Analysis of INFOSYS Placement Papers

Analysis of Infosys Placement Papers

Quantitative Aptitude/Logical Reasoning

Corporate & International Relations

Amrita Vishwa Vidyapeetham

Analysis of Infosys Placement Papers Quants Funda

2 Corporate & International Relations Amrita Vishwa Vidyapeetham

TABLE OF CONTENTS

Introduction

Number System Questions taken from students’ forum

Theory, Examples & Explanations

LCM & HCF Questions taken from students’ forum


Averages Theory, Examples & Explanations

Percentages Questions taken from students’ forum


Linear Equations /Word Problems/Problems on ages Questions taken from students’ forum


Profit & Loss Questions taken from students’ forum


Ratio Proportion Questions taken from students’ forum


Time, Speed & Distance/ Boats and streams Questions taken from students’ forum


Time & Work Questions taken from students’ forum


Heights & Distances Theory, Examples & Explanations

Sets Theory, Examples & Explanations

Functions Theory, Examples & Explanations

Sequence & Series Theory, Examples & Explanations

Clocks & Calendar Questions taken from students’ forum


Geometry (2D & 3D)/ Coordinate Geometry Questions taken from students’ forum


Counting Techniques and Probability Questions taken from students’ forum


Logical Reasoning Questions taken from students’ forum


Data Interpretation/Data Sufficiency Questions taken from students’ forum


Miscellaneous Questions taken from students’ forum

Formula Booklet



The book “Analysis of Infosys

Placement Papers” is an attempt to

provide you with a mirror to previous

years’ papers. Only practicing with

questions is not enough because of its

unpredictability. Learning concepts

can make you tackle all sorts of

problems with ease. The Review of

topics is designed to familiarize you

with the mathematical skills and

concepts likely to be tested. This

material includes many definitions

and examples with solutions. Note,

however, this review is not intended

to be comprehensive. It is assumed

that certain basic concepts are

common for all examinees. Emphasis

is, therefore, placed on the more

important skills, concepts, and

definitions, and on those particular

areas that are frequently asked. If

any of the topics seem especially

unfamiliar, we encourage you to

consult appropriate mathematics

texts for a more detailed treatment

of those topics.

All the Best



INFOSYS

Quantitative Aptitude

Number System (Prime Numbers/Roman Numbers)

Questions taken from students’ forum

1. Three friends divided some bullets equally. After, all of them shot four bullets the total number of bullets remaining is equal

to the bullets each had after division. Find the original number divided. Ans: 18

2. The average of 11 prime numbers is a prime number (which is also a prime number).

What is the least prime number among them?

What is the second least prime number among them?

What is the largest prime number among them?

How many prime number are there which are greater than 20?

3. Find the value of MCMLXXXVIII? (Using I =1, V=5, X = 10, L=50, C=100, D = 500, M=1000)

4. There are three societies A, B, C. A lent some tractors to B and C as many as they had already. After some time B gave as

many tractors to A and C as many as they have. After sometime c did the same thing. At the end of this transaction, each one

of them had 24 tractors. Find the number of tractors each originally had.

Ans: A 39, B 21 and C 12

5. Make 120 with five zeros.

Ans: (0!+0!+0!+0!+0!)! = 120

6. Find a 9-digit number such that first 9 digits is divisible by 9, first 8 digits is divisible by 8, first 7 divisible by 7, and so on.

7. Find the 3-digit number whose last digit is the square root of the first digit and second digit is the sum of the other two

digits. Ans: 462

8. A person travels in a car with uniform speed. He observes the milestone, which has a two-digit number. After one hour, he

observes another milestone with same digits reversed. After another hour, he observes another milestone with same two

digits separated by 0. Find the speed of the car? Ans: 45

9. In an educational tour to a Zoo, a student told to his friend about the size and color of a snake he has seen in Zoo. It is one

of the colors brown/black/green and one of the sizes 35/45/55 units.

If it were not green or if it were not of length 35, its length is 55.

If it were not black or if it were not of length 45, it is 55.

If it were not black or if it were not of length 35, it is 55. Back to Table of Contents



What is the color and length of the snake? Ans: brown & 55

10. There is a square cabbage patch. He told his sister that I have a larger patch than last year and hence produced 211 more

cabbages than the previous year. Find the number cabbages in the field? Ans: 106*106=11236

11. The product of five different temperatures in a city is 12. If all of them are integers, then find all the temperatures.

Ans: -2, -1, 1, 2, 3

12. I have a peculiar a five-digit number. The fifth digit is one fourth of the third digit and one-half of the fourth digit. Third

digit is one- half of the first digit. Second digit is 5 more than the fifth digit. Find the number? Ans: 86421

13. There are three towns attacked by three dragons, namely X, Y, Z. Number of days X attacking town “A” is equal to number

of days Y attacking another town “B”. Number of days X attacking town “A” is equal to half the square root of number of days

Z attacking town “C”. Number of days Y attacking the town “B” is twice the square root of Number of days Z attacking town

“C”. Find the number days each dragon attacking the town.

14. John had decided to divide his RS.1000/- for his four children according to their ages. The elder child should be a RS.20/-

extra than his younger child. What will be the share of Mahesh who is the youngest?

15. A person while going through a book, he began to write the page numbers. After reading the whole book, he found that he

altogether used digit 3 for 61 times. How many pages did the book have? Ans: 300

16. Mr. Charlie comes to Mr. Wilson’s office to meet him. Mr. Charlie says, “So there must be 30 clerks working in your

office!” Mr. Wilson replies “Not in the near future. Only if there were twice the number of girls or thrice the number of men,

the strength would be 30”. Find the number of clerks presently working. Ans: 18 CLERKS. (Girls: 12, Men: 6)

17. There were two systems ‘A’ and ‘B’. The temperature 140 in ‘A’ is equivalent to 360 in system ‘B’. In addition, 1330 in ‘A’ is

equivalent to 870 in ‘B’. What is the temperature where they both r equal? Ans: 51.25

18. Find the number of four digit numbers whose middle two-digit is 15 and are divisible by 15.

19. Find a five-digit number which satisfying the following conditions:

Three pairs of digit with sum eleven each.

Last digit is 3 times the first one.

3rd digit is 3 less than the second.

4th digit is 4 more than the second one. Ans: 25296.

20. There are five thieves, each loot a bakery one after the other such that the first one took half of the total number of the

breads plus half of bread. The second one took half of the remaining plus half of bread. Similarly 3rd, 4th and 5th also did the

same. Finally, three breads left in the bakery itself. Find the number of number of breads were there. Ans: 127

21. Two dice are rolled. Each time the score is calculated as a product of the numbers appeared on the surface. The score for

the second roll is six more than that of the 1st roll, the score for 3rd roll is 6 less than the 2nd roll, the score for 4th role is 11

more than the 3rd roll, the score on 5th roll is five more than the 4th roll. Find the scores of 1,2,3,4 rolls. Ans: 4, 10, 4, 15, 20.

Back to Table of Contents



22. Three persons A, B, C have Rs.165, Rs.180, and Rs.150. The person having the highest amount gives 1/5th of his amount to

two other persons equally. The game ends when each one has contributed. (Assume that A starts the game)

What was the amount left with C at the end of game?

What was the amount left with B at the end of game?

What was the amount left with A at the end of game?

What was the amount left with A after his contribution (i.e. after 2nd step)?

In how many steps the game ends.

23. Every day in his business, a merchant had to weigh amounts from 1 kg to 121 kg, to the nearest kg. What are the minimum

numbers of weights required to check all weights from 1 kg to 121 kg? Ans: 5/ 1, 3, 9, 27 & 81 kg


Number Theory

Finding Remainders of a product (derivative of remainder theorem)

If ‘a1‘is divided by ‘n’, the remainder is ‘r1’ and if ‘a2’ is divided by n, the remainder is r2. Then if a1+a2 is divided by n, the

remainder will be r1 + r2

If a1 – a2 is divided by n, the remainder will be r1 – r2

If a1 × a2 is divided by n, the remainder will be r1 × r2

Ex. If 21 is divided by 5, the remainder is 1 and if 12 is divided by 5, the remainder is 2.

Then if (21 + 12 = 33) is divided by 5, the remainder will be 3 (1 + 2).

If 9(21 – 12) is divided by 5, the remainder will be 1 – 2 = – 1.

But if the divisor is 5, – 1 is nothing but 4. 9 = 5 × 1 + 4.

So, if 9 is divided by 5, the remainder is 4 and 9 can be written as 9 = 5 × 2 – 1.

So here – 1 is the remainder. So – 1 is equivalent to 4 if the divisor is 5. Similarly – 2 is equivalent to 3.

If 252(21 × 12) is divisible by 5, the remainder will be (1 × 2 = 2).

If two numbers ‘a1’ and ‘a2‘ are exactly divisible by n. Then their sum, difference and product is also exactly divisible by n.

i.e.,

If ‘a1’ and ‘a2’ are divisible by n, then

a1 + a2 is also divisible by n

a1 – a2 is also divisible by n and If a1 × a2 is also divisible by n.

Ex.1 12 is divisible by 3 and 21 is divisible by 3. Back to Table of Contents



Sol. So, 12 + 21 = 33, 12 – 21 = – 9 and 12 × 21 = 252 all are divisible by 3.

Finding Remainders of powers with the help of Remainder theorem:

Ex.2 What is the remainder if 725 is divided by 6?

Sol. If 7 is divided by 6, the remainder is 1. So if 725 is divided by 6, the remainder is 125 (because 725 = 7× 7 × 7… 25 times.

So remainder = 1 × 1 × 1…. 25 times = 125).

Ex.3 What is the remainder, if 363 is divided by 14.

Sol. If 33 is divided by 14, the remainder is – 1. So 363 can be written as (33)21.

So the remainder is (– 1)21 = – 1. If the divisor is 14, the remainder – 1 means 13. (14 – 1 = 13)

By pattern method

Ex.4 Find remainder when 433 is divided by 7.

Sol. If 41 is divided by 7, the remainder is 4. (41 = 4 = 7 × 0 + 4)

If 42 is divided by 7, the remainder is 2 (42 = 16 = 7 × 2 + 2)

If 43 is divided by 7, the remainder is 1 (43 = 42 × 4, so the Remainder = 4× 2 = 8 = 1)

If 44 is divided by 7, the remainder is 4 (44 = 43 × 4, so the Remainder = 1× 4 = 4)

The remainders of the powers of 4 repeats after every 3rd power. So, as in the case of finding the last digit, since the

remainders are repeating after every 3rd power, the remainder of 433 is equal to the remainder of 43 ( since 33 is exact

multiple of 3) = 1. (OR) If 43 is divided by 7, the remainder is 1. So 433 = (43)11 is divided by 7, the remainder is 11

Application of Binomial Theorem in Finding Remainders

The binomial expansion of any expression of the form

(a + b)n = nC0 an + nC1 an – 1 × b1 + nC2 × an – 2 × b2 + ..…. + nCn – 1 × a1 × bn – 1 + nCn × bn

Where nC0, nC1, nC2,…are all called the binomial coefficients.

In general,

There are some fundamental conclusions that are helpful if remembered, i.e.

a. There are (n + 1) terms.

b. The first term of the expansion has only a.

c. The last term of the expansion has only b.

d. All the other (n – 1) terms contain both a and b.

e. If (a + b)n is divided by a then the remainder will be bn such that bn < a.

Ex.5 What is the remainder if 725 is divided by 6?

Sol. (7)25 can be written (6 + 1)25. So, in the binomial expansion, all the first 25 terms will have 6 in it. The 26th term is (1)25.

Hence, the expansion can be written 6x + 1. 6x denotes the sum of all the first 25 terms. Since each of them is divisible by 6,

their sum is also divisible by 6, and therefore, can be written 6x, where x is any natural number. Back to Table of Contents



So, 6x + 1 when divided by 6 leaves the remainder 1. (OR) When 7 divided by 6, the remainder is 1. So when 725 is divided by

6, the remainder will be 125 = 1.

Wilson’s Theorem

If n is a prime number, (n – 1)! + 1 is divisible by n.

Let take n = 5

Then (n – 1)! + 1 = 4! + 1 = 24 + 1 = 25 which is divisible by 5.

Similarly if n = 7

(n – 1)! + 1 = 6! + 1 = 720 + 1 = 721 which is divisible by 7.

Corollary

If (2p + 1) is a prime number (p!)2 + (– 1)p is divisible by 2p + 1.

e.g If p = 3, 2p + 1 = 7 is a prime number

(p!)2 + (– 1)p = (3!)2 + (– 1)3 = 36 – 1 = 35 is divisible by (2p + 1) = 7.

Property

If “a” is natural number and P is prime number then (ap – a) is divisible by P.

e.g. If 231 is divided by 31 what is the remainder?

(231/31) = (231-2+2)/31

= So remainder = 2

Fermat’s Theorem

If p is a prime number and N is prime to p, then (N)p-1 – 1 is a multiple of p.

Corollary

Since p is prime, p – 1 is an even number except when p = 2.

Therefore (N(p-1/2)+1)(N(p-1/2)-1) = M(p)

Hence either N(p-1/2)+1 or N(p-1/2)-1 is a multiple of p, that is N(p-1/2) = kp+1 or kp-1, where, k is some positive integer.

Base Rule and Conversion

This system utilizes only two digits namely 0 & 1 i.e. the base of a binary number system is two. e.g. 11012 is a binary number,

to find the decimal value of the binary number, powers of 2 are used as weights in a binary system and is as follows:

1 × 23 = 8

1 × 22 = 4

0 × 21 = 0

1 × 20 = 1

Thus, the decimal value of 11012 is 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 13.

Conversion from decimal to other bases Back to Table of Contents



We will study only four types of Base systems,

1. Binary system (0, 1)

2. Octal system (0, 1, 2, 3, 4, 5, 6, 7).

3. Decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

4. Hexa-decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C D, E, F) where A = 10, B = 11 … F = 15.

Let us understand the procedure with the help of an example

Ex.6 Convert 35710 to the corresponding binary number.

Sol. To do this conversion, you need to divide repeatedly by 2, keeping track of the remainders as you go. Watch below:

As you can see, after dividing repeatedly by 2, we end up with these remainders:

These remainders tell us what the binary number is! Read the numbers outside the division block, starting from bottom and

wrapping your way around the right-hand side and moving upwards. Thus,

(357)10 convert to (101100101)2. This method of conversion will work for converting to any non-decimal base. Just don't forget

to include the first digit on the left corner, which is an indicator of the base. You can convert from base-ten (decimal) to any

other base.

Conversion from other bases to Decimal

We write a number in decimal base as

345 = 300 + 40 + 5 = 3 × 102 + 4 × 101 + 5 × 100

Similarly, when a number is converted from any base to the decimal base then we write the number in that base in the

expanded form and the result is the number in decimal form.

Ex.7 Convert (1101)2 to decimal base

Sol. (1101)2 = 1 × 23 + 1 x 22 + 0 × 21 + 1 × 20

= 8 + 4 + 1 = 13

So (1101)2 = (13)10

Ex.8 Convert the octal no 3456 in to decimal number.

Sol. 3456 = 6 + 5 × 8 + 4 × 82 + 3 × 83

= 6 + 40 + 256 + 1536

= (1838)10 Back to Table of Contents



Ex.9 Convert (1838)10 to octal.

Sol.

= (3456)8

Ex.10 What is the product of highest 3 digit number & highest 2 digit number of base 3 system?

(1) (21000)3 (2) (22200)3 (3) (21222)3 (4) (21201)3 (5) None

Sol. The highest 3 digit & 2 digit numbers are 222 & 22

222 = 2 + 2 × 3 + 2 × 32 = 26

22 = 2 + 2 × 3 = 8

∴ Product = 26 × 8 = 208

Convert back to base (21201)3

Answer: (4)

Ex.11 What is the remainder, if 2429 + 3429 is divided by 29?

