)                 So, there are 13 cards of each suit Clubs, Diamonds, Hearts and Spades.

 

ii)                There are 4 Aces, 4 Jacks, 4 Queens and 4 Kings.

iii)              There are 26 red and 26 black cards.

iv)               There are 12 face cards

 

Examples:

1.    A coin is tossed and a single 6-sided die is rolled. Find the probability of getting the head side of the coin and getting a 3 on the die

Solution: Probability of getting a head when a coin is tossed <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>2</mn></mfrac></math>

Probability of getting a 3 when a die is rolled = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>6</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Now, the required probability that both occurs at the same time&#xA0;</mtext><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>&#xD7;</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>12</mn></mfrac></math>

 

2.     A Mathematics teacher conducted two tests in her class. 25% of the students passed both tests and 42% of the students passed the first test. What per cent of the students passed the second test given that they have already passed the first test?

 

Solution:

 

This problem describes a conditional probability, since it asks us to find the probability that the second test was passed given that the first test was passed.

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;This can be solved bv multiplication rule.&#xA0;</mtext><mi>P</mi><mfenced separators="|"><mfrac><mi>B</mi><mi>A</mi></mfrac></mfenced><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mtext>&#xA0;and&#xA0;</mtext><mi>B</mi><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mfrac></math>

According to the formula,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>(</mo><mtext>&#xA0;second/first&#xA0;</mtext><mo>)</mo><mo>=</mo><mfrac><mrow><mi>P</mi><mo>(</mo><mtext>&#xA0;first and second&#xA0;</mtext><mo>)</mo></mrow><mrow><mi>P</mi><mo>(</mo><mtext>&#xA0;first&#xA0;</mtext><mo>)</mo></mrow></mfrac><mo>=</mo><mfrac><mn>0.25</mn><mn>0.42</mn></mfrac><mo>=</mo><mn>0.60</mn><mo>=</mo><mn>60</mn><mi mathvariant="normal">%</mi></math>

3.    A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a king or a club?

 Sol. There are 4 kings in a standard deck and 13 club cards.

Also 1 king is of club, so probability of getting a king = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>4</mn><mn>52</mn></mfrac></math>

Probability of getting a club = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>13</mn><mn>52</mn></mfrac></math>

Probability of getting a king club = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mn>52</mn></mfrac></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore, required probability of getting a king or club&#xA0;</mtext><mo>=</mo><mfrac><mn>4</mn><mn>52</mn></mfrac><mo>+</mo><mfrac><mn>13</mn><mn>52</mn></mfrac><mo>&#x2212;</mo><mfrac><mn>1</mn><mn>52</mn></mfrac><mo>=</mo><mfrac><mrow><mn>4</mn><mo>+</mo><mn>13</mn><mo>&#x2212;</mo><mn>1</mn></mrow><mn>52</mn></mfrac><mo>=</mo><mfrac><mn>16</mn><mn>52</mn></mfrac><mo>=</mo><mfrac><mn>4</mn><mn>13</mn></mfrac></math>

4.    A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability that it is yellow or green?

Solution: Total marble = 1 + 3 + 2 + 4 = 10, i.e., n(s) = 10

Now, probability of getting a yellow marble = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>4</mn><mn>10</mn></mfrac></math>

Probability of getting a green marble = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>3</mn><mn>10</mn></mfrac></math>

 

Since, the events are mutually exclusive

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Therefore,&#xA0;</mtext><mi>P</mi><mtext>&#xA0;(yellow or green&#xA0;</mtext><mo>)</mo><mo>=</mo><mi>P</mi><mo>(</mo><mtext>&#xA0;yellow&#xA0;</mtext><mo>)</mo><mo>+</mo><mi>P</mi><mo>(</mo><mtext>&#xA0;green&#xA0;</mtext><mo>)</mo><mo>=</mo><mfrac><mn>4</mn><mn>10</mn></mfrac><mo>+</mo><mfrac><mn>3</mn><mn>10</mn></mfrac><mo>=</mo><mfrac><mn>7</mn><mn>10</mn></mfrac></math>

5.    A person can hit a target 4 out of 7 shots. If he fixes 10 shots, what is the probability that he hit the target twice?

 

Solution: Here, n = 10 and r = 2

<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>&#xA0;Success,&#xA0;</mtext><mi>p</mi><mo>=</mo><mfrac><mn>4</mn><mn>7</mn></mfrac><mspace linebreak="newline"/><mtext>&#xA0;Failure,&#xA0;</mtext><mi>q</mi><mo>=</mo><mn>1</mn><mo>&#x2212;</mo><mfrac><mn>4</mn><mn>7</mn></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>7</mn></mfrac><mspace linebreak="newline"/><mi>P</mi><mo>(</mo><mtext>&#xA0;hit the target twice&#xA0;</mtext><mo>)</mo><msup><mo>=</mo><mn>10</mn></msup><msub><mi>C</mi><mn>2</mn></msub><msup><mfenced separators="|"><mfrac><mn>4</mn><mn>7</mn></mfrac></mfenced><mn>2</mn></msup><msup><mfenced separators="|"><mfrac><mn>3</mn><mn>7</mn></mfrac></mfenced><mn>8</mn></msup><mspace linebreak="newline"/><mo>=</mo><mfrac><mrow><mn>10</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo><mo>(</mo><mn>10</mn><mo>&#x2212;</mo><mn>2</mn><mo>)</mo><mo>!</mo></mrow></mfrac><msup><mfenced separators="|"><mfrac><mn>4</mn><mn>7</mn></mfrac></mfenced><mn>2</mn></msup><msup><mfenced separators="|"><mfrac><mn>3</mn><mn>7</mn></mfrac></mfenced><mn>8</mn></msup><mspace linebreak="newline"/><mo>=</mo><mn>45</mn><mfrac><mrow><mo>(</mo><mn>4</mn><msup><mo>)</mo><mn>2</mn></msup><mo>(</mo><mn>3</mn><msup><mo>)</mo><mn>8</mn></msup></mrow><msup><mn>7</mn><mn>10</mn></msup></mfrac></math>

 


1 Venn Diagram
N/A

A Venn diagram is a diagram that helps us visualize the logical relationship between sets and their elements.

