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:
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.
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.
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.
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.
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.
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.
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:
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.
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.
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.
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.
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.
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.
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:
Solution:
2.
Solution: 393 x 197 + 5600
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?
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
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)
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:
Solution: Let the required price be $ x.
Then, less toys, less cost (Direct Proportion)
Therefore, 6: 5:: 264.37: x
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
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:
Therefore, 1 * 40 * x = 40 * 1 * 40
x = 40
Solution:
5. If of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?
Solution:
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:
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.
Dividend is paid annually as per share or as a percentage.
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
NOTE:
Examples:
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
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
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
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:
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
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:
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 =
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
According to the formula, Average speed =
So, the average speed of the person for the whole journey is 24 miles/h
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
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:
Solution: Given
... A can give (100 - 75) = 25 points to B
Solution:
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
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:
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:
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
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:
Solution:
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:
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:
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
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:
Important Things:
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:
Examples:
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
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
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
Solution:
Since Bob is moving upwards, 90e will be added
Since Bob is moving downwards, 150e will be subtracted
By equating the number of steps,
60s + 60e = 180s -180e
s = 2e
N=80a+80x= 80a+16a=96a Time taken when the escalator is stationary= 96a/a= 96seconds.
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
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:
If two objects are moving with speeds X and Y, their relative speed is X – Y, if they are moving in the same direction
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:
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
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
Solution: In an hour, the earlier racer covers a distance of 5 km.
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.
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
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
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
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
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.
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
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:
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
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
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:
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
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
Solution: here ten’s digit of base is even and exponent is odd
So, last two digits of 145157 will be 25
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
Number of factors of a number:
If a number N can be represented as 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.
544/8 = 68, divisible by 2, 4 and 17.
Take the highest power of each prime
68/ (4 * 17) = 1.
544 = 8 * 4 * 17 = 25 * 17
Number of factors = 6 * 2 = 12
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.
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 =
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
Sum of odd factors of a number
Number of even factors of a number
Let a number is of the form, N = 2a x ...
For e.g., 1080 = 23 * 33 * 5
Number of even factors of 1080 = 3 * (3 + 1) * (1 + 1) = 24
Sum of even factors of a number
For e.g., 1080 = 23 * 33 * 5
Sum of even factors of 1080 = (21 + 22 + 23) * (30 + 31 + 32 + 33) * (50 + 51) = 3360
Other conditions
What if we are asked to find the sum and number of factors of 1080 which are divisible by 15?
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:
= (20 + 21+ 22+ 23) * (31+ 32+ 33) * (51)
= 15 * 39 * 5 = 2730
2. Find the sum of factors of 544 which are perfect squares
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
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
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
Total sum – sum of divisors which are perfect squares = 2340 – 130 = 2210
Total factors – number of divisors which are perfect squares = 16 – 4 = 12
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
100 = 22 * 52
We need to remove all factors that cannot yield a 22 from two’s bracket and also remove all factors that cannot yield a 52 from 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
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:
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 =
Here, first number = 4 and last number = 31
Therefore, required average =
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 =
Here, the last odd number = 45
Therefore, Average of 1 to 45 odd numbers =
Again, according to the formula,
Average of 1 to n even numbers =
Here, the last odd number = 44
Therefore, Average of 1 to 44 odd numbers =
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
And average weight of students of sections B and C=60kg
50b + 72c = 60b + 60c
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
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.
Conversion of fraction into Percentage:
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
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,
Formula to calculate Per Cent:
Some quick Results:
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,
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%.
Now, the salary is reduced by 5% after the increase.
Therefore, required change is a decrease i.e., 100 – 99.75 = 0.25
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%
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:
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.
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:
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%
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%
Now, CP = $ 2500 and gain = 20%
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
Therefore, Cost Price of 100 eggs = 2.5 x 100 = $ 250
Now, let the SP of 100 eggs be $ u.
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.