Sol. an + bn is always divisible by a + b, if is odd.

2429 + 3429 is always divisible by 24 + 34 = 58.

So, it is always divisible by 29. So, the remainder is 0.

Ex.12 What is the remainder, if 12243 is divided by 10?

Sol. 12243

The remainder repeats after every 4th power.

So, the required answer is the remainder of 123 is divided by 10. i.e. 8

Ex.13 What is the value of (FBA)16 in binary system?

Sol. A = 10, B = 11, F = 15

Since 24 = 16,

While converting each digit of the decimal, can be written as 4 digit binary no:

A = 1010, B = 1011, F = 1111

(FBA)10 = (111110111010)2

Ex.14 Convert (721)8 to binary. Back to Table of Contents



Sol. Since 23 = 8, write each digit of octal no. as 3 digits of binary which gives equivalent value.

7 = 111, 2 = 010, 1 = 001

∴ (721)8 = (111 010 001)2

CYCLICITY

At times there are questions that require the students to find the units digit in case of the numbers occurring in

powers. If anyone asks you to find the unit digit of 33, you will easily calculate it also you can calculate for 35 but if any one ask

you the unit digit of 17399, it will be hard to calculate easily.

But it’s very simple if we understand that the units digit of a product is determined by whatever is the digit at the

units place irrespective of the number of digits. E.g. 5 × 5 ends in 5 & 625 × 625 also ends in 5. Now let’s examine the pattern

that a number generates when it occurs in See the last digit of different numbers.

Unit Digit Chart

Power

From the above table we can conclude that the unit digit of a number repeats after an interval of 1, 2 or 4. Precisely we can

say that the universal cyclicity of all the numbers is 4 i.e. after 4 all the numbers start repeating their unit digits.

Therefore, to calculate the unit digit for any exponent of a given number we have to follow the following steps

Step 1: Divide the exponent of the given number by 4 and calculate the remainder.

Step 2: The unit digit of the number is same as the unit digit of the number raise to the power of calculated remainder.

Step 3: If the remainder is zero, then the unit digit will be same as the unit digit of N4. Back to Table of Contents



Let us consider an example

Ex.1 Find the last digit of (173)99.

Sol. We notice that the exponent is 99. On dividing, 99 by 4 we get 24 as the quotient & 3 as the remainder. Now these 24

pairs of 4 each do not affect the no. at the units place so, (173) 99 ≈ (173)3. Now, the number at the units place is 33 = 27.

Factors

A factor is a number that divides another number completely. e.g. Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.

Number of Factors

If we have a number, N = pa × qb × rc Where p, q, and r are prime numbers and a, b, and c are the no. of times each

prime number occurs , then the number of factors of n is found by (a + 1) (b + 1)(c

Example:

Find the number of factors of 24 × 32.

Number of factors = (4 + 1) (2 + 1) = 5(3) = 15

Number of Ways of Expressing a Given Number as a Product of Two Factors

When a number is having even number of factors then it can be written as a product of two numbers in

(a+1)(b+1)(c+1)/2 ways. But if a number have odd number of factors then it can be written as a product of two different

numbers in [(a+1)(b+1)(c+1)-1]/2 ways and can be written as a product of two numbers (different or similar) in

[(a+1)(b+1)(c+1)+1]/2 ways.

Examples:

1. 148 can be expressed as a product of two factors in 6/2 or 3 ways. {Because (p + 1) (q + 1) (r + 1) in the case of 148 is equal

to 6}.

2. 144 (24.32) can be written as a product of two different numbers in [(4 + 1)( 2 + 1) −1]/2 i.e. 7 ways

Sum of the factors of a number:

If a number N is written in the form of N = ap.bq.cr ,where a, b & c are prime numbers and p, q & r are positive

integers, then the sum of all the factors of the number are given by the formula

Sum of factors = Back to Table of Contents



Factorial

Factorial is defined for any positive integer. It is denoted by L or !. Thus “Factorial n” is written as n! Or

n! is defined as the product of all the integers from 1 to n. Thus n! = 1.2.3. …. n. (n! = n(n – 1)!)

Finding the Highest power of the number dividing a Factorial

Ex.2 Find the largest power of 3 that can divide 95! without leaving any remainder.OR Find the largest power of 3

contained in 95!.

Sol. First look at the detailed explanation and then look at a simpler method for solving the problem. When we write 95!

in its full form, we have 95 × 94 × 93 ….. × 3 × 2 × 1. When we divide 95! by a power 3, we have these 95 numbers in the

numerator. The denominator will have all 3’s. The 95 numbers in the numerator have 31 multiples of 3 which are 3, 6, 9….90,

93.Corresponding to each of these multiplies we can have a 3 in the denominator which will divide the numerator completely

without leaving any remainder, i.e. 331 can definitely divide 95!

Further every multiple of 9, i.e. 9, 18, 27, etc. after canceling out a 3 above, will still have one more 3 left. Hence for every

multiple of 9 in the numerator, we have an additional 3 in the denominator. There are 10 multiples of 9 in 95 i.e. 9, 18….81,

90. So we can take 10 more 3’s in the denominator.

Similarly, for every multiple of 33 we can take an additional 3 in the denominator. Since there are 3 multiples of 27 in 91 (they

are 27, 54 and 81), we can have three more 3’s in the denominator.

Next, corresponding to every multiple of 34 i.e. 81 we can have one more 3 in the denominator. Since there is one multiple of

81 in 95, we can have one additional 3 in the denominator.

Hence the total number of 3’s we can have in the denominator is 31 + 10 + 3 + 1, i.e., 45. So 345 is the largest power of 3 that

can divide 95! without leaving any remainder.

The same can be done in the following manner also.

Divide 95 by 3 you get a quotient of 31. Divide this 31 by 3 we get a quotient of 10. Divide this 10 by 3 we get a

quotient of 3. Divide this quotient of 3 once again by 3 we get a quotient of 1. Since we cannot divide the quotient any more

by 3 we stop here. Add all the quotients, i.e. 31 + 10 + 3 + 1 which gives 45 which is the highest power of 3.

Add all the quotients 31 + 10 + 3 + 1, which give 45.

{Note that this type of a division where the quotient of one step is taken as the dividend in the subsequent step is called

“Successive Division”. In general, in successive division, the divisor need not be the same (as it is here). Here, the number 95

is being successively divided by 3. Please note that this method is applicable only if the number whose largest power is to be




found out is a prime number. If the number is not a prime number, then we have to write the number as the product of

relative primes, find the largest power of each of the factors separately first. Then the smallest, among the largest powers of

all these relative factors of the given number will give the largest power required.

Ex.3 Find the largest power of 12 that can divide 200!

Sol. Here we cannot apply Successive Division method because 12 is not a prime number. Resolve 12 into a set of prime

factors. We know that 12 can be written as 3 × 4. So, we will find out the largest power of 3 that can divide 200! and the

largest power of 4 that can divide 200! and take the LOWER of the two as the largest power of 12 that can divide 200!.

To find out the highest power of 4, since 4 itself is not a prime number, we cannot directly apply the successive division

method. We first have to find out the highest power of 2 that can divide 200!. Since two 2’s taken together will give us a 4,

half the power of 2 will give the highest power of 4 that can divide 200!. We find that 197 is the largest power of 2 that can

divide 200!. Half this figure-98-will be the largest power of 4 that can divide 200!.

Since the largest power of 3 and 4 that can divide 200! are 97 an 98 respectively, the smaller of the two, i.e., 97 will be the

largest power of 12 that can divide 200! without leaving any remainder.

Ex.4 What is the last digit of 234 × 334 × 434

Sol. Given = (24)34

Last digit of 4n is 6, if n is even.

Answer 6

Ex.5 What is the right most non zero digit of (270)270

Sol. The required answer is the last digit of 7270.

Last digit of 7 powers repeat after every 4.

So, the last digit of 7270 is the last digit of 72 = 9.

Ex.6 How many factors do 1296 have?

Sol. 1296 = 4 × 324

= 4 × 4 × 81

= 24 × 34

Number of factors = (4 + 1) (4 + 1) = 25.

Ex.7 If x is the sum of all the factors of 3128 and y is the no of factors of x and z is the number of ways of writing ‘y’ as a

product of two numbers, then z = ?

Sol. 3128 = 4 × 782

= 4 × 2 × 391

= 23 × 17 × 23




= 15 × (17 + 1) (23 + 1)

= 3 × 5 × 9 × 2 × 8 × 3

= 24 × 34 × 5

y = (4 + 1) (4 + 1) (1 + 1)

= 2 × 52

z = 1/2 { (1 + 1) (2 + 1) } = 3

Ex.8 How many cofactors are there for 240, which are less than 240?

Sol. 240 = 16 × 15

= 24 × 3 × 5

Number of co primes to N, which are less than N

if N = ab × bq × - - - - (a, b, - - - - are Prime no.s)

= 64

Ex.9 What is the sum of all the co primes to 748? Which are less than N?

Sol. 748 = 4 × 187

= 22 × 11 × 17

Sum of all the co primes to N. which are less than N is N/2 (number of co primes to N, which are less than N. Sum

= 320 = 119680

Ex.10 In how many ways 5544 can be written as a product of 2 co primes?

Sol. If N = ap × bq × - - - -, where a, b, - - - - are prime numbers

N can be written as a product of two co primes in 2n-1 ways, where n is the number of prime factors to N.

5544 = 11 × 504

= 11 × 9 × 56

= 11 × 9 × 8 × 7

= 23 × 32 × 7 × 11

Answer: = 24-1 = 23 = 8. (Because, 2, 3, 7 & 11 are four different prime factors).

Ex.11 If n! have 35 zeroes at the end. What is the least value ‘n’ will take?

(1) 110 (2) 120 (3) 130 (4) 140 (5) 145

Sol. Since the number of zeroes are 35, 535 should exactly divide n! by trail & error, take n = 140.




So, there are 34 zeroes.

The answer should be 145.

Answer: (5)

Ex.12 ‘N’ is a five digit number. The last digit of N35 is 2. What is the last digit of N?

(1) 2 (2) 3 (3) 7 (4) 8 (5) Cannot be determined

Sol. The last digit repeats after every 4th power.

Since the last digit of N35 is 2

The last digit of N3 is 2, which is possible only for 8.

Answer: (4)

Ex.13 What is the right most non zero digit in 4040/2020

Sol. =

The required answer is the last digit of 260 = 6

LCM & HCF


1. Seven friends in an office visited an exhibition in following pattern. The first friend visited every day. The second visited

every second day. The third visited every third day and so on up to 7th friend on every 7th day. When will they all meet

together? Ans: 420

3. Ms. Geetha was playing with her brother using 55 blocks. She gets bored playing and starts arranging the blocks such that

the number of blocks in each row is one less than that in the lower row. Find how many were there in the bottom most row?

Ans: 10

4. Three friends Rahul, Arul and Suresh, they divide the total 770 chocolates in the following manner. When Rahul takes 4




Arul takes 3 and when Rahul take 6 Suresh takes 7. Find the share of each one. Arul-198, Rahul-264,Suresh-308

5. A shopkeeper likes to arrange his collection of stamps. He arranges them sometimes in pair, sometimes in bundle of three,

sometimes in bundle of fours, occasionally in bundle of fives and sixes. Every time, he is left with one stamp in hand after

arrangement in bundles. However, if he arranges in the bundle of seven, he is not left with any stamp. How many stamps

does a shopkeeper have? Ans: 301


LCM & HCF

Highest Common Factor

Every number has some factors but if two or more numbers taken together can have one or more common factors.

Out of those common factors the greatest among them will be the highest common divisor or highest common factor of those

numbers. Such as 12 and 18 have 1, 2, 3 and 6 as common factors but among them 6 is the highest common factor. So H.C.F.

of 12 and 18 is 6.

Least Common Multiple

When we write the multiples of any two or more numbers taken together, we find that they have some common

multiples. Out of those common multiples the smallest among them will be the least common multiple of those numbers.

Such as 12 and 18 have 36, 72, 108, 144….. as common multiples but 36 is the least among them. So L.C.M. of 12 and 18 is 36.

Methods of finding L.C.M

(a) Factorisation Method: Resolve each one of the given numbers into prime factors. Then the product of the highest power

of all the factors gives the L.C.M.

Methods of finding H.C.F

(a) Factorisation method:

Express each number as the product of primes and take the product of the least powers of common factors to get the H.C.F.

(b) Division Method:

Divide the larger number by smaller one. Now divide the divisor by the remainder. Repeat the process of dividing the

preceding divisor by the remainder last obtained, till the remainder zero is obtained. The last divisor is the required H.C.F.

Ex.1 Find the L.C.M of 2, 4, 6, 8, 10.

Sol. Write the numbers as the product of primes.

2, 22, 2× 3, 23, 2× 5

Take the highest powers of all the primes. i.e., 23 × 3 × 5 = 120.

Ex.2 Find the H.C.F. of 2, 4, 6, 8, 10 using factorization method. Back to Table of Contents



Sol. For H.C.F, take the common prime factors. i.e., 2. (This is only the prime factor common in all the numbers).

Funda

i) Product of two numbers = L.C.M. × H.C.F.

ii) Product of n numbers = L.C.M of n numbers × Product of the HCF of each possible pair Or

If the HCF of all the possible pairs taken is same then we will have Product of n numbers = L.C.M of n numbers × (H.C.F of

each pair)(n – 1)

iii) If ratio of numbers is a : b and H is the HCF of the numbers Then

LCM of the numbers = H × a × b = HCF × Product of the ratios.

vi) H.C.F of fractions =

vii) L.C.M. of fractions =

viii) If HCF (a, b) = H1 and HCF (c, d) = H2, then HCF (a, b, c, d) = HCF (H1, H2).

Important Results

Type of Problem Approach of Problem

Find the GREATEST NUMBER that will exactly divide x, y, z.

Required number = H.C.F. of x, y, and z (greatest divisor).

Find the GREATEST NUMBER that will divide x, y and z

leaving remainders a, b and c respectively.

Required number (greatest divisor) = H.C.F. of (x – a), (y –

b) and (z – c).

Find the LEAST NUMBER which is exactly divisible by x, y and

z.

Required number = L.C.M. of x, y and z (least divided).

Find the LEAST NUMBER which when divided by x, y and z

leaves the remainders a, b and c respectively.

Then, it is always observed that (x – a) = (z – b) = (z – c) = K

(say).

Required number = (L.C.M. of x, y and z) – K.



Find the LEAST NUMBER which when divided by x, y and z

leaves the same remainder ‘r’ each case.

Required number = (L.C.M. of x, y and z) + r.

Find the GREATEST NUMBER that will divide x, y and z

leaving the same remainder in each case.

Required number = H.C.F of (x – y), (y – z) and

(z – x).

Ex.3 What is the greatest number which exactly divides 110, 154 and 242?

Sol. The required number is the HCF of 110, 154 & 242.

110 = 2 × 5 × 11

154 = 2 × 7 × 11

242 = 2 × 11 × 11

HCF = 2 × 11 = 22

Ex.4 What is the greatest number, which when divides 3 consecutive odd numbers produces a remainder of 1.

Sol. If x, y, z be 3 consecutive odd numbers, then the required number will be the HCF of x – 1, y – 1 and z – 1. Since x-1,

y-1 & z-1 are 3 consecutive even integers, their HCF will be 2. So answer is 2.

Ex.5 What is the highest 3 digit number, which is exactly divisible by 3, 5, 6 and 7?

Sol. The least no. which is exactly divisible by 3, 5, 6, & 7 is LCM (3, 5, 6, 7) = 210. So, all the multiples of

210 will be exactly divisible by 3, 5, 6 and 7. So, such greatest 3 digit number is 840. (210 × 4).

