Ever feel like your mind is a flickering candle in a vast, intellectual ocean? Do you long to unlock your full cognitive potential, but feel lost in the sea of test prep options? Well, ditch the rusty anchors and set your sails! IQ Foundations Prep Course is your compass, guiding you on a thrilling voyage to unleash the hidden power within your mind.
Forget dusty textbooks and endless drills. This course is your cognitive playground, where learning is an adventure, not a chore. We'll take you on a journey of discovery, illuminating the fundamental building blocks of intelligence and equipping you with the tools to:
Unravel intricate puzzles, dismantle fallacies with a smile, and become a champion of clear thinking.
Transform your mind into a lightning-fast processor, effortlessly absorbing and recalling information with laser-like precision.
Learn to dissect arguments, identify hidden biases, and uncover the truth like a seasoned detective.
Unleash your inner wordsmith, express yourself with articulate eloquence, and impress everyone with your vocabulary prowess.
Tackle any challenge with a strategic mind, approach obstacles with creative solutions, and leave everyone in awe of your ingenuity.
IQ Foundations Prep Course is more than just test scores. It's about:
Ace those exams, impress your professors, and unlock your full potential as a learner.
Think smarter, communicate more effectively, and stand out in the competitive job market.
Make informed decisions, analyze information with clarity, and gain a deeper understanding of the world around you.
Believe in your intellectual abilities, embrace your potential, and shine brightly in any intellectual arena.
This course is your launchpad for a brighter, more empowered future. You'll experience:
★ Interactive exercises and quizzes: Test your progress, identify areas for improvement, and solidify your learning through active engagement.
★ Personalized study plans: Chart your own course based on your needs and goals, maximizing your time and effort.
★ A supportive community: Connect with fellow learners, share insights, and motivate each other on your journey to cognitive excellence.
Ready to set your mind ablaze with the fire of intellectual growth? Enroll in IQ Foundations Prep Course today and embark on a transformative journey of self-discovery. Remember, the only limit is the one you set for yourself. Let's unlock your hidden potential, one brilliant spark at a time!
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What is BobPrep?
Unsure how to get begin your journey? Hesitant about spending hundreds if not thousands of dollars on 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 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 tutors is now available to all.
Foundations:
Foundations is for students who need to review the mechanics of the 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 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 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 score of up to 550.
Ever feel like your mind is a flickering candle in a vast, intellectual ocean? Do you long to unlock your full cognitive potential, but feel lost in the sea of test prep options? Well, ditch the rusty anchors and set your sails! IQ Foundations Prep Course is your compass, guiding you on a thrilling voyage to unleash the hidden power within your mind.
Forget dusty textbooks and endless drills. This course is your cognitive playground, where learning is an adventure, not a chore. We'll take you on a journey of discovery, illuminating the fundamental building blocks of intelligence and equipping you with the tools to:
Unravel intricate puzzles, dismantle fallacies with a smile, and become a champion of clear thinking.
Transform your mind into a lightning-fast processor, effortlessly absorbing and recalling information with laser-like precision.
Learn to dissect arguments, identify hidden biases, and uncover the truth like a seasoned detective.
Unleash your inner wordsmith, express yourself with articulate eloquence, and impress everyone with your vocabulary prowess.
Tackle any challenge with a strategic mind, approach obstacles with creative solutions, and leave everyone in awe of your ingenuity.
IQ Foundations Prep Course is more than just test scores. It's about:
Ace those exams, impress your professors, and unlock your full potential as a learner.
Think smarter, communicate more effectively, and stand out in the competitive job market.
Make informed decisions, analyze information with clarity, and gain a deeper understanding of the world around you.
Believe in your intellectual abilities, embrace your potential, and shine brightly in any intellectual arena.
This course is your launchpad for a brighter, more empowered future. You'll experience:
★ Interactive exercises and quizzes: Test your progress, identify areas for improvement, and solidify your learning through active engagement.
★ Personalized study plans: Chart your own course based on your needs and goals, maximizing your time and effort.
★ A supportive community: Connect with fellow learners, share insights, and motivate each other on your journey to cognitive excellence.
Ready to set your mind ablaze with the fire of intellectual growth? Enroll in IQ Foundations Prep Course today and embark on a transformative journey of self-discovery. Remember, the only limit is the one you set for yourself. Let's unlock your hidden potential, one brilliant spark at a time!
Enroll today and join the revolution!
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
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
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:
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
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
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
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
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
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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