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
Therefore, profit percentage = 20%
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
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
Basic Formulae Related to Discount:
Examples:
Sol: Given r1 = 10% and r2 = 5%
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
Now, second discount = 1200 – 1104 = $ 96
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
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
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
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
Basic Formulae Related to Simple Interest:
Where, SI = Simple Interest, P = Principal, R = Rate of interest, T = Time, A = Amount.
Things to remember:
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
Where, A = Total amount paid;
x = Value of each instalment.
Where, P is the principal n is the number of instalments R is the rate of interest
Examples:
Sol: Let the sum be $100
Now, principal becomes 100 + 5 = 105
Hence, amount at the end of 1 year = 105 + 5.25 = $ 110.25
Therefore, Effective SI = 110.25 – 100 = $ 10.25%
Sol: Given, t = 3 yr, r = 7%, P = $ 8750
SI = $ 1837.50
Sol: According to the question,
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)
Case II:
P = $ p, T = 5 yr
R = R%, SI = $(1040 – p)
From Eq (1) and (2), we get
Therefore, p = $ 800
Sol: Let the rate of interest allowed by bank be r%
According to the question,
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
Compound interest = Amount – Principal
5. If rates of interest are R1%, R2% and R3% for 1st, 2nd and 3rd years respectively, then
Instalments:
When a borrower pays the sum in parts, then we say that he/she is paying in instalments.
x = value of each instalment
n = Number of instalments
Examples:
Sol: Given, R = 4%, n = 2 years and A = $ 169 and P =?
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,
New rate = 5 + 5 = 10%
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?
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,
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
Population at the end of 3rd year = $ 5720
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:
2. If rate of interest is R% and money due for amount is A after T yr, then
3. If money due A, rate of interest R% and time T are given, then
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
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
Examples:
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 =?
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
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 =?
TD = A – PW = 4840 – 4000 = $ 840
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.
Type of Ratio:
The different types of ratio are explained as under
For example: Duplicate ratio of 3 :4 = 3 :4 =9:16
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
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.
Note:
Comparison of Ratios:
Rules used to compare different ratios are as follows
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:
3. Mean proportional between a and b is . If mean proportional is x, then a: x :: x: b
Examples:
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.
Sol: Let Y’s salary = 100
Therefore, X’s salary = 80
Therefore, required ratio = 80: 100: 96 = 20: 25: 24
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
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
Where, x = total amount a = 3, b = 50
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.
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:
Solution: 40% sugar is in 600g of sugar solution.
Let x g sugar be added.
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.
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,
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
x = 60 L
Sol. Let the quantity of milk and water in initial mixture be 7x and 5x L.
Therefore, quantity of water in initial mixture = 5 x 5 = 25 L
And quantity of water in new mixture = 25 + 15 = 40 L
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:
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
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:
Solution: P’s share: Q’s share = Ratio of their investments
= 13000: 25000
= 13: 25
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
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
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
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:
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,
Note:
Mostly questions on ages can be solved with the use of linear equations.
Examples:
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,
Therefore, Kevin’s present age = 5 x 10 = 50 yr and Steve’s present age = 10 yr
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
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
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
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
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
For example:
The time taken by a vehicle in covering a certain distance is inversely proportional to the speed of the vehicle.
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
Solution: Let the required number of days be ‘d’.
More men, less days (Indirect proportion)
Less hours, more days (Indirect proportion)
Therefore, Required number of days = 30
Solution: Let the required number of days be ‘d’.
Less work, Less days (Direct proportion)
Solution: For 50, food is sufficient for 45 days.
Therefore, for 1 student food is sufficient for 45 x 50 days
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
Solution: Let number of toys be x.
More hours, more toys (Direct proportion)
2: 80:: 1: x
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:
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,
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
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
According to the formula, M1 D1 W2 = M2 D2 W1
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?
After 2 day’s A left the work
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:
Examples:
Solution: Time taken by Alex = 6 days
time taken by Kevin = 5 days
Total amount earned = $660
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?
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
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?
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
Now, they worked for 10 days.