Ex.6 In a farewell party, some students are giving pose for photograph, If the students stand at 4 students per row, 2

students will be left if they stand 5 per row, 3 will be left and if they stand 6 per row 4 will be left. If the total

number of students are greater than 100 and less than 150, how many students are there?

Sol. If ‘N’ is the number of students, it is clear from the question that if N is divided by 4, 5, and 6, it produces a

remainders of 2, 3, & 4 respectively. Since (4 – 2) = (5 – 3) = (6 – 4) = 2, the least possible value of N is LCM (4, 5, 6) – 2

= 60 – 2, = 58. But, 100 < N < 150. So, the next possible value is 58 + 60 = 118.

Ex.7 There are some students in the class. Mr.X brought 130 chocolates and distributed to the students equally, then

he was left with some chocolates. Mr Y brought 170 chocolates and distributed equally to the students. He was

also left with the same no of chocolates as MrX was left. Mr Z brought 250 chocolates, did the same thing and left

with the same no of chocolates. What is the max possible no of students that were in the class?

Sol. The question can be stated as, what is the highest number, which divides 130, 170 and 250 gives the same

remainder, i.e. HCF((170 −130),(250 −170),(250 −130)). I.e. HCF (40, 80, 120) = 40.





Averages

Definition

Simple Average (or Mean) is defined as the ratio of sum of the quantities to the number of quantities.

By definition, Average =

Putting in symbols,

Here represent the n values of quantity under consideration and is the mean. Average or Mean is said to be a

measure of central tendency.

Ex.1 If a person with age 45 joins a group of 5 persons with an average age of 39 years. What will be the new average

age of the group?

Sol. Total age will be 45 + 5× 39 = 240. And there will be 6 persons now.

So the average will be 240/6 = 40.

(or) Since 45 is 6 more than 39, by joining the new person, the total will increase by 6 and so the average will increase

by 1. So, the average is 39 + 1 = 40.

Ex.2 Two students with marks 50 and 54 leave class VIII A and move to class VIII B. As a result the average marks of the

class VIII A fall from 48 to 46. How many students were there initially in the class VIII A?

Sol. The average of all the students of class VIII A is 46, excluding these two students.

They have 4 and 8 marks more than 46. So with the addition of these two students, 12 marks are adding more, and

hence the average is increasing 2. There should be 6 students in that class including these two. This is the initial

number of students,

Ex.3 The average of x successive natural numbers is N. If the next natural number is included in the group, the average

increases by:-

(1) Depends on x (2) Depends on the starting number of the series

(3) Both (1) and (2) (4) (5) None of these

Sol. The average of consecutive numbers is the middle number. If one more number is added to the list, the middle

number moves 0.5 towards right. So the answer is (4).




Weighted Mean

If some body asks you to calculate the combined average marks of both the sections of class X A and X B, when both

sections have 60% and 70% average marks respectively? Then your answer will be 65% but this is wrong as you do not know

the total number of students in each sections. So to calculate weighted average we have to know the number of students in

both the sections. Let N1, N2, N3, …. Nn be the weights attached to variable values X1, X2, X3,…….. Xn respectively. Then the

weighted arithmetic mean, usually denoted by

For any two different quantities taken in different ratios. The weighted average is just like a see-saw. More the ratio of a

quantity more will be the inclination of the average from mid value towards the value with more ratios.

Ex.4 The average marks of 30 students in a section of class X are 20 while that of 20 students of second section is 30.

Find the average marks for the entire class X?

Sol. We can do the question by using both the Simple average & weighted average method. = = 24

By the weighted mean method, Average = =24

Real Facts about average

1. If each number is increased / decreased by a certain quantity n, then the mean also increases or decreases by the same

quantity.

2. If each number is multiplied/ divided by a certain quantity n, then the mean also gets multiplied or divided by the same

quantity.

3. If the same value is added to half of the quantities and same value is subtracted from other half quantities then there will

not be any change in the final value of the average.

Average Speed

Average Speed =

If d1, d2 are the distances covered at speeds v1 and v2 respectively and the time taken are t1 and t2 respectively, then the

average speed over the entire distance (X1 +X2) is given by

If both the distances are equal i.e. d1 = d2 = d then, Average Speed =

But if the time taken are equal i.e. t1 = t2 = t then, Average Speed =

Ex.5 The average of 10 consecutive numbers starting from 21 is:

Sol. The average is simply the middle number, which is the average of 5th & 6th no. i.e, 25 & 26 i.e. 25.5.

Ex.6 There are two classes A and B., each has 20 students. The average weight of class A is 38 and that of class B is 40. X

and Y are two students of classes A and B respectively. If they interchange their classes, and then the average weight of

both the classes will be equal. If weight of x is 30 kg, what is the weight of Y?




Sol. Total weight of class A = 38 × 20, and class B = 40 × 20, if X & Y are interchanged, then the total ages of both the

classes are equal.

38 × 20 – x + y = 40 × 20 – y + x . 2(y – x) = 2 × 20,

y = x + 20 = 50 (OR)

Since both the classes have same number of students, after interchange, the average of each class will be 39. Since

the average of class ‘A’ is increasing by 1, the total should increase by 20.

So, x must be replaced by ‘y’, who must be 20 years elder to ‘x’. So, y must be 50 years old..

Ex.7 The average weight of 10 apples is 0.4 kg. If the heaviest and lightest apples are taken out, the average is 0.41 kg. If

the lightest apple weights 0.2 kg, what is the weight of heaviest apple?

Sol. Total weight of the apples is 0.4 × 10 = 4 kg.

Weight of apples except heaviest & lightest = 0.41 × 8 = 3.28 kg

Heaviest + lightest = 4 – 3.28 = 0.72 kg. It is given lightest = 0.2 kg.

Heaviest is 0.72 – 0.2 = 0.52 kg.

Ex.8 While finding the average of ‘9’ consecutive numbers starting from X; a student interchanged the digits of second

number by mistake and got the average which is 8 more than the actual. What is X?

Sol. Since the average is 8 more than the actual, the second no will increase by 72 (9 × 8) by interchanging the digits.

If ab is the second no, then 10a + b + 72 = 10b + a, 9(b – a) = 72. ∴ b – a = 8.

The possible number ab is 19. Since the second no is 19.

The first no is 18. X = 18

Ex.9 There are 30 consecutive numbers. What is the difference between the averages of first and last 10 numbers?

Sol. The average of first 10 numbers is the average of 5th & 6th no. Where as the average of last 10 numbers is the average

of 25th & 26th no. Since all are consecutive numbers, 25th number is 20 more than the 5th number. We can say that the

average of last 10 nos is 20 more than the average of first 10 nos. So, the required answer is 20.

Instructions for next 3 examples:

There are 60 students in a class. These students are divided into three groups A, B, C of 15, 20 & 25 students each. The

groups A & C are combined to form group D.

Ex.10 What is the average weight of the students in group D?

1) More than the average weight of A.

2) More than the average weight of C.

3) Less than the average weight of C.

4) Less than the average weight of B.

5) Cannot be determined. Back to Table of Contents



Sol. We know only the no of students in each class, but we don’t know the average weight of any class, we can’t find the

answer.

Ex.11 If one student from group A is shifted to group B, which of the following is necessarily true?

1) The average weight of both groups increases

2) The average weight of both groups decreases.

3) The average weight of class remains the same.

4) The average weight of group A decreases and that of group B increases.

5) None of these

Sol. Options (1) & (2) are not possible. Average of both cannot increase or decrease. Option (4) can be eliminated because

we are not sure, whether the average of A increases & B decreases or A decreases & B increases or both remains

unchanged. It depends on the weight of the student, who shifted from A to B. Option (3) is always true because even

the student shifts from one group to other. The average weight of the whole class does not change. Answer: (3)

Ex.12 If all the students of the class have the same weight, which of the following is false?

1) The average weight of all the four groups is same.

2) Total weight of A & C is twice that of B.

3) The average weight of D is greater than that of A.

4) The average weight of class remains same even the students shifts from one group to other.

5) None of these.

Sol. Since each student has same weight (1) & (4) are right. Since the number of students in group

A & C together is 15 + 25 = 40. Where as in B, there are only 20 students. So option (2) is also correct. But (3) is false,

because, the average weight of each group is same, since all the students have same weight.

Answer: (3)

Percentages


1. Out of 30 questions, the three persons A, B and C answered 45% correct answers, B answered 55% of A. B and C together

answered 25 % more of what A answered. Find number of questions each answered.





Percentages

What is the meaning of percent?

The terms percent means “for every hundred”. A fraction whose denominator is 100 is called ercentage and the numerator of

the fraction is called the rate percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained

Rs. 20 for every hundred rupees he invested in the business, i.e. 20/100 rupees for each Rupee.

The abbreviation of percent is p.c. and it is generally denoted by %.

Ex.1 84% of a particular total is 630 marks. What is 90% equal to?

(1) 750 (2) 675 (3) 450 (4) 550 (5) None of these

Sol. The required answer is = 675.

Answer: (2)

Ex.2 Two numbers are greater than the third number by 25% and 20% respectively. What percent of first number is the

second number?

(1) 92% (2) 94 % (3) 96 % (4) 98 % (5) None of these

Sol. Assume the third number is 100. So the first number is 125 and the second number is 120.

So the required answer is = 96%.

Answer: (3)

Ex.3 A is earning 20% more than B, who earns 20% less than C. By what percent A earns more or less than C?

Sol. A is 4% less than C.

(b) Multiplying Factor

While dealing with %age increase or decrease picture the following scale in your mind with reference as 100% ( = 1) in the

center. So we can say that multiplying factor (M.F) of 10% increase is 1.1 and that of 15% decrease is 0.85. An increase by x%

implies the value lies on the right hand side of 100% & vice versa. Let’s start with a number X

1. X increased by 10% would become X + 0.1 X = 1.1X

2. X increased by 1% would become X + 0.01 X = 1.01X

3. X increased by 0.1 % would become X + 0.001 X = 1.001X

4. X decreased by 10% would become X – 0.1X = 0.9X

5. X decreased by 1% would become X – 0.01 X = 0.99X

6. X decreased by 0.1% would become X – 0.001 X = 0.999X Back to Table of Contents



7. X increased by 200% would become X + 2X = 3X

8. X decreased by 300% would become X – 3X = − 2X

Ex.4 Coconut oil is now being sold at Rs. 27 per kg. During last month its cost was Rs. 24 per kg. Find by how much

percent a family should reduce its consumption, to keep the expenditure the same.

Sol. Assume the consumption last year is 1 kg, and then it cost Rs. 24.

But now for Rs.24, only kg of oil will come.

So the % reduction in consumption =

(c) Successive Percentage change:

The population of a city increases by 10% in one year and again increases by 10% in the next year, then what is the net

increase in the population in two years. The very common answer is 20% which is wrong. Why? Let us see

If Original population = P

After 1st year =

After 2nd year =

I.e increases by 21% of the original value.

This successive change in the percentage can be calculated in the shortcut way as explained below:

Let us consider a product of two quantities A = a x b. If a & b change (increase or decrease) by a certain percentage say x & y

respectively, then the overall percentage change in their product is given by the formula

Ex.5 If the volume of a milk and water solution is increased by 25% by pouring only water. By what percentage does the

concentration of milk reduce?

Sol. Assume initially, there is 100 lts of solution, out of which x lts is milk.

So the concentration of milk is . Now it is

So the percentage decrease =

This formula also holds true if there are successive changes as in the case of population increase or decrease. But care has to

be taken when there are either more than 2 successive changes or there is a product of more than 2 quantities as in the case

of volume. In these cases we have to apply the same formula twice.

If there is successive increase of x% and y%, then the net change will be

If there is successive discount of x% and y%, then the total discount will be

If there is x% increase and then x% decrease, then the net change

If the values are different, then net change Back to Table of Contents



Ex.6 If A is increased by 20% and B is decreased by 20%, then both the quantities will be equal. What percentage of B is

A.

Sol. If A is increased by 20%, it will be come 1.2 A. and B is decreased by 20%, it will become 0.8 B.

It is given 1.2 A = 0.8 B, A = B.

A = 0.66 B.

A is 66.66% of B.

Ex.7 If A is 3 times to B then B is what percentage of A.

Sol. A = 3B

Ex.8 If ‘x’ is increased by 20% & 25% successively, then its value increases by 30. What is the value of X?

Sol. If X is increased by 20%, it will become 1.2X or X

If it is again increased by 25%, its value becomes =

It is given that 30.

Ex.9 A man spends 30% of his salary for food and 20% of the remaining on rent and 20% of the remaining on other

expenses. If he saves Rs. 8960, what is his salary?

Sol. Let his salary be ‘K’. If he spent 30% on food, he will be left with 70% of K i.e. 0.7K or on this, he spends

20% on rent and 80 left with 80%. i.e

After spending 20% of this on other expenses he will be left with 80% of this.

i.e.

It is given as Rs.8960

K = 20000

Ex.10 Last year an employee used to save 40% of his salary. But now his salary is increased by 50% and expenses also

increases by 20%. What is his percentage savings now?

Sol. Assume ‘X’ was the salary of the employee last year. Since his savings were 40%, his expenses were 60%

i,e, 0.6X. At present, his salary is 1.5X and expenses are 1.2(0.6X) = 0.72X

Savings = 1.5X – 0.72X = 0.78X

% Savings = Back to Table of Contents



Ex.11 A man earns X% on the first Rs. 2000 and Y% on the rest of income. If he earns Rs. 700 from Rs.4000 and Rs. 900

from Rs. 5000 income; find X%.

Sol. We can form two equations from the above information as

Equation (1) 3 – Equation (2) 2

X =15%

Linear Equations /Word Problems/Problems on ages


1. Three friends A, B and C with A and B have their ages in reverse order and the difference in their ages is twice that of C’s

age. B is ten times as old as C. Find the present ages of A, B, C.

2. Ramesh sits around a round table with some other friends. He has one rupee more than the person sitting on his right

person and this person in turn has 1 rupee more than the person to his right and so on. Ramesh decided to give 1 rupee to his

right and he in turn gives 2 rupees to his right, this person gives 3 rupees to his right and so on. This process went on until

someone has n o money to give to his right. It is given that at the end Ramesh left with 4 times the money of his right. How

many men are there along with Ramesh and what is the money that the poorest fellow had at the beginning.

3. Venkat has one son and two daughters. The product of ages of children is 72. The sum of their ages gives the door number

of Venkat’s house. Moreover, it is known that son is elder of three. Can you tell the ages of all the three?

4. A boy asks his father, "what was the age of grandfather?” Father replied " He was x years old in x2 years", and also said, "I

am talking about 20th century". What is the year of birth of grandfather?

5. There are two persons each having same amount of marbles in the beginning. After that 1st person gain 20 more from 2nd

person and he eventually lose two third of it during the play and the 2nd person now has 4 times marble of what 1st person is

having now. Find out how much marble did each have in the beginning. Ans: 100 each

6. Father says, “My son is five times older than my daughter”. My wife is 5 times older than my son, I am twice old as my

wife. Altogether, sum of our ages is equal to my mother’s age and she is celebrating her 81st birthday. So what is my son's

age? Ans: 5 years




7. If the digits of my present age are reversed then I will get the age of my son. One year ago my age was twice as that of my

son. Find my present age. Ans: Father-73, Son-37

8. Father’s age is reverse of son`s age. one year back father’s age was twice of son`s age. Find the fathers current age. Ans: 73

9. In a Zoo, number of animals is 11 more than the number of birds. If you interchanged the number of animals and birds, then

the number of legs will be reduced by of the original numbers. Find number of animals and birds present in the Zoo.