It shows logical relations between two or more sets

Venn diagrams are also called logic or set diagrams

They are widely used in set theory, logic, math, teaching, business data science and statistics.

 A Venn diagram typically uses circles other closed figures can also be used to denote the relationship between sets.

In general, Venn diagrams shows how the given items are similar and different

In Venn diagram 2 or 3 circles are most used one, there are many Venn diagrams with larger number of circles (5, 6, 7, 8, 10….).

Union: When two or more sets intersect, all different elements present in sets are collectively called as union.

It is represented by U

Union includes all the elements which are either present in Set A or set B or in both A and B

i.e., A ∪ B = {x: x ∈ A or x ∈ B}.

The union of set corresponds to logical OR

For example: If we have A = {1, 2, 3, 4, 5} and B = {3, 5, 7}

A U B = {1, 2, 3, 4, 5, 7}

Intersection: When two or more sets intersect, overlap in the middle of the Venn diagram is called intersection.

This intersection contains the common elements in all the sets that overlap.

It is denoted by ∩

All those elements that are present in both A and B sets denotes the intersection of A and B. So, we can write as A ∩ B = {x: x ∈ A and x ∈ B}.

The intersection of set corresponds to the logical AND

For example: If we have A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}

A ∩ B = {4, 5}

Cardinal Number of Set:

The number of different elements in a finite set is called its cardinal number of a set

It is denoted as n(A)

A = {1, 2, 3, 4, 5, 7}

n(A) = 6

Formula:

  • n (A ∪ B) = n(A) + n(B) – n (A ∩ B)
  • n (A ∪ B ∪ C) = n(A) + n(B) + n(C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C)

Examples:

  1. Out of 20 boys, 5 boys have ice-cream only and 10 boys have chocolate only, 3 boys have both chocolate and ice-cream, how many boys has only one of ice-cream or chocolate.

Solution: Given total boys= 20

Number of boys having ice-cream = 5

number of boys having chocolate = 10

number of boys who has only one of ice-cream or chocolate = 5 + 10 = 15

  1. In a class of 60 students, 21 play Tennis, 13 play Cricket and 14 play Basketball. 6 plays both Tennis and Cricket, 5 play Cricket and Basketball and 7 play Tennis and Basketball. If 22 students do not play any of these given sports, how many students play exactly two of these sports?

Solution: Given Total = 60;

 T = 21, C=13, and B=14; 

T ∩ C=6, C ∩ B = 5, and T ∩ B = 7.

Neither=19.

[Total] = Tennis + Cricket + Basket Ball – (TC+CB+TB) + (All three) + (Neither)

55 = 21 + 13 + 14 - (6+5+7) + (All three) + 22

(All three) = 3;

Students play only Tennis and Cricket are 6-3=3;

Students play only Cricket and Basketball are 5-3=2;

 Students play only Tennis and Basketball are 7-3 = 4;

 Hence, 3 + 2 + 4 = 9 students play exactly two of these sports.

 3. Last month 30 students of a certain college travelled to Egypt, 30 students travelled to India, and 36 students travelled to Italy. Last month no students of the college travelled to both Egypt and India, 10 students travelled to both Egypt and Italy, and 17 students travelled to both India and Italy. How many students of the college travelled to at least one of these three countries last month?

Solution: Given

Students travelled to Egypt n(A) = 30

Students travelled to India n(B) = 30

Students travelled to Italy n(C) = 36

Egypt and India travellers n (A∩ B) = 0

Egypt and Italy travellers n (A ∩ C) = 10

India and Italy travellers n (B ∩ C) = 17

From all the information we can determine that 0 people travelled to all 3 countries because 0 people travelled to both Egypt and India.

To know how many students travelled to at least one country,

Total travellers = Egypt + India + Italy - sum of (travelled exactly two countries) - 2 times (travelled all three countries)

Total travellers = 30 + 30 + 36 - (10 + 17 + 0) - 2(0)

Total travellers = 96 - 17 - 0 = 69

Thus, 69 people travelled to at least one country.

 4. Each person who attended a conference was either a client of the company, or an employee of the company or both. If 56 percent of these who attended the conference were clients and 49 percent were employees. What percent were clients, who were not employees?

Solution: Total = Stockholders + Employees - Both;

100 = 56 + 49 – Both

 Both = 5;

Percent of clients, who were not employees is: Clients - Both = 56 - 5 = 51.

 5. There are 45 students in PQR College. Of these, 20 have taken an accounting course, 20 have taken a course in finance and 12 have taken a marketing course. 7 of the students have taken exactly two of the courses and 1 student has taken all three of the courses. How many of the 40 students have taken none of the courses?

Solution:  Given Total= 45; Let P = 20, Q = 20, and R = 12; sum of EXACTLY 2 - group overlaps = 7;

P ∩ Q ∩ R =1;

Total = P + Q + R – (sum of exactly 2 – group overlaps) – 2 * P ∩ Q ∩ R + None

45 = 20 + 20 + 12 – 7 – (2 * 1) + none

None = 2


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