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:
Examples:
Solution: Let the leak empties full tank in x h, then part emptied in 1 h by leak =
Therefore, leak will empty the full tank in 20 h.
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
Therefore, Tap B can fill the tank in 120 min.
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
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.
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
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.
6. Angle covered by hour hand in 1 min.
7. From point 5 and 6, we can say that the minute hand goes ahead by 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:
Solution: Difference of time between 4.00 am on Monday to 7.00 pm Tuesday
24 + 12 + 3 = 39 h
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
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
4. At what point of time after 3 O' clock, hour hand and the minute hand are at right angles for the first time?
Given that, n = 3
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.
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.,
Odd Days
Day Gain/Loss
Ordinary Year (± 1 day)
For example: 9th August 2013 is Friday, then 9th August 2014 has to be Friday +1 = Saturday.
Leap Year (+ 2 days)
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.
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:
Solution: Number of days in x weeks = 7x + x
... Total number of days is x week x days = 7x + x = 8x days
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.
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.
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
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.
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:
Examples:
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.
Solution:
Given that,
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
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:
According to the question,
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:
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.
Rules of inequalities:
Rule1:
Rule2:
Rule3:
Rule4:
Rule5:
Having minus in front of a and b changes the direction of the inequality.
Rule6:
The reciprocal of both a and b changes the direction of the inequality.
When a and b are both positive or both negative.
Rule7:
Taking a square root will not change the inequality.
Solving inequalities:
To solve an inequality, the following steps:
Important points:
It is an inequality with an absolute value symbol in it and can be solved by two methods.
The absolute value inequalities are of two types:
Formulae
Formulae to solve the inequality:
If |x|< a => -a < x < a
If |x|≤ a => -a ≤ x ≤ a
Formulae to solve the inequality:
If |x|> a => x < - a or x > a
If |x|≥ a => x ≤ - a or x ≥ a
|x|< −a or |x|≤ −a ⇒ No Solution
|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 +
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
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:
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:
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
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
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:
7.54 x 1022 < 8.432 x 1024
7.54 x 1012 > 7.54 x 1010
7.54 x 1012 > 7.54 x 1010
Important points to remember:
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:
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
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
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:
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,
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:
Inconsistent System:
The above system will be inconsistent, if and do not have any solution. It represents a pair of parallel lines.
Examples:
2x – 3y + 1 = 0, 3x + 4y – 5 = 0
Solution: Given 2x – 3y + 1 = 0, 3x + 4y – 5 = 0
By cross multiplication method,
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.
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
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.
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
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
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
Example: If a = 12 and b =11, then (122 -112) = (12 + 11) (12-11) = 23
Important Relations:
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.
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:
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
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
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.
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 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
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
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)
Properties of Cubes and Cube Roots
Example Problems:
2. What least number should be subtracted from 6862, so that 19 be the cube root of the result from this subtraction?
Solution: Given,
Solution:
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
Therefore, cube root of 1259712 = 108
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:
Surds
Numbers which can be expressed in the form √p + √q, where p and q are natural numbers and not perfect squares.
Note
1. All surds are irrational numbers
2. All irrational numbers are not surds
Order of Surds
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
Example: 2√2, 3√2, 4√2
Example: 2√2, 2√3, 2√5
Properties of Surds
Example: a + b ≠ √c or √a – b ≠ √c
To Arrange the Surds in Increasing or Decreasing 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:
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:
Solution: Given, ax = b, by = c and xyz = 1
Let us take b = ax
Solution: LCM of 4, 6, 12 = 12
5. If 3x – 3x-1 = 18, then xx is equal to?
Solution: Given 3x – 3x-1 = 18
Therefore, xx = 33 = 27
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.
Note:
Important Points Related to Quadratic Equations:
=> (x – p) (x – q) = 0
Examples:
Solution:
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
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
According to the formula, a3 + b3 + c3 = 3abc, if a + b + c = 0
From this p3 + q3 + r3 = 3pqr
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
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
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
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
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.
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
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
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
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
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
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.