Ans: Birds-11, Animals-22

10. Two truck drivers, one with 64 barrels of wine, other with 20 barrels of wine travelling from Mumbai to Calcutta. They do

not have enough money to pay duty for the same. First driver pays 40 francs and gives 5 barrels. Second driver gives 2 barrels

but gets 40 francs in exchange. What is value of each barrel, and duty for each barrel?

Ans: 120 francs/barrel, Duty on each-10 francs

11. Two girls went for shopping. In the first shop, they spent half of what they have plus Rs.2. In the next shop, they spent half

of remaining plus Rs.5. The remaining Rs.10 spent on coffee. How much did they start with? Ans: 64

12. A woman gave Rs.1 more than half she has to the 1st beggar, Rs.2 more than half she has to the 2nd beggar and Rs.3 more

than the half she has to the 3rd. Now she is left with Rs.1. How much did she originally have? Ans: 42 rupees

13. There are two barbers in a village, both take same time for haircut and shave. It is known that 15 minutes required for hair

cut and 5 minutes required for shave. What is the possible minimum time if there are three customers in their shop for hair

cut as well as shave? Ans: 30 minutes


Linear Equations /Word Problems/Problems on ages

General Theory of Equations

Polynomial equation: A polynomial which is equal to zero is called a polynomial equation. For example, 2x + 5 = 0, x2 – 2x + 5

= 0, 2x2 – 5x2 + 1 = 0 etc. are polynomial equations.

Linear Equation: A linear equation is 1st degree equation. Its general form is ax + b = 0

Quadratic Equation: A quadratic equation is 2nd degree equation. The general form of a quadratic equation is

ax2 + bx + c = 0.

Root of a polynomial equation:

If f (x) = 0 is a polynomial equation and f (α) = 0, then α is called a root of the polynomial equation f (x) = 0.

Theorem: Every equation f (x) = 0 of nth degree has exactly n roots.




If f(x) is linear the number of roots will be one, if it is quadratic, then number of roots is two. If we draw the graph of the

expression y = x2 – 2x + 1, then we would note that the graph cuts the x-axis at two points. This is because it has two real

roots.

Factor Theorem:

If α is a root of equation f (x) = 0, then the polynomial f (x) is exactly divisible by x – α (i.e., remainder is zero). For example, x2

– 5x + 6 = 0 is divisible by x – 2 because 2 is the root of the given equation.

Remainder Theorem:

If we divide polynomial f(x) by (x – α ) then f( α ) is the remainder.

For example, f(x) = x2 – 5x + 7 = 0 is divided by x – 2 the remainder is f(2) where f(2) = 22 – 5 × 2 + 7 = 1.

Linear Equation

A linear equation is 1st degree equation. It has only one root. Its general form is a x + b = 0 and root is – .

LINEAR EQUATION OF ONE AND TWO VARIABLES

Linear Equation: The equation which when reduced to its simplest form contains only the first power of the variable is called

linear equation or simple equation. In other words, a linear equation in one variable is an equation of the type ax + b = 0, or ax

= c, where a, b, c are constants (real numbers), a ≠ 0 and x is a variable.

We know that the solution of ax + b = 0, a ≠ 0 is x = - .

We say x = - is a root of the equation ax + b = 0.

Linear Equation in two variables: A general linear equation in two variables x and y is usually written in the following forms:

ax + by + c = 0 where a ≠ 0, b ≠ 0; a, b, c are real constants

ax + by = d where a ≠ 0, b ≠ 0; a, b, d are real constants

Any pair of values of x and y which satisfies ax + by + c = 0 is called its solution.

A pair of linear equations in two variables say x and y is said to form a system of simultaneous linear equations in two

variables.

The general form of a system of linear equations in two variables x and y is

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0




Set of Equations

Consistent System (means have solution) Inconsistent System (means have no solution)

Dependent System (has infinitely many solutions)

Condition

Independent System (Only one/Unique Solution)

Condition

Word Problems

Questions based on word problems can be solved with the help of linear questions:

Examples:

1. Ten years ago, the average age of a family of four members was 25 years. Two children having been born, the average

age of the family is the same today. What is the age of the youngest child if they differ in age by 2 years?

(A) 3 years (B) 2 years (C) 4 years (D) 8 years (E) 5 years

Sol. Present total age = 25 × 6 = 150;

Total age of 4 members 10 years ago = 25 × 4 = 100.

Total age of 4 members now = 100 + 40 = 140

Sum of ages of children = 150 – 140 = 10 years

Let x be the age of youngest child.

x + x + 2 = 10

x = 4. Answer: (C)

2. If the average age of three persons is 20 years. With the addition of one more person, average age of all four

becomes 21 years. Find the age of the fourth member.

(A) 24 years (B) 22 years (C) 21 years (D) 25 years (E) None of these

Sol. Sum of ages of three members = 20 × 3 = 60 years.

Let’s the age of fourth members = X

Now 60 + X = 21 × 4 = 84.

So age of fourth member (X) = 24 years.

Answer: (A)




Quadratic Equation

Quadratic Equation in “x” is one in which the highest power of “x” is 2. The equation is generally satisfied by two

values of “x”, but these values may be equal to each other. The quadratic form is generally represented by ax2 + bx + c = 0

where a ≠ 0, and a, b, c are constants.

e.g.

x2 – 6x + 4 = 0

3x2 + 7x – 2 = 0

A quadratic equation in one variable has two and only two roots, which are and

Nature of Roots

The term (b2 – 4ac) is called the discriminant of the quadratic equation ax2 + bx + c and is denoted by D.

Rules:

1. if D > 0, x1 & x2 are real and unequal.

2. if D = 0, x1 & x2 are real and equal.

3. if D is a perfect square, x1 & x2 are rational and unequal.

4. if D < 0, x1 & x2 are imaginary, unequal, and conjugates of each other.

If x1 and x2 are the two roots of ax2 + bx + c = 0, then sum of roots = x1 + x2 = – b/a and product of roots = x1 x2 = c/a.

Formation of equation from roots:

1. If x1 and x2 are the two roots, then (x – x1) (x – x2) = 0 is the required equation.

2. If (x1 + x2) and x1 × x2 are given then equation is x2 – (x1 + x2) x + x1 x2 = 0.

x2 – Sx + P = 0 where S = sum of roots, P = product of roots.

Profit & Loss


1. A man sold two pens. Initial cost of each pen was Rs.12. He sold the two pens together, one with 25% profit and another

with 20% loss. Find the amount of loss or gain.

2. A farmer sold a pair of cows for Rs.210. He sold, one with a profit of 10% and another with a loss 10%. Altogether, he made

a profit of 5%. How much did each cow originally cost him? Ans: Rs.150 & Rs.50

3. A bargain hunter bought some silver plates for $ 1.30 on Saturday, when 2cents was marked off at each on each article .On




Monday; he went to return them at regular prices and bought some cups and saucers without paying any extra amount. The

normal price of one silver plate is equal to the price of 1 cup and 1 saucer. In total, he bought 16 items more. It is given that

each saucer costs 3 cents and he bought 10 saucers more than cups. Find the number of cups and saucers he bought.


Profit, Loss & Discount

Definition

Cost Price: C.P. is the price at which one buys anything

Selling Price: S.P. is the price at which one sells anything

Profit/Loss: This is the difference between the selling price and the cost price. If the difference is positive it is called the

profit and if negative it is called as loss.

Profit/Loss %: This is the profit/loss as a percentage of the C.P.

Margin: Normally is in % terms only. This is the profit as a percentage of S.P.

Marked Price: This is the price of the product as displayed on the label.

Discount: This is the reduction given on the marked price before selling it to a customer. If the trader wants to make a

loss he can offer a discount on the cost price as well

Markup: This is the increment on the cost price before being sold to a customer.

Formulae

• Gain = (S.P. – C.P.), Loss = (C.P. – S.P.)

• Gain % = (Gain × 100)/C. P, Loss % = (Loss × 100)/C. P.

• Given the cost & the gain percent, S.P. = (100 + gain %) × C. P. / 100

• Given the cost & the loss percent, S.P. = (100 – loss %) × C. P. / 100

• Given the M.P. & the discount, C.P. = (100 – Discount %) x M.P / (100 + gain %)

• Given the M.P. & the discount, C.P. = (100. – Discount %) x M.P / (100 – loss %)

Ex.1 A person sells 36 oranges for one rupee and suffers a loss of 4%. Find how many oranges per rupee to be sold to

have a gain of 8%?

Sol. Let ‘X’ is the cost price of each orange.

Since he is giving 36 apples for one rupee, the selling price of an orange is 1/36 rupee.

Since he got 4% loss, the selling price of each orange is 0.96X = 1/36

To get 8% gain he has to sell it for 1.08X= (1.08/0.96)*(1/36) = 1/32 rupee. Back to Table of Contents



So for one rupee, he has to give 32 oranges to get a gain of 8%.

Discount

You always come across different offers attracting the customers such as “Buy 1 get 2 Free” or “Buy 3 get 5 Free” or

“SALE 50% + 40%”.

Can you calculate the discount offered to you?

Most of us are not aware about the offer given to us. The percentage of the discount offered in the first case is not

200% but it is 66.66% only. The discount is always on the number of items sold, not on the number of items purchased. In case

of successive discounts we can treat the problem as the problem of successive percentage change and can use the formula

Net discount =

E.g.: 40% + 30% discount = = (70 – 12) % = 58%.

Markup Price

It is also known as list price or Tag price which is written on the item. The markup price written is always greater than the

actual C.P of the item and the percentage rise in the markup price is on the C.P of the item. Percentage increase in the Markup

price =

Ex.2 The price of a trouser is marked 50% more than its cost price and a discount of 25% is offered on the marked price

of the trouser by the shopkeeper. Find the percentage of profit/loss.

Sol. M.P = 1.5 C.P

S.P = 0.75× 1.5 C.P = 1.125 C.P

So profit percentage = 12.5 %.

Ex.3 After allowing a discount of 11.11%, a trader still makes a profit of 14.28%. At how much percent above the cost

price does he mark on his goods?

Sol. Discount of 11.11% means a discount of 1/9 and 14.28% means 1/7

So selling price =

So MP = CP. So it is = 28.56% more than the CP.

Two different articles sold at same selling price

Overall % loss = where x is the percent profit or loss on the transaction.

Ex.4 Each of the two horses is sold for Rs. 1875. The first one is sold at 25% profit and the other one at 25% loss. What

is the % loss or gain in this deal?

Sol. It is loss of = 6.25 % loss.

Ex.5 What is the total loss or gain (in rupees) in the above example? Back to Table of Contents



Sol. Since he got 6.25 % loss means (1/16)th loss.

So, his selling price should be (15/16)th of the CP

So loss is (1/15)th of the SP = (1/15)(1875+1875) = Rs. 250.

Faulty Balance

Sometimes traders may sell their products at the rate at which they purchased or even less than the actual cost

incurred to them. Even in this transaction they make profit by cheating on volume. If the weighing balance of a shopkeeper

reads 1000 grams for every 900 grams, what is the profit or loss the shopkeeper is making?

On the other hand if the faulty balance reads 900 grams for every 1000 grams, is he still making profit? If not why?

Ex.6 Instead of a meter scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling

the same cloth. If he offers a discount of 20% on the cash payment, what is his overall profit percentage?

Sol. When the merchant is buying he is using a scale of 120 cm instead of 100 cm thus multiplying factor for him in this

transaction = 120/100 = 6/5 …….(1)

When selling the cloth the merchant is measuring 80 cm for every 100 cm , so multiplying factor of this transaction is

= 100/80 = 5/4 = …….(2)

For the discount offered by the merchant the multiplying factor = 80/100 = 4/5 ……. (3)

Net profit = .

Hence making a profit of 20% in the whole transaction

Ex.7 Mr. A purchased an article and sold it to Mr. B at 20% profit. Mr. B sold it to Mr. C at 20% profit. If Mr. C paid Rs.

2880 to Mr. B, what is the profit in rupees earned by Mr. A?

Sol. Assume that Mr. A purchased the article for Rs. X,

Then he sold it to Mr. B for Rs. 1.2X and Mr. B sold it to Mr. C for Rs. 1.2 × 1.2X = 1.44X.

This is given as Rs. 2880. So X = Rs. 2000.

Profit earned by A is 20% of X = Rs. 400.

Ex.8 A cloth merchant allows 25% discount on a saree and still makes 20% profit. By selling a saree, he gained Rs. 160.

What is the marked price of that saree?

Sol. He makes a profit of 20% and it is given as Rs. 160.

20% of C.P = 160

So C.P = Rs. 800.

So S.P = 800 + 160 = Rs. 960.

And 0.75 M.P = 960 (since he is giving 25% discount)

M.P = Rs. 1280




Ex.9 A man purchased some chocolates at 80 per Rs 100 and same number of chocolates of other type at 120 per Rs.

100. He sold each chocolate per 1 rupee each, what is his profit/loss percentage?

Sol. The first type of chocolate costs Rs. 100/80 rupees = 5/4 rupee.

The second type costs Rs. 100/120 = 5/6 rupee

Since he purchased both the chocolates in equal number, the

Average cost per chocolate is

But he is selling each chocolate at Re. 1. So he gets a loss of

Ratio Proportion


1. There are three containers A, B, C. Half of the container A contains wine. One fourth of B, which is twice the size of A,

contains wine. Now the two containers are filled with water and are together poured into third vessel C. Find the ratio

between wine and water in C.


Ratio Proportion

Concept

If a and b (b ≠ 0) are two quantities of the same kind, then Ratio is the relation which one quantity bears to another of the

same kind in magnitude.

Now in two quantities a and b the fraction a/b is called the ratio of a to b. It is usually expressed as a: b,

a and b are said to be the terms of the ratio. The former (numerator) ‘a’ is called the Antecedent of the ratio and latter

(denominator) ‘b’ is called consequent.

Basics of Ratio

(i) As ratio is a relation between two quantities so ratio is independent of the concrete units employed in the quantities

compared.

(ii) Ratio exists only between two quantities; both the quantities must be in the same units.




Composition of Ratio

I. Compounded Ratio: When two or more ratios are multiplied term wise, the ratio thus, obtained is called their compounded

ratio. Compounded ratio of (a: b, c: d and e: f)

II. Reciprocal Ratio

For any ratio a : b, the reciprocal ratio will be

The application of it is in calculating the ratio of the wages distributed among the workers, which is equal to the reciprocal

ratio of number of days taken by them to complete the work.

Proportion

When two ratios are equal, the four terms involved, taken in order are called proportional, and they are said to be in

proportion.

1. The ratio of a to b is equal to the ratio of c to d i.e. if , we write a : b : : c : d

2. If a : b : : c : d Then we have ad = bc i.e. the terms a and d are called Extremes and, the terms b and c are called the Means.

The term a, b, c and d are known by the name 1st, 2nd, 3rd and the 4th proportion respectively.

Continued Proportion:

Three quantities are said to be in continued proportion, if the ratio of the first to the second is same as the ratio of the second

to the third.

Thus a, b and c are in continued proportion if a : b = b : c (or)

Mean Proportion:

If a, b, c are in continued proportion then second quantity ‘b’ is called the mean proportional between ‘a’ and ‘c’ and a, b and

c are known as 1st, 2nd and 3rd proportion respectively.

Then b2 = ac (or) b =

Ex.1 What is the mean proportional to

Sol. Let the mean proportion of and be x

Ex.2 What should be added to each of the numbers 19, 26, 37, 50 so that the resulting number should be in proportion?

(1) 2 (2) 3 (3) – 2 (4) – 5

Sol. Let the required number be x.

According to question, Back to Table of Contents



19 + x, 26 + x, 37 + x and 50 + x are in proportion

19 + x : 26 + x = 37 + x : 50 + x

(19 + x) (50 + x) = (37 + x) (26 + x) x2 + 69x + 950 = x2 + 63x + 962

x2 – x2 + 69x – 63x = 962 – 950

Hence the resulting number = 2.

Basics of Proportion

(i) Let us take four quantities a, b, c & d such that they are in proportion i.e. a : b : : c : d then,

This operation is called componendo and Dividendo.

(ii) If then

Partnership & Share

If there is profit in the business run by two partners A and B then,

Ex.3 Saman begins business with a capital of Rs. 50,000 and after 3 months takes Manu into partnership with a capital

of Rs. 75000. Three months later Amandeep joins the firm with a capital of Rs. 1, 25,000. At the end of the year, the

firm makes a profit of Rs. 99,495. How much of this sum should Amandeep receive?

Sol. Money invested by Saman for 12 months = Rs. 50,000

Money invested by Manu for 9 months = Rs. 75000

Money invested by Amandeep for 6 months = Rs. 1,25,000

Share of Saman: Manu: Amandeep

= 50,000 × 12 : 75,000 × 9 ; 1,25,000 × 6

= 6,00,000 : 6,75,000 : 7,50,000 = 600 : 675 : 750 = 8 : 9 : 10

Total profit = Rs. 99,495.

Profit of Amandeep = = Rs.36, 850

PROPORTIONS

Direct Proportion

If A is directly proportional to B, then as A increases B also increases proportionally or in other words the proportional

change occurs in the same direction. In general when A is directly proportional to B, then = Constant.




Examples of Direct proportion are Time & Distance problems; time taken to travel a distance is directly proportional to the

distance traveled when the speed is constant. This means the distance-traveled doubles if the time taken doubles provided

speed remains constant.

Remember: If X is directly proportional to Y,

Direct Relation

In these types of cases an increase in one causes increase in the other but the increase is not proportional as in the

case of direct proportion.

For example:

With simple interest the amount increases with increase of the number of years but not proportionally while on the

other hand interest doubles or triples after 2nd and 3rd years. So the increment in amount is in direct relation while increment

in interest is direct proportion.

Remember: If X is in direct relation to Y then, X = k1+yk2

Ex.4 If 6 men can lay 8 bricks in one day, then how many men are required to lay 60 bricks in the same time?

(1) 45 men (2) 40 men (3) 60 men (4) 50 men

Sol. Since the time is same so to do more work we need more persons. Hence this is the problem of direct

proportion, i.e.

Ex.5 The cost of New Year party organized in TCY is directly related to the number of persons attending that party. If 10

persons attend the party the cost per head is Rs 250 and if 15 people attend, the cost per head is Rs. 200. What will

be the total cost of the party if 20 persons attend it?

Sol. This is the problem of direct relation

Let the total cost of party is

Cost = K1 + K2 N (where K1 & K2 are fixed and variable costs and N is number of persons)

250 x 10 = K1 + 10K2 ………(1)

200 x 15 = K1 + 15K2 ………(2)

Solving them we get K1 = 1500 and K2 = 100

So total cost for 20 persons = 1500 + 20 x 100 = Rs. 3500

Indirect/Inverse Proportion

A is in indirect proportion to B if as A increases, B decreases proportionally i.e. the proportional change occurs in the

opposite direction.

In general if A is in indirect proportion to B, then AB = constant.

Examples of Indirect proportion are Price & Quantity (expenditure remaining same), Number of men required & rate of work

done (amount of work remaining same), Time & Speed problems for same distance. Back to Table of Contents



Ex.6 If 6 men can build a wall in 9 days, then 60 men can build a similar wall in ______ days?

Sol. Work = Men x Days

Ex.7 A can do a piece of work in 12 days, B is 60% more efficient than A. Find the number of days required for B to do

the same piece of work.

Sol. Ratio of the efficiencies is A : B = 100 : 160 = 5 : 8. Since efficiency is inversely proportional to the number of days, the

ratio of days taken to complete the job is 8 : 5. So, the number of days taken by B =

Ex.8 Ten years ago, the ratio of ages of A and B is 3 : 4, now, it is 4 : 5. What is the present age of A?

Sol. 10 years ago, let their ages be 3k and 4k.

So, present ages are 3k + 10 and 4k + 10.

The ratio is given as 4: 5.

k = 10

Percentage of A = 3k + 10 = 40 years.

Ex.9 Three friends A, B, C earn Rs. 2000 together. If they want to distribute this money such that ‘A’ should get Rs. 300

more than B and Rs. 100 more than C, in what ratio, they have to distribute the money?

Sol. A = 300 + B = 100 + C

So, A = 300 + B and C = 200 + B

A + B + C = 2000

B = 500 A = 800 and C = 700 So, the required ratio = 8 : 5 : 7

Ex.10 Four friends A, B, C, D share Rs. 10,500 in the ratio , how much more money A and C together get than B

and D together?

Sol. The given ratio is ,

Multiply with 24 = 8 : 6 : 4 : 3.

Take A = 8k, B = 6k, C = 4k, D = 3k.

Total = 21k = 10,500

k = 500

Required answer = (A + C) – (B + D)

= (8k + 4k) – (6k + 3k) = 3k = Rs. 1500

Ex.11 xyz has to distribute Rs. 3500 such that, for every 2 rupees x takes, y will take 3 rupees and for every 4 rupees y

takes, z will take 5 rupees. How much money z will get?

Sol. We can write from the given information that, Back to Table of Contents



x : y = 2 : 3 ………(1)

and y : z = 4 : 5 ………(2)

Equation (1) × 4 x : y = 8 : 12

Equation (2) × 3 y : z = 12 : 15

x : y : z = 8 : 12 : 15

z will get

Time, Speed & Distance/ Boats and streams


1. Two trains start from stations A and B, which is 50 km apart at the same time. As the trains start, a bird flies from one train

to the other and reaching the second train, it flies back to the first train. This is repeated until the trains collide. If the speeds

of the trains are 25 km/hr, 35 km/hr respectively and that of the bird is 100km/hr. Find the distance travelled by the bird.

2. Mr. Ram starts walking down on the stairs and when he is on the 7th step, he sees his professor coming from ground floor.

When the professor reaches the top, he is having four steps to take to reach the floor. In addition, given that when the

professor covers two steps, he will cover one step. Find the total number of steps.

3. When I travelled a distance from A to B, my car got punctured. It took 10minutes to fix that. Because of that, I drove the

rest of the journey at 30miles/hr and reached 45minutes late. Had the puncture taken place after 20 minutes from the place

where it actually happened and traveled the remaining distance at 30miles/hr, I would have been only 30 minutes late. Find

the total distance covered.

4. There are two balls touching each other circumferentially. The radius of the big ball is 4 times the diameter of the small ball.

The smaller ball rotates in anticlockwise direction circumferentially over the bigger one at the rate of 16 rev/sec. The bigger

wheel also rotates anticlockwise at N rev/sec. Find the value of 'N' for the horizontal line from the centre of small wheel

always is horizontal.

5. There are two guards Mahinder and Johinder walking on the side of a wall of a warehouse (12m X 11m) in opposite

directions. They meet at a point and Mahinder says to Johinder “See you again in the other side". After a few moments of

walking Mahinder decides to go back for a smoke but he changes his direction again to his previous one after 10 minutes of

walking in the other(opposite) direction remembering that Johinder will be waiting to meet. If Mahinder and Johinder walk 8

and 11 feet respectively, how much distance, they would have travelled before meeting again.




6. A man starts his walking at 3:00 PM from point A; he walks at the rate of 4km/hr in plains and 3km/hr in hills to reach the

point B. During his return journey, he walks at the rate of 6km/hr in hills and 4km/hr in plains and reaches the point A at 9:00

PM. Find the distance between A and B? Ans: 12km

7. A boy travels in a cycle after covering of the distance, the wheel got punctured. Then he covered the remaining distance

by walk. In addition, it is given that walking time is twice that of the riding time ( of the distance). Find the ration between

riding speed and walking speed.

8. A van has maximum speed of 100km/hr. The van can accommodate a maximum of seven persons. Moreover, the speed of

the van is 88km/hr, when it has three persons. Calculate the speed of the van with six persons.

9. There are two persons, namely A and B walking through an escalator. A takes 50 steps and B takes 75 steps while the

escalator is moving down. Given that the time taken by A to cover one step is equal to time taken by B to cover three steps.

Find the number of steps in the escalator. Ans: 100

steps

10. Every day Mr. Rajesh travels certain distance from home to market by walk. If he walks with 30 miles/hr, then he will reach

1 hour before the normal time. If he walks with 20 miles/hr, then he will reach 1 hour later than the normal time. Find the

distance between home and market.

11. A sportive young hare and tortoise raced in opposite directions around a circular track that was 100 yards in diameter.

They started at the same spot, but the hare did not move until the tortoise had a start of one-eighth of the distance. Then

hare started running in opposite direction and met tortoise. At the time, hare had gone one-sixth of the distance. How many

times, the hare should run faster than the tortoise in order to win the race. Ans: 85/4

12. A man was travelling to a place 30 miles away from his house. He was travelling at 60 miles/hr. While returning to his

house, his car broke down and half an hour was wasted in repairing that. Altogether, he took 1 hr for return journey. Find the

average speed of the whole journey.

13. There is truck scheduled to reach the destination at 11 o’clock. If it travels at 30 mph speed then it will reach the

destination 1 hour before the normal time and if it travels at 20 mph speed then it will reach the destination 1hour later than

the normal time. Find the distance the truck has to cover. Ans: 120 miles

14. A girl took part in a cycling game with many others in a circular closed circuit. After cycling for several minutes, he found

that of the cyclists ahead of her and of the cyclists behind her together formed the total number of participants. How

many were participating in the race? Ans: 13

15. There are 20 poles with a constant distance between each pole. A car takes 24 second to reach the 12th pole. How much

time will it take to reach the last pole? Ans: 41.45 seconds

3. A ship went on a voyage. After it had travelled 180 miles, a plane started with 10 times the speed of the ship from the

shore. Find the distance from starting point when they met. Ans: 200miles





Time, Speed & Distance

1. The fundamental relationship between Distance (s), time (t) and speed (v) is given by: s = v x t.

2. Let v1 and v2 be the velocity of the two vehicles and let v1 > v2

If both the vehicles are moving in the same direction then their Relative Velocity = R.V. = v1 – v2

If both the vehicles are moving in the opposite direction then their Relative velocity = R.V. = v1 + v2

3. Average velocity =

If x1 & x2 are the distances covered at velocities v1 & v2 respectively then the average velocity over the entire distance (x1 + x2)

is given by

4. A man travels first half of the distance at a velocity v1, second half of the distance at a velocity v2 then,

Average velocity =

5. If the distance is covered in three equal parts with different speeds v1, v2 and v3 then,

Average velocity =

6. For the same distance, the time is inversely proportion to the speed of the object. These types of problems can be solved as

the problems of percentage.

7. When time is constant the ratio of speeds of the object is equal to the ratio of the distance covered by them i.e.


Ex.1 If I decrease my speed by 20% of original speed, I reach office 7 minute late. What is my usual time and new time of

reaching office?

Sol. Since speed is decreased by 20% i.e. 1/5 of the original. New speed will become 4/5 of the original speed. For the

same distance, the time will become 5/4 of original time. Therefore new time increase by 1/4 of the original. This is

given equal to 7 minutes. So Usual time = 7 × 4 = 28 minutes and New time = 28 + 7 = 35 minutes.




Ex.2 A train leaves Calcutta for Mumbai, a distance of 1600 kms at the same time a train leaves Mumbai to Calcutta. The

trains meet at Nagpur which is at a distance of 700 kms from Mumbai. What is the ratio of the speeds of the trains?

Sol. From the problem, it is clear that when the first train travels a distance of 1600 – 700 = 900 km, the second train

travels a distance equal to 900. So, the ratio of their speeds is 9: 7.

Trains

(i) When a train approaches a stationary object (a tree, a stationary man, a lamp-post; we assume the length of the object to

be infinitely small; provided its length isn’t mentioned)

Time taken by the train to cross Pole =

(ii) However, when a train approaches a platform, the time taken by the train to cross the platform is same as the time taken

by the train to cross a distance equal to its own length plus the length of the platform at its own velocity.

Time taken to cross the platform =

(iii) For two trains having lengths l1 & l2 and traveling in the same direction with speeds v1 & v2 respectively (v1 > v2).

Time taken to cross each other completely =

(iv) Similarly, for two trains traveling in the opposite direction:

Time taken to cross each other completely =

(v) If two trains/Object start at the same time from two points X & Y and move towards each other and after crossing they

take a & b hrs respectively to reach opposite points Y and X, then

Ex.3 A train 110 m long travels at 60 kmph. How long does it take?

(a) To pass a telegraph post by the side of the track?

(b) To pass a man running at 6 kmph in the same direction as the train?

(c) To pass a man running at 6 kmph in the opposite direction?

(d) To pass a station platform 240 m long?

(e) To pass another train 170 m long, running at 40 kmph in the same direction?

(f) To pass another train 170 m long, running at 60 kmph in the opposite direction?

Sol. (a) Speed of train =

Time taken to cross the telegraph post =

(b) Speed of man =

Time taken to pass the man = Length of the train/ Relative velocity =




(c) Time = Length of the train/ Relative velocity =

(d) Time = (Length of the train + Length of platform) / Relative velocity =

(e) Speed of the second train = mps

Time = Sum of the length of the two trains/ Relative velocity =

(f) Time = Sum of the length of the two trains/Relative velocity =

Boats & Streams

Let Speed of boat in still water = b km/hr

Speed of stream = w km/hr

Speed of boat with stream (Down Stream), D = b + w

Speed of boat against stream (Up stream), U = b

Speed of boat in still water, b = (D + U)

Speed of stream, w = (D – U)

Ex.4 A man can row 4.5 km/hr in still water. It takes him twice as long to row upstream as to row downstream. What is

the rate of the current?

Sol. Speed of boat in still water (b) = 4.5 km/hr.

It is given upstream time is twice to that of down stream.

Downstream speed is twice to that of upstream.

So b + u = 2(b – u)

u = = 1.5 km/hr.

Circular Motion

Circular Motion with two people

Ex.5 Sachin and Saurav, as a warm-up exercise, are jogging on a circular track. Saurav is a better athlete and jogs at

18km/hr while Sachin jogs at 9 km/hr. The circumference of the track is 500 m (i.e. ½ km). They start from the same

point at the same time and in the same direction. When will they be together again for the first time?

Sol. Method 1: Since Saurav is faster than Sachin, he will take a lead and as they keep running, the gap between them will

keep widening. Unlike on a straight track, they would meet again even if Saurav is faster than Sachin.

The same problem could be rephrased as “In what time would Saurav take a lead of 500 m over Sachin”?




Every second Saurav is taking a lead of over Sachin in Therefore; they would meet for

the first time after 200 sec. In general, the first meeting if both are moving in the same direction and after both have

started simultaneously occurs after

Time of first meeting =

Method 2: For every round that Sachin makes, Saurav would have made 2 rounds because the ratio of their speeds is

1 : 2. Hence, when Sachin has made 1 full round, Saurav would have taken a lead of 1 round. Therefore, they would

meet after [Here, is Sachin’s speed.]

Ex.6 Suppose in the earlier problem when would the two meet for the first time if they are moving in the opposite

directions?

Sol. If the two are moving in the opposite directions, then

Relative speed = 2.5 + 5 = 7.5 m/s.

[Hence, time for the first meeting = Circumference / Relative speed =

Ex.7 If the speeds of Saurav and Sachin were 8 km/hr and 5 km/hr, then after what time will the two meet for the first

time at the starting point if they start simultaneously?

Sol. Let us first calculate the time Saurav and Sachin take to make one full circle.