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
Cylinder
Solid Cylinder:
Hollow Cylinder:
If cylinder is hollow, then
where, R = External radius of base, r = Internal radius of base and h = Height
Examples:
Volume of iron rod = 0.88 m3
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?
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
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
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.
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.
Frustum of Cone:
Sphere:
Hollow Sphere or Spherical Shell:
Hemisphere:
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.
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
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.
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.
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.
Prism:
Pyramid:
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
The sides of base are 5s = 5 x 2 = 10 cm, 12s = 12 x 2 = 24 cm, 13s = 13 x 2 = 26 cm.
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)
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
On putting, these values in formula,
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
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
It can be broadly divided into two parts
Plane Geometry
Solid Geometry
Point:
A figure of which length, breadth and height cannot be measured is called a point. It is infinitesimal.
Line:
Plane:
It is a flat surface having length and breadth both but no thickness. It is a 2-dimensional figure.
Parallel Lines:
Transversal Lines:
Angle:
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:
Internal angle bisector:
Here, two angles are formed ∠AOC and ∠BOC. Both angles are equal (θ) because OB is the internal angle bisector.
External angle bisector:
Here, ∠A'OB and ∠BOC are equal and external bisector is OB.
Examples:
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 = 3∠p, then ∠p is equal to
∠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
Triangle:
A triangle is a three - sided closed plane figure which is formed by joining 3 non-collinear points.
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:
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.
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.
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.
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.
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.
Similarity of Triangles
Two triangles are said to be similar if they satisfy the following conditions.
Properties of Similar Triangle:
Some Important Terms Related to Triangles:
Angle Bisector:
In ,AD, BE and CF are angle bisectors and meet at incentre J.
Perpendicular Bisector:
Median:
Examples:
Solution: Let the angle measure (3x) °, (5x) ° and (7x) °
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
∠CBD = ∠CDB…(i)
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:
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
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.
(i) AC2 + 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.
(i) The 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,
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
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.
(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
(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.
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
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.
2. Trigonometrical Identities:
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:
Solution:
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)
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
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.
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
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
Properties of similar triangles:
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
Examples:
Solution: A(ΔPQR)/A(ΔDEF) = PQ2/DE2 = 36/49
Solution: Given Coordinates of the vertices
Calculate the ratios of the lengths of corresponding sides
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.
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
Area:
Perimeter:
Triangle:
Equilateral Triangle:
It has all three sides equal and each angle equal to 60o
Isosceles Triangle:
It has any two sides and two angles equal and altitude drawn on non-equal side bisects it.
where, a = Each of two equal sides b = Third side
Scalene Triangle:
It has three unequal sides.
Right Angled Triangle:
It is a triangle with one angle equal to 90°
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.
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
10. The area of the triangle formed by joining the mid-points of the sides of a given triangle is of the area of the given triangle
Examples:
Sol. Given, perimeter of an equilateral triangle is 51 cm.
Let each side of triangle be a cm, then sum of sides = 51 cm
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
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
Hence it is a right-angled triangle
So, the triangular filed is right angled at B.
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
= 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
Quadrilateral:
Square:
It is a parallelogram with all 4 sides equal and each angle is equal to 90°.
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
4. If area of two squares is in the ratio of A1: A2 then ratio of their perimeter is given
Rectangle:
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.
Note: Opposite angles are equal in a parallelogram but they are not right angle.
Properties of Parallelogram:
Trapezium
It is a quadrilateral with any one pair of opposite sides parallel
Where a and b are parallel sides and h is the height or perpendicular distance between a and b.
Rhombus:
Where, a = side, d1 and d2 are diagonals.
Properties of Rhombus:
Examples:
Solution: Given area of square = 289 sq m.
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 =?
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
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.
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?
perimeter of rhombus = 10 cm
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 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.
Circular Ring:
(i) Area = R2 – r2
(ii) Difference in circumference of both rings =
Examples:
Solution: According to the question, Area of semi-circle = 77 m
Solution: Let original radius be r.