Time taken by Saurav =

Time take by Sachin =

Hence, after every 225 sec, Saurav would be at the starting point and after every 360 sec Sachin would be at starting

point. The time when they will be together again at the starting point simultaneously for the first time, would be the

smallest multiple of both 225 and 360 which is the LCM of 225 and 360. Hence, they would both be together at the

starting point for the first time after LCM (225, 360) = 1800 sec = 0.5 hr. Thus, every half an hour, they would meet at

the starting point.

Note: From the solution you could realise that it is immaterial whether they move in the same direction or in the opposite.

Circular motion with three people:

If three persons A, B, and C are running along a circular track of length d meters with speeds Va ,Vb ,Vc respectively. To

find the time when all the three will meet for the first time, we have to calculate the relative time of the meeting of any one

(A or B or C) among the three with other two runners and then calculate the LCM of these two timings. This will be the time

when all the three runners will meet for the first time.




And to calculate when they all meet for the first time at the starting point, we have to take the LCM of the timings taken by all

the runners separately to cover one full circular motion.

Ex.8 Let us now discuss the cases of circular motion with three people: Laxman joins Saurav and Sachin, and all of them

run in the same direction from the same point simultaneously in a track of length 500 m. Laxman moves at 3

km/hr, Sachin at 5 km/hr and Saurav at 8 km/hr. When will all of them be together again?

a. for the first time?

b. for the first time at the starting point?

Sol. (a) Break the problem into two separate cases.

In the first case, Saurav moves at the relative speed of (8 – 5) = 3 km/hr with respect to Sachin.

At a relative speed of 3 km/hr, he would meet Sachin after every

In the second case, Saurav moves at the speed of (8 – 3) km/hr = 5 km/hr with respect to Laxman.

At a relative speed of 5 km/hr, he would meet Laxman after every

If all the three have to meet, they would meet after every [LCM (10, 6)] min = 30 min or ½ hour.

Hence, they would all meet for the first time after 30 min.

(b) If we need to find the time after which all of them would be at the starting point simultaneously for the first time,

we shall use the same method as in the case involving two people.

At a speed of 8 km/hr, Saurav takes 225 sec. to complete one circle.

At a speed of 5 km/hr, Sachin takes 360 sec. to complete one circle.

At a speed of 3 km/hr, Laxman would take 600 sec. to complete one circle.

Hence, they would meet for the first time at the starting point after LCM (225, 360, 600) sec. = 1800sec.

Ex.9 A thief is spotted by a policeman from a distance of 200 m. When the policeman starts a chase, the thief starts

running. Speed of thief is 10 Kmph and that of policeman is 12 kmph. After how many hours will the policeman

catch the thief?

Sol.

Ex.10 A man steals a car at 1 : 30 pm & drives at 40 kmph. At 2 pm the owner starts chasing his car at 50 kmph. At what

time does he catch the man?

Sol. Distance covered by the thief in 1h = 40 km Distance covered in

Now, time taken to catch the thief = Back to Table of Contents



Relative velocity = 50 – 40 = 10 kmph ( Both are moving in same direction)

Time = 4 p.m. (2 p.m. + 2)

Ex.11 A and B started moving simultaneously from P towards Q and their respective speeds are 36 kmph and 15 m/s

respectively. What is the distance between them after moving for 2 minutes after starting from P?

Sol. Speed of A = 36 kmph

Speed of B = 15 m/s

So in 1 sec, B covers 5 m extra than A and so, the distance between them will be 5 m.

So in 2 mins = 120 sec, the distance between will be 120 × 5 m = 600 m.

Ex.12 Two trains are moving in opposite directions with the respective speeds of 36 kmph and 45 kmph. They will cross

each other in 20 seconds. If they are moving in the same direction at the same speed, how much time will they take

to cross each other?

Sol. Speeds = 36 kmph, 45 kmph = 10 m/s, 12.5 m/s

If they are moving in opposite directions, relative velocity = 10 + 12.5 = 22.5m/s.

And they take 20 seconds to cross each other.

If they move in the same direction, relative velocity = 12.5 – 10 = 2.5

Since the velocity in the second case is times the velocity in the first case.

So, they will take 9 times more time to cross each other. i.e., 20 × 9 = 180 sec = 3 min.

Ex.13 Two trains are moving in the same direction. The speed of the faster train is twice the speed of the slower train.

The faster train takes 60 sec to overtake the slower train. If they move in opposite directions, how much time will

they take to cross each other? Back to Table of Contents

Sol. Assume ‘v’ is the speed of the slower train, so 2v is the speed of the faster train.

If they move in the same direction, relative speed = 2v – v = v.

If they move in opposite directions, relative velocity = 2v + v = 3v.

Since, the velocity is three times, the time required is

i.e.

Ex.14 Train A starts from city P to city Q with a velocity of 40 kmph. Train B starts at city P towards city Q, 1 hour after

train A with a speed of 60 kmph. If both the trains reach station Q simultaneously, what is the distance between

cities, P and Q?

Sol. Assume, the distance as d.

(Since train B takes 1 hour less than that of train A) Back to Table of Contents



Therefore d =120 km

Ex.15 Two friends A and B are moving towards Q and P respectively from P and Q respectively. The distance between P

and Q is 600 m and the speeds of the friends, A and B are 6 m/s and 8 m/s respectively. How much time after A

starts for P, does B have to start for Q so that they meet at the exact midpoint of P and Q?

Sol. The midpoint means exactly 300 m from both the sides.

A can walk this 300 m at a speed 6 m/s speed in 50 sec.

Whereas B can cover this distance in

So B has to wait for (50 – 37.5) = 12.5 sec, if they want to reach the midpoint simultaneously

Ex.16 A river flows at a speed of 1.5 kmph and a boatman who can row his boat at a speed of 2.5 kmph in still water,

takes hours to go a certain distance up stream and return to the starting point. What is the total distance

covered by the boatman?

Sol. The speeds of the boatman upstream and downstream are 1 kmph and 4 kmph respectively.

If the distance covered each way = x km, then total time taken to go upstream and downstream

The distance covered both ways is 12 km.

Races & Games of Skill

Races:

Any contest of speed in running, riding, driving, sailing or rowing is called a race. The path on which the contests are

held is called a race course. The point from which the race begins is called the starting point. The point where the race ends is

called the winning post or the goal. The person who first reaches the winning post is called the winner. If all the contestants

reach the goal at the same time then the race is called a dead heat race.

There are two types of races

(a) Handicap Race:

In this type of races 1st runner gives 2nd runner benefit of running for some time before 1st start the race. e.g. A can give B a

handicap of 3 sec. mans they started from the same point but A start 3 sec after.

(b) Headway/Start race:

This means that runner A has started from the “K” distance behind runner B but at the same time. e.g. A gives B a start of 10

m means before starting A is on starting point and B is 10 meter forward from A and they start the race at the same time. In a

contest with 2 participants, one is the winner and the other is the loser.




(i) The winner can give/allow the loser a start of t seconds or x metres, i.e. start distance = x metres and start time = t seconds.

(ii) The winner can beat the loser by t seconds or x metres, i.e. beat distance = x metres and beat time = t sec

Interpretation from the statements given in the problems concerned with race

(1) A beats B by X meter:

This means that the winner of the race is A and B is X meters behind A when he crosses the finishing line.

(2) A and B start together at P:

This means both the runners have started from the same starting point. The ratio of the speed of the runners is the ratio of

the distance covered by them in the time, in which winner reaches the finishing line.

(3) A gives B a start of X metre:

This means that A stands at starting point and B is X meter forward of the starting line at the starting of the race.

(4) A beats B by t seconds

This states that after coming of A at the finishing point B will take more t seconds to cover the left distance. If we know that

distance and time taken we can cover the speed of the looser?

Winner's (A) time = Loser's (B) time – t

A and B starts together at P, A finishes at Q, but t seconds before B finishes.

Ex.17 A can run 330 metres in 41 seconds and B in 44 seconds. By how many seconds will B win if he has 30 metres start?

Sol. B runs 330 metres in 44 seconds.

B runs (330 – 30) metres in

But A runs 330 metres in 41 seconds

So, B wins by (41 – 40) seconds, i.e. 1 second.

Ex.18 In one kilometer race, A beats B by 36 metres or 9 seconds. Find A's time over the course.

Sol. Here A is the winner and B is the loser.

The time taken by B to cover the distance of 36 metres = 9 seconds.

Time taken by B to cover 1 kilometer = (1000 × 9)/36 = 250 sec.

Therefore time taken by A to cover 1km = 250 – 9 = 241 seconds.

Alternative method:

Using formula:

Escalator

Let us understand this concept with the help of an example. Back to Table of Contents



Ex.19 In an escalator Rajesh covers 4 steps with 3 steps of Suresh in the same time in a static escalator. When the

escalator is moving Rajesh takes 24 steps while Suresh takes 21 steps to reach the top of the escalator. What are

the total numbers of steps in the escalator?

Sol. Assume that S be the number of steps of escalator that helped Rajesh and Suresh for the same time.

Then, For every 4 steps of Rajesh, Suresh will take 3 steps.

So for 24 steps Rajesh take, Suresh will take 18 steps.

But Suresh has taken 21 steps which are of the original steps.

These extra steps taken are the result of extra time consumed by him and helped by the escalator. So, the equation

will become

Total number of steps in the escalator = 24 + S = 24 + 18 = 42 steps.

Ex.20 In a 100m race, A can beat B by 10m, and C by 20m. If B and C run with the same velocity, By how many meters can

B beat C in a 900 m race?

Sol. From the first line of the question we can understand that when A runs 100m, B runs 90 m and C runs 80 m. So B can

beat C by 10m in a 90 m race. So in a 900m race, B can beat C by 100m.

Ex.21 A can beat B by 44 metres in a 1760 meter race, while in a 1320 metres race, B can beat C by 30 metres. By what

distance (in meters) will A beat C in a 880 meter race?

Sol. When A runs 1760 metres, B runs 1716 metres.

When B runs 1320 metres, C runs 1290 metres.

When B runs 1716 metres, C runs 1290 (1719/1320) = 1677 metres.

A B C

1760 1716 1677

When A runs 880 metres, C runs 1677 (880/1760) = 838.5 metres.

A beats C by (880 – 838.5) = 41.5 metres.

Ex.22 Both A and B run a 2 km race. A gives B a start of 100 m and still beats him by 20 seconds. If A runs at a speed of 20

km per hour, find B's speed in kilometres per hour.

(1) 17 (2) 18 (3) 19 (4) 19.5

Sol. A covers a distance of 2 km. in (2/20) hour i.e. 360 sec.

B covers the distance of (2000 – 100) i.e. 1900 m in 360 + 20 i.e. 380 sec.

B's speed = (1900/380) = 5 m/s =

Ex.23 J is times as fast as K. If J gives K a start of 150 m, how far must be the winning post so that the race ends in a

dead heat? (1) 100 m (2) 440 m (3) 550 m (4) 200 m

Sol. Race ends in a dead heat, i.e. times taken by J and K are the same. Back to Table of Contents



Time & Work


1. In a field, 40 cows can graze a patch in 40 days or 30 cows in 60 days. If grass is growing at constant rate, then find in how

many days will 20 cows finish the field?

2.If 5/2 artists make 5/2 paintings using 5/2 canvases in 5/2 days, then how many artists are required to make 25 paintings

using 25 canvases in 25 days?

3. 'A' says, “If we sell seventy-five chickens, then the feed will last for extra 20 days”. However, 'B' says, “if we buy additional

100 chickens then, then the feed will get over 15 days earlier”. Find the number of chickens. Ans: 300


Time & Work

To start with, let’s take an example. If 2 men take 10 days to build a wall, 1 man will take 20 days (not 5 days) to build

the same wall. The question arises why? The reason is simple that man and time are inversely proportional to each other.

When men will decrease the number of days required to complete the work will increase. Here men become half, so time will

double and will complete the work in 20 days. -------- (1), Where M & D are number of men and number of days

respectively. If we further break number of days to hours, then total hours = DH, where H are number of hours per day. Now,

our formula becomes

To elaborate it further, let’s say M men take D days to build a room. Now if work is doubled (they have to build two

rooms of same size) in same D no. of days, obviously they have to double their strength Or we can say that no. of men are

directly proportional to the work to be finished. In mathematics, we can write it as

M W, -------------------------------------- (2)

Where W is the work to be finished.

By combining (1) and (2), we get (or) , Where K is constant of proportionality (or)



This is our general formula to solve time & work problems. It is also known as Work Equivalence Method. Majority of

time & work questions can be divided into two types.

Type I: Where efficiency of individual’s is not mentioned (as in the above example): These are the cases where the man-days

concept is utilized. Here the Rule of Fractions or the proportion (direct or indirect) concept can also be used.

Type II: Where the efficiency is mentioned. Here the case of unit day’s or one day’s concept is utilized.

Concept of unit time

If A does a work in 10 days, then what is the one-day work done by A? The answer is one-tenth part of the total work. We can

say that Unit timework =

Work done in unit time is known as efficiency of the worker or we can say, that if a worker takes less number of days (than

second worker) to finish a work, he will be more efficient.

LCM Approach

In this type of approach the total work is considered as the LCM of the time taken by the individuals and then unit time work is

calculated. Here the work done is not 1(unity) but the LCM of the time taken by number of persons. Let us understand this

with the help of an example.

Ex.1 If Manni and Gopi finish a work in 10 and 15 days respectively, what will be number of days taken by both of them

to complete the work when both work together?

Sol. Let the total work is equal to the LCM of number of days of Manni & Gopi taken to do the work respectively.

Work = LCM of 10 & 15 = 30 units

One day work of Manni =

One day work of Gopi =

One day work of both = 5 units

So number of days taken =

Simultaneous Working Problems

It is always a good way if we try to solve such type of questions with the help of LCM approach. Let us understand the concept

with the help of an example.




Ex.2 A, B and C can finish a work independently in 10, 12 and 15 days. C starts the work and after 1 day, B joins him.

After 1 day of B, A also joins them but leave 3 days before completion of the work, while B left two days before

completion of work. What will be the total number of days taken to complete the work?

Sol. Let total work is LCM (10, 12, 15) = 60 units

One-day work:

Let total days taken = N

Days of A = (N – 5), Days of B = (N – 3), Days of C = N

According to the question we have,

6(N – 5) + 5(N – 3) + 4N = 60

Solving we have N = 7 days

Alternate Days concept

In these, type of problems, we see the work done by two or more men on alternate days or hours.

Ex.3 A can build a wall in 20 days while B can build the same wall in 30 days. If they work on alternate days in how many

days, will the wall be completed if A start the job?

Sol. Let total work = 60 units

1 day work of A = 3 units and 1 day work of B = 2 units

2 days work = 3 + 2 = 5 units

To do 60 units, days required =

Negative Work Concept

These types of problems revel about the problems of pipe and cistern in which one is inlet and other is outlet. Let us solve a

puzzle “A monkey wants to climb a pole which is 24 m tall. In 1 minute, he can climb up by 3 m and in the next minute, he slips

by 2 m. In how many minutes will the monkey reach the top of the pole?

Ex.4 If A build the wall in 20 days and B can destroy that wall in 30 days and work on alternate days. What will be the

number of days required to build the wall for the first time?

Sol. A can build 1/20 th of the wall in 1 day. Where as B will destroy 1/30 th of the wall in 1 day and since they are working

on alternate days, So in 2 days, 1/20-1/30 = 1/60 th of the wall will be constructed. So, in 57 × 2 = 114 days, 57/60 th

of the work will be completed and on the 115th day, A will come and completes the remaining 3/60 = 1/20 th work. So

it takes 115 days to construct the wall.




Pipes and Cisterns

Pipes and Cisterns problems are almost the same as those of Time and Work problems. Thus, if a pipe fills a tank in 6

hrs, then the pipe fills 1/6 th of the tank in 1 hour. The only difference with Pipes and Cisterns problems is that there are

outlets as well as inlets. Thus, there are agents (the outlets) which perform negative work too. The rest of the process is

almost similar.

Inlet:

A pipe connected with a tank (or a cistern or a reservoir) is called an inlet, if it fills the tank. The work done by this is

taken as positive work.

Outlet:

A pipe connected with a tank is called an, outlet, if it empties the tank and the work done by this is taken as the

negative work.

Ex.5. Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened

simultaneously, how much time will be taken to fill the tank?

Sol. Part filled by A alone in 1 hour = 1/36

Part filled by B alone in 1 hour = 1/45

Part filled by (A + B) in 1 hour = 1/36+1/45 = 9/180 = 1/20

Hence, both the pipes together will fill the tank in 20 hours.

Ex.6. Pipe A can fill a tank in 36 hours and pipe B can empty it in 45hours. If both the pipes are opened simultaneously,

how much time will be taken to fill the tank?

Sol. Total Volume of the tank = 180 units (LCM of 36 and 45)

1 hour work of A = 5 units

1 hour work of B = – 4 units

1 hour work of A + B = 1 units

So, Total time taken = 180 hours

Ex.7. A can do a work in 30 days, and B can do in 40 days. If A and B work together for 10 days and A left, then C joined

with B and completed the work in 10 days. How many days C alone can complete the work?

Sol. A work only for 10 days, B work for 20 days and C work for 10 days. Let us assume C require X days to complete the

work. So, in 1 day, A can do 1/30th of the work. B can do 1/40 th of the work, and c can do 1/X of the work.

So, 10/30 + 20/40 + 10/X = 1. So X = 60 days.

So, C can do the work in 60 days.

Ex.8. Four men can do work in 15 days. If a man left the work after 5 days and again joined after 5 more days, the

remaining three work continuously till the end of work. How many more days than the estimated, it takes to

complete the work? Back to Table of Contents



Sol. After 15 days, the work that is left is equal to the amount of work that can be done by 1 man in 5 days.

This work can be done in 5/4 days by the 4 men. So it takes 5/4 more days to complete the work.

Ex.9. One man started a work on first day, the second day one more joined with him. The next day one more joined.

Every day one new person joined until the work gets completed. If the work is completed in 15 days, how many

days it takes for 10 men to complete the same work, if they work regularly?

Sol. Let us assume one man can do x amount of work in 1 day.

So the amount of work, that can be completed

On 1st day = x

2nd day = 2x

……………..

15th day = 15x

Total = Σ15 x = 120 x

But 10 men can do 10x work in 1 day.

So, 10 men take 12 days to complete the work.

Ex.10. B is twice as efficient as ‘A’ and ‘C’ is 50% more efficient than B. If B and C together can complete a work in 10 days,

how much time it takes for A and B to complete the work, if they work on alternate days starting with ‘A’?

Sol. If A can do x work in 1 day.

B can do 2x

C can do 3x

B and C together takes 10 days Total work = 50 x.

If A and B work on alternate days,

In 2 days x + 2x = 3x work will completed.

In 16 × 2 = 32 days 16 × 3x = 48x will be completed.

33rd day A will come and do x work. So, 48x + x = 49x will be completed.

34th day B will come and complete the work in 1/2 day.

So, answer = 33 1/2 days.





Heights & Distances

Angles and their relationship

Angles are measured in many units’ viz. degrees, minutes, seconds, radians, gradients. Where 1 degree = 60 minutes, 1 minute

= 60 seconds, π radians = 180° = 200g

1 radian = 180°/π and 1 degree = π/ 180 radians.

The angle at the centre is of 1 radian

Basic Trigonometric Ratios

In a right triangle ABC, if θ be the angle between AC & BC

If θ is one of the angle other then right angle, then the side opposite to the angle is perpendicular (P) and the sides containing

the angle are taken as Base ( B) and the hypotenuse (H). In this type of triangles, we can have six types of ratios. These ratios

are called trigonometric ratios.

Important Formulae

For any angle θ:

1. sin2θ + cos2θ = 1 [Note sin2θ = (sin θ)2 and not (sin θ2)]

2. 1 + tan2 θ = sec2θ

3. 1 + cot2θ = cosec2θ

Range of Values of Ratios

If 0 ≤ θ ≤ 360o, then the values for different trigonometric ratios will be as follows. Back to Table of Contents



1. – 1 ≤ Sin θ ≤ 1

2. – 1 ≤ cos θ ≤ 1

3. – ∞ ≤ tan θ ≤ ∞

4. – ∞ ≤ cot θ ≤ ∞

5. – ∞ ≤ Sec θ ≤ -1 & 1 ≤ Sec θ ≤ ∞

6. – ∞ ≤ Cosec θ ≤ – 1 & 1 ≤ Cosec θ ≤ ∞

Sign of Trigonometric ratios

We divide the angle at a point (i.e. 360°) into 4 parts called quadrants. In the first quadrant all the trigonometric ratios are

positive

Values of trigonometric Ratio for some special angles:

00 300 450 600 900

Sin 0

1

Cos 1

0

Tan 0

1

Properties of Triangle

Sine rule

In any triangle ABC if AB, BC, AC be represented by c, a, b respectively then we have

R Circum – Radius =

Cosine Rule

In a triangle ABC of having sides of any size, we have the following rule;

Area of Triangle

Area =

Where S = semi Perimeter





Sets

Defining a Set

"Set" is synonymous with the words "collection", "aggregate", "class" and is comprised of lements/objects/members.

A set is a well-defined collection of elements. By well-defined elements it means that given a set and an element, it

must be possible to decide whether or not the element belongs to the set. A set can be described in any one of the following

ways. For example, the set of beautiful Actress of Bollywood or the set of Good Players of cricket in India are not sets as the

world beautiful and Good are not well defined. What are the criteria of choosing an actress as beautiful and which player is

said to be a good player of cricket. But on the other hand if we say Set of good player of cricket in India those has played at

least 25 international games, will be a set as the word Good is now well defined.

Representation of a Set

A set can be described in two different ways.

1. Roaster Form:

A set is described by listing elements, separated by commas, within brackets. For example, A = {a, e, i, o, u} is a set of

vowels of English alphabets and a finite set and N = {2, 4, 6,…} is a set of even natural number and is a infinite set. Also,

repetition of an element has no effect. For example {1, 2, 3, 2} is the same set as {1, 2, 3}.

2. Set Builder Form:

A set can also be described by a characterizing property P(x) of its elements x. In such a case the set is described by {x

| P (x) holds} or {x: P(x) holds}, which is read as 'the set of all x such that P(x) holds'. For example,

A = {x | x is the root of x2 + 5x – 6 = 0}

Cardinal number of a set

The number of elements contained in a set is known as the cardinal number of that set. On the basis of cardinal

number a set can be an Empty Set or a Singleton Set. A set is said to be empty or void or null set if it has no elements and its

cardinal number is 0. It is denoted by the symbol φ or { } such as Set of odd numbers divisible by 2. And a set is singleton set if

the cardinal number is 1 i.e. it contains only one element, such as Set of lady Prime-minister of India.

If a set A has 5 elements then its cardinal number is written as n(A) = 5, and read as “Cardinal number of set A is 5”.

Subsets

Let A and B be two sets. If every element of A is contained in B, then A is called the subset of B. If A is subset of B, we

write A B, which is read as "A is a subset of B" or "A is contained in B". Back to Table of Contents



If A B then a A a B. (The symbol stands for "implies" and stands for “belongs to”). Let P = {a, b, c}, then

subsets of P are φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} i.e. it has 8 number of subsets and Empty set is the subset of every.

Also every set is the subset of itself and it is improper subset of the set. Every set has one improper subset and other subsets

as proper subsets.

If A B and also B A, then we can say every element of A is present in B and every element of B is present in A.

These types of sets having same number and identical elements are known as Equal sets.

Power Set

The set of subsets of a set is the power set of the set. If A is a set then its Power set is denoted as P(A). If A = {a, b, c},

then P(A) = {φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} The cardinal number of any set A is 2n, where n = number of elements

present in set A.

Universal Set

In any discussion in theory, there happens to be a set U that contains all sets under consideration. Such a set is called

the universal set. Thus a set that contains all sets in a given context is called the universal set. For example, in plane geometry

the set of all points in the plane is the universal set.

Some Important Universal Sets:

N = Set of all natural numbers = {1, 2, 3, 4,…}

Z or I Set of all integers = {…– 3, – 2, – 1, 0, 1, 2, 3,..}

Z+ = Set of all positive integers = {1, 2, 3,…} = N

Z– = Set of all negative integers = {– 1, – 2, – 3,..}

W = Set of all whole numbers = {0, 1, 2, 3,…}

Z0 = The set of all non-zero integers = {± 1, ± 2, ± 3,..}

Q = The set of all rational numbers.

=

R = The set of all real numbers.

R –Q =The set of all irrational numbers.

e.g , , , ……. , e, log 2 etc… are all irrational numbers.

Operations of sets

Union of Sets:

The union of two sets A and B, is a set containing all the elements present in set A or in set B. The union of set A and B

is represented as A U B. So, A U B = {x | x A or x B}

For example:

A = {5, 7, 8, 9, 11} and B = {car, house, ball, sofa), then A U B = {5, 7, 8, 9, 11, car, house, ball, sofa} Back to Table of Contents



Intersection of Sets:

The intersection of two sets A and B, is a set containing all the common elements present in set A or in set B. The

intersection of set A and B is represented as A B. So, A B = {x | x A and x B}

For example

A = {5, 7, 8, 9, 11} and B = {3, 4, 5, 6, 7, 8}, then

A B = {5, 7, 8}

Difference of sets:

A – B is the difference of set A from set B and is defined as the set of elements present in set A but not in set B.

So, A – B = {x | x A and x B}

Complement of a set:

The complement of set A in U the set of those elements which are present in Universal set but not present in set A.

Compliment of A is denoted by A' The shaded part represents A'. So, A‘= {x | x U and x A}

(a) Also the Compliment of set A is defined as the difference of Universal set U and set A. i.e.⇒ A‘ = U – A

Facts and Rules:

i. A A = A & A A = A

ii. A φ = A & A φ = φ

iii. A U = U & A U = A

iv. A B = B A & A B = B A

v. A A’ = & A A’ = φ

vi. n (A B) = n (A) + n (B) – n (A B)

vii. n (A B C) = n (A) + n (B) + n (C) – n (A B) – n (B C) –n (C A) + n (A B C)

viii. n (A – B) = n (A) – n (A B)

Ex.1 In a certain city only two newspapers A & B are published. It is known that 25% of the city population read A & 20%

read B while 8% read both A & B. It is also known that 30% of those who read A but not B, look into advertisements

and 40% of those who read B but not A, look into advertisements while 50% of those who read both A & B look

into advertisements. What % of the population reads an advertisement?

Sol. Let A & B denote sets of people who read paper A & paper B respectively and in all there are 100 people, then

n (A) = 25, n(B) = 20, n (A B) = 8.

Hence the people who read paper A only i.e. n (A – B) = n (A) – n (A B) = 25 – 8 = 17. And the people who read paper

B only i.e. n (B – A) = n (B) – n (A B) = 20 – 8 = 12. Now percentage of people reading an advertisement = [(30 % of

17) + (40% of + 12) + (50% of 8)] % = 13.9 %. Back to Table of Contents



Venn Diagrams

Venn diagram is the pictorial representation of the set and also the operation involved in the sets. We often use

circles to represent the sets and overlapping of the circles to represent the common elements in two or more sets. The

universal set U is represented by interior of a rectangle and its subsets are represented by interior of circles within the

rectangle.

Maximum and Minimum elements in a set

Let us consider the formula, n (A B) = n (A) + n (B) – n (A B) and imagine that the value of n (A B) is not given

and asked to find the value of n (A B), then we will get the range of the values. Let us understand the concept of maximum

and minimum values with the help of an example.

Ex.2 If in a Survey, Organized by any N.G.O on the cold drinks after effects found that 80% of the total people like Coca -

Cola and 70% like Limca. What can be the minimum and maximum number of people who drink both the drinks?

Sol. Limca Coca Cola

80 - X X 70 - X

Here n (L C) ≤ 100%

Using the formula, n (L) + n (C) – n (L C) = n (L C) ≤ 100%

80 + 70 – X ≤ 100

Solving we get X ≥ 50%, Also X ≤ 70%

Minimum value = 50% and Maximum value = 70%.


Functions

Definition

Suppose, if we take any system, the output will be a function of input. That means a function is a relation between

input and output. For Example, there is a system, which finds the square of the given input. That means the output is a square

of the given input. This can be represented by output= (input) 2 or f(x) = x2. Where x is input and f(x) is output. Here f is called

the function of x, which is defined as f(x) = x2. So, if f(x) = x 2 , f(1) = 12 = 1, and f(2) = 22 = 4.

In general, if f(x) = x2, f(a) = a2. Back to Table of Contents



Ex.1 (a) If f(x) = 2x2 – 2x + 1, find f( – 1). (b) If f(t) = 3t – 1, find f(a2)

Sol. (a) We substitute – 1 in place of x. f( – 1) = 2( – 1)2 – 2 ( – 1) + 1 = 5

We substitute a2 in place t. f(a2) = 3(a2) – 1 = 3a2 – 1.

Odd and Even Functions

Odd function:

A function f is said to be odd if it changes sign when the sign of the variable is changed. i.e. If f(– x) = – f (x). For

example: f (x) = sin x ; 0 ≤ x ≤ 2π is a odd function.

Even function:

A function f is said to be an even function if it doesn’t change sign when the sign of the variable is changed. i.e. if f(– x)

= f (x).For example f (x) = x4 + x2 and g (x) = cos x are even functions.

NOTE:

There are many functions which are neither odd nor even i.e it is not necessary for a function to be either even of to

be odd. E.g: g (x) = 3x3 + 4x2 – 9 is a function in x which is neither even nor odd.

Composite Functions

A composite function is the function of another function. If f is a function from A in to B and g is a function from B in

to C, then their composite function denoted by ( g o f) is a function from A in to C defined by (g o f) (x) = g [ f(x)]

e.g. if f(x) = 2x, and g(x) = x + 2, Then (gof)(x) = g [ f(x)] = g (2x) = 2x + 2

(fog)(x) = f [g(x)] = f(x + 2) = 2(x + 2) = 2x + 4

FACT: This shows that it is not necessary that (fog)(x) = (g of)(x).

Ex.2 Let a function fn+1 (x) = fn (x) + 3 . If f2(2) = 4. Find the value of f6(2).

Sol. We have, fn+1(x) = fn(x) + 3

So, f3(2) = f2(2) + 3 = 7

f4(2) = f3(2) + 3 = 10

f5(2) = f4(2) + 3 = 13

f6(2) = f5(2) + 3 = 16

Alternate Method:

Since the function is increasing with constant value.

So, f6(2) = f2(2) + 3( 6 – 2) = 4 + 12 = 16

Ex.3 If f(x) = 2x2 – 3, then f(2) = ?

Sol. f(2) = 2(2)2 – 3 = 5

Ex.4 f(x) = if f(2) + f(3) = f(5), then K =

Sol. f(2) + f(3) = = Back to Table of Contents



f(5) =

Ex.5 f (n + 1) = 2f (n) + 4 & f(1) = 0. Then f (5) =

Sol. f(2) = 2f(1) + 4 = 0 + 4 = 4

f(3) = 2 × 4 + 4 = 12

f(4) = 2 × 12 + 4 = 28

f(5) = 2 × 28 + 4 = 60

Ex.6 f(x) = x + 5, g(x) = x2 – 3. Then gofog(3) is :

Sol. gofog(3) = gof(32 – 3) = gof(6) = g(6 + 5)

= g(11) = 112 – 3 = 118.