Therefore, original radius of the circle = 3 cm
3. The ratio of the areas of the circumcircle and the incircle of a square is
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
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
Volume:
Surface Area:
Cube:
where, a is Side (edge) of the cube
Cuboid:
where, = Length, b - Breadth and h – Height
Some other cube or cuboidal shaped object are as follows.
Room
where, = Length, b = Breadth and h = Height
Box
= 2 x (l+ b) x h + l x b
Note For calculation of any of the parameter, length, breadth and height should be in same unit.
Examples:
Solution:
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
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
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
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.
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:
Solution. Let the number be y.
Then, according to the question,
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
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
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 =
Hence, number of pieces of chocolate left with Peter is 3
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:
Examples:
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
Solution: Given expression
= 6 – [9 – {18 – (15 – 12 + 9)}]
= 6 – [9 – {18 – 12}]
= 6 – [9 – 6]
= 6 – 3
= 3
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.
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.
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:
Examples:
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
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
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
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
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
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:
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).
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
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
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
Solution:
Required number = ((555+ 445) *2*110) + 30 = 220030
Hence option (d) is the answer
Dividend = (divisor * quotient) + Remainder.
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.
(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.
Theory:
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, ...}
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:
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.
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
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:
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.
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.
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
Examples:
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
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)
Solution: Required H.C.F = (H.C.F OF 9, 12, 18, 21)/ (L.C.M OF 10, 25, 35, 40)
= 3/2800
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.
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.
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:
Example: ½, 21/43.
Example: 17/13, 18/14,
Example: 1/ (7/9), (13/11)/17,
Example: let given function is 3/7, then inverse function is 7/3.
ii. Operations on Simple Fractions
Addition of Simple Fractions
Example: (1/4) + (2/4) = (1+2) (1/4) = (3/4)
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
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.
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.
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)
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).
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).
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:
Basic Formulae:
Examples:
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).
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.
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.
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.
Since, there is only 1 repeated digit. Therefore, only single 9 is placed in denominator.
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… =
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.
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 . 17). Now, 204 . 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 . 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.
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….
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).
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:
Median(M) = If n is odd, then
If n is even, then
Mode = The value which occurs most frequently
Standard Deviation(S) =
Where, x = observations given
= Mean
= Total number of observations
Solution: Here N = 4
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:
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:
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?
COMBINATION & PERMUTATION RULES EXPLAINED WITH EXAMPLES:
Ans-> 10!
Ans-> 9! -> There are 9 totally different books
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
Ans-> 3! There is no differentiation between the books, except the type of book they are
Ans-> 2!
Ans -> 3!
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
If it is 5 tosses = = 32 is the total sample space
Total = 8/8 = 1
For five tosses:
If it is 5 tosses = = 32 is the total sample space
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:
Pick any 3 cars
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
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.
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
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.
4. Arrangement 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:
Solution: There are five numbers and number of places to be filled up = 4
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 = = 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 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 =
Number of arrangements, so that the three vowels are never together = 20160 - 2160= 18000
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
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.
It signifies number of groups formed from n different things, when r things are taken into consideration.
Important Points:
Cases of Combination
There are several cases of Combination,
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.
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
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:
Solution: (i) When two particular members are included then, we have to select 5—2 = 3 members out of 10 – 2 = 8
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?
= 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
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.
Probability means the chances of happening/occurring of an event.
Examples:
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
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
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.
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
Similarly, P(B/A) = Probability of occurrence of B, if A has already occurred and P(B) ≠ 0
Examples:
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
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:
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
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
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
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 probability of complement of an event can be found by subtracting the given probability from
Mutually Exclusive and Exhaustive events:
Two events, A and B, are said to be mutually exclusive if the occurrence of A prohibits the occurrence of B (and vice versa)
Dependent Events:
Independent Events
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:
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.
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
Probability of getting a 3 when a die is rolled =
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.
According to the formula,
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 =
Probability of getting a club =
Probability of getting a king club =
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 =
Probability of getting a green marble =
Since, the events are mutually exclusive
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
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:
Examples:
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
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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