Ex.7 f(x) = 3x2 – 2x + 4 then f(1) + f(2) + - - - - + f(20) =

Sol. =

= = 8270


Sequence & Series

A set of numbers whose domain is a real number is called a SEQUENCE and sum of the sequence is called a SERIES. If

a1, a2, a3, a4… an … is a sequence, then the expression a1 + a2 + a3+ a4+ a5 + …+ an + … is a series.

Those sequences whose terms follow certain patterns are called progressions.

For example

1, 4, 7, 10, 13 …….

7, 4, 1, – 2, – 5………

1, 2, 4, 8, 16………

8, 4, 2, 1, ½….……

Also if f (n) = n2 is a sequence, then f(1) = (1)2 = 1, f(2) = 22 = 4, f(3) = (3)2 = 9, f (10) = 102 = 100 and so on.

The nth term of a sequence is usually denoted by Tn

Thus T1 = first term, T2 = second term, T10 = tenth term and so on.

There are three different progressions Back to Table of Contents



Arithmetic Progression (A.P)

Geometric Progression (G.P)

Harmonic Progression (H.P)

Arithmetic Progression (A.P.)

It is a series in which any two consecutive terms have common difference and next term can be derived by adding

that common difference in the previous term. Therefore Tn+1 – Tn = constant and called common difference (d) for all n N.

Examples:

1. 1, 4, 7, 10, ……. is an A. P. whose first term is 1 and the common difference is d = (4 – 1) = (7 – 4) = (10 – 7) = 3.

2. 11, 7, 3, – 1 …… is an A. P. whose first term is 11 and the common difference d = 7 – 11 = 3 – 7, = – 1 – 3 = – 4.

If in an A. P. a = first term, d = common difference = Tn – Tn-1

Tn = nth term (Thus T1 = first term, T2 = second term, T10 tenth term and so on.)

l = last term,

Sn = Sum of the n terms.

Then a, a + d, a + 2d, a + 3d... are in A.P.

nth term of an A.P.

The nth term of an A.P is given by the formula Tn = a + (n – 1) d

Note: If the last term of the A.P. consisting of n terms be l, then l = a + (n – 1) d

Sum of n terms of an A.P

The sum of first n terms of an AP is usually denoted by Sn and is given by the following formula:

Where ‘l ’ is the last term of the series.

Ex.1 Find the series whose nth term is Is it an AP series? If yes, find 101st term.

Sol. Putting 1, 2, 3, 4….. We get T1, T2, T3, T4…..

= 1, 5/2, 4, 11/2……

d1 = 3/2, d2 = 3/2, d3 = 3/2

As the common differences ae equal

Therefore the series is an AP

T101 = a+ 100d = 1+100*(3/2) = 151

Ex.2 Find 8th, 12th and 16th terms of the series; -6, -2, 2, 6, 10, 14, 18…..

Sol. Here a = -6 and d = -2 –(-6) = 4

Therefore T8 = – 6 + 7 × 4 = 22 [T8 = a + 7d] Back to Table of Contents



T12 = a + 11d = – 6 + 11 × 4 = 38 [T12 = a + 11d]

T16 = a + 15d = – 6 + 15 × 4 = 54 [T16 = a + 15d]

Properties of an AP

I. If each term of an AP is increased, decreased, multiplied or divided by the same non-zero number, then the resulting

sequence is also an AP.

For example: For A.P. 3, 5, 7, 9, 11…

If you add constant let us say 1 in each

term, you get

4, 6, 8, 10, 12...... This is an A.P. with common difference

2

If you multiply by a constant let us say

2 each term, you get

6, 10, 14, 18, 22….. Again this is an A.P. of common

difference 4

II. In an AP, the sum of terms equidistant from the beginning and end is always same and equal to the sum of first and last

terms.

III. Three numbers in AP are taken as a – d, a, a + d.

For 4 numbers in AP are taken as a – 3d, a – d, a + d, a + 3d

For 5 numbers in AP are taken as a – 2d, a – d, a, a + d, a + 2d

IV. Three numbers a, b, c are in A.P. if and only if 2b = a + c

Ex.3 The sum of three numbers in A.P. is – 3, and their product is 8. Find the numbers.

Sol. Let the numbers be (a – d), a, (a + d). Then,

Sum = – 3 (a – d) + a + (a + d) = – 3 3a = – 3 a = – 1

Product = 8

(a – d) (a) (a + d) = 8

a (a2 – d2) = 8

(–1) (1 – d2) = 8

d2 = 9

d = ± 3

If d = 3, the numbers are – 4, – 1, 2. If d = – 3, the numbers are 2, – 1, – 4.

Thus, the numbers are – 4, – 1, 2 or 2, – 1, – 4.

Ex.4 A student purchases a pen for Rs. 100. At the end of 8 years, it is valued at Rs. 20. Assuming that the yearly

depreciation is constant. Find the annual depreciation.

Sol. Original cost of pen = Rs. 100 Back to Table of Contents



Let D be the annual depreciation.

Price after one year = 100 – D = T1 = a (say)

Price after eight years = T8 = a + 7 (– D) = a – 7D

= 100 – D – 7D = 100 – 8D

By the given condition 100 – 8D = 20

8D = 80 D = 10.

Hence annual depreciation = Rs. 10.

Geometric Progression

A series in which each preceding term is formed by multiplying it by a constant factor is called a Geometric

Progression G. P. The constant factor is called the common ratio and is formed by dividing any term by the term which

precedes it. In other words, a sequence, a1, a2, a3… an… is called a geometric progression

If = constant for all n N

The General form of a G. P. with n terms is a, ar, ar2,…ar n –1

Thus if a = the first term

r = the common ratio,

Tn = nth term and

Sn = sum of n terms

General term of GP = Tn = ar n –1

Ex.5 Find the 9th term and the general term of the progression.

Sol. The given sequence is clearly a G P with first term a = 1 and common ratio = r =(-1/2)

Now T9 = ar8 =

And Tn = arn-1 = (-1/2) n-1

Sum of n terms of a G.P:

Where r>1

Where r<1

Sum of infinite G.P:

If a G.P. has infinite terms and –1 < r < 1 or < 1, then sum of infinite G.P is

Ex.6 The inventor of the chess board suggested a reward of one grain of wheat for the first square, 2 grains for the

second, 4 grains for the third and so on, doubling the number of the grains for subsequent squares. How many

grains would have to be given to inventor? (There are 64 squares in the chess board).

Sol. Required number of grains Back to Table of Contents



= 1 + 2 + 22 + 23 + ……. To 64 terms = 1 =

Recurring Decimals as Fractions

If in the decimal representation a number occurs again and again, then we place a dot (.) on the number and read it as that

the number is recurring. e.g., 0.5 (read as decimal 5 recurring).

This mean 0. 5 = 0.55555…….

0. = 0.477777……

These can be converted into fractions as shown in the example given below

Ex.7 Find the value in fractions which is same as of 0.4

Sol. We have 0.4 = 0.4373737……….

= 0.4 + 0.037 + 0.00037 + 0.0000037 + ………..

= +…………

=

=

Properties of G.P

I. If each term of a GP is multiplied or divided by the same non-zero quantity, then the resulting sequence is also a GP.

For example: For G.P. is 2, 4, 8, 16, 32…

If you multiply each term by constant

let say 2,you get

4, 8, 16, 32, 64.. This is a G.P

If you divide each term by constant

let say 2,you get

1, 2, 4, 8, 16 .. This is a G.P

II. SELECTION OF TERMS IN G.P.

Sometimes it is required to select a finite number of terms in G.P. It is always convenient if we select the terms in the

following manner:

No. of terms Terms Common ratio

3

4

5

If the product of the numbers is not given, then the numbers are taken as a, ar, ar2, ar3, …. Back to Table of Contents



III. Three non-zero numbers a, b, c are in G.P. if and only if b2 = ac. b is called the geometric mean of a & c

IV. In a GP, the product of terms equidistant from the beginning and end is always same and equal to the product of first and

last terms as shown in the next example.

Harmonic Progression (H.P.)

A series of quantities is said to be in a harmonic progression when their reciprocals are in arithmetic progression.

e.g. 1/3, 1/5, 1/7, ….. and 1/a, 1/ (a+d), 1/(a+2d)….. are in HP as their reciprocals

3, 5, 7 ….. and a, a+d, a+2d……. are in AP

nth term of HP

Find the nth term of the corresponding AP and then take its reciprocal.

If the HP be 1/a, 1/ (a+d), 1/(a+2d)…..

The corresponding AP is a, a+d, a+2d…….

Tn of the AP is a + (n-1)d

In order to solve a question on HP, one should form the corresponding AP

A comparison between AP and GP

Description AP GP

Principal Characteristic Common Difference (d) Common Ratio (r)

nth Term

Mean

Sum of First n Terms

Mean

Arithmetic - Geometric Progression

a + (a+d)r + (a+2d)r2 + (a+3d)r3 + ……. Is the form of Arithmetic Geometric Progression (A.G.P)

One part of the series is in Arithmetic progression and other part is a Geometric progression.

The sum of n terms series is –

The infinite term series sum is

Arithmetic Geometric Series can be solved as explained in the example below:

Relation between AM, GM and HM:

A =Arithmetic mean =




G = Geometric mean =

H = Harmonic mean =

Multiplying A and H, we get AH =

=

Ex.8 Find the sum of 1 +2x +3x2 + 4x3 + …..

Sol. The given series is an Arithmetic – Geometric series whose corresponding A.P and G.P are 1, 2, 3, 4…

And 1, x, x2, x3… respectively. The common ratio of the G.P is x.

= 1 +2x+ 3x2+4x3+….. ------------- (i)

x = x+2x2+3x3+…… -------------- (ii)

Subtracting (ii) from (i), we get

x = 1 +[x+x2+x3+…… ]

x) = 1 +

=

Ex.9 If the first item of an A>P is 12, and 6th term is 27. What is the sum of first 10 terms?

Sol. a = 12, t6 = a +5d = 27 d =3

Therefore S10 = (10/2)[2*12 +(10 -1)3] = 255

Ex.10 If the fourth and sixth terms of an A.P are 6.5 and 9.5. What is the 9th term of that A.P?

Sol. a+3d = 6.5 and a+5d = 9.5

a =2 and d =1.5

Therefore t9 = a+8d = 14

Ex.11 What is the arithmetic mean of first 20 terms os an A.P. Whose first term is 5 and 4th term is 20?

Sol. a = 5, t4 = a+3d = 20

d =5

A.M is the middle number = average of 10th and 11th number = 50+55/2 = 52.5 Back to Table of Contents



Ex.12 The first term of a G.P is half of its 4th term. What is the 12th term of that G.P, if its sixth term is 6

Sol. t1 = t4

a = ar3

r3 = 2

t6=ar5 = 6

t12 = ar11 = ar5 r6 = 24

Ex.13 If the 1st and 5th terms of a G.P are 2 and 162. What is the sum of these five terms?

Sol. a = 2

ar4 = 162

r = 3

S5 = 2(35 – 1)/(3-1) = 242

Ex.14 What is the value of r +3r2 + 5r3 + …….

Sol. Assume S = r +3r2 + 5r3 + ……. ----------(1)

rS = r2 + 3r3 + 5r4+ ……. -----------(2)

(1) – (2)

S(1 –r) = r + 2r2 + 2r3 + …….

= r + 2r2 + 2r3 + ……..

S =

Ex.15 The first term of a G.P. 2 and common ratio is 3. If the sum of fist n terms of this G.P is greater than 243 then the

minimum value of ‘n’ is

Sol. > 243

> 243

> 244

n >5

So, minimum possible value of n is 6.

Ex.16 + + + ……. + is

Sol. =

=

= Back to Table of Contents



=

Ex.17 an = 2n + 1 then (a1 + a2 + …….. + a20) – a21 is:

Sol. a1 + a2 + …….. + a20 = 21 + 1 + 22 + 1 + 23 + 1 …….. + 220 + 1

= 2(220 – 1) + 20 = 221 + 18

Therefore Answer = 17

Clocks & Calendar


1. There are two friends P and Q, talking to each other about their office timing. P asked Q, “Why u did come to office

yesterday”. Q replied, “I came but I returned immediately”. P asked, “Can u tell me what time u reached office”. Q replied,

“When I started from my house between 9 AM and 10 AM and I returned between 10 AM and 11 AM, with the interchanged

position of minute and hour hand. Find the correct when he left the house.

2. One quarter of the time until now from midnight and half of the time remaining from now up to midnight, add to the

present time. What is the present time?

3. In a particular clock, minute and hour hand were meeting after every 65 minutes. Does the clock lose or gain the time and

by what amount it will gain or loss per hour?

4. At 6 o’clock clock ticks for 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12'o clock.

Ans: 66 sec

5. Three friends A, B and C are talking about a party. About the date of the party, they made the following statements:

A said, “It was on May 8th, Thursday”

B Said, “It was on May 10th, Tuesday”

C said, “It was on June 5th, Friday”

Among the three statements only one is completely true (Month, Date and Day). In the other two statements one told the day

right and the other said date right. Moreover, it is known that April 1st was Tuesday.

What is correct date, month and day? Ans: May 10th Sunday

6. Given that fifty minutes ago, it was four times as many minutes past three o'clock. How many minutes is it to six o'clock?

7. If a clock takes 7seconds to strike 7, how long will the same clock take to strike 10?

Ans: The clock strikes for the first time at the start and takes 7 seconds for 6 intervals thus for one interval time taken=7/6.

Therefore, for 10 seconds there are 9 intervals and time taken is 9*7/6 seconds.





Clocks & Calendar

The dial of a clock is a circle whose circumference is divided into 12 parts, called hour spaces. Each hour space is

further divided into 5 parts, called minute’s spaces. This way, the whole circumference is divided into 12 × 5 = 60 minute

spaces. If we consider clock as a circular track and the two hands of clock minute hand and hour hand are just like two players

running in the same direction. Let total length of the track is 360o and minute hand completes one full round in 1 hour while

hour hand cover full round in 12 hours.

Speed of hour hand = =

Speed of minute hand =

Since they are moving in the same direction, so the relative speed of both the hands with respect to each other =

Time taken by minute hand to overtake hour hand =

There are 4 types of problems on clocks:

1. To calculate the angle between the two hands when time is given.

2. To calculate the time when both the hands will be at some angle.

3. Concept of slow and fast clocks.

4. Overall gain/loss

Calculating the angle

The angle between the two hands is given by the following formula

Formula for the angle between the hands =

Where H ---- Hour Reading & M---Minute Reading




Ex.1 Calculate the angle between the two hands of clock when the clock shows 5: 25 p.m.

Sol. Given time = 5: 25 p.m. Hence H = 5 and M = 25

We can apply the following direct formula to find the angle between the hands

=

Required angle = =

Alternative Method:

Since at 5: 25 the minute hand will be at 5 and the angle between them will be same as the distance covered in

degree by the hour hand in 25 minutes. Required angle = distance of hour hand = speed × time = =

Calculating the time

To calculate the time when both the hands will be at some angle

In one minute the net gain of minute hand over hour hand =

If the gain is then the time is 1 min.

If the gain is then the time is

If the gain is then the time is

Note:

If between H and (H+1) o’clock, the two hands are together at an angle then required time =

minutes, where H is reading of hour.

Ex.2 At what time between 4 and 5 o’clock are the hands of the clock together?

Sol. Method 1: 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 minutes over the hour hand. Now, we know that 55 min. are

gained in 60 min.

20 min are gained in =

Therefore, the hands will be together at

Alternate method: Using the formula:

Required time =

Here = (Hands of clock are together) and H =4

Required time = = 21 min

Therefore, the hands will be together at 21 min past 4. Back to Table of Contents



Another formula:

Between H and (H+1) o’clock, the two hands will be together at 5 min past H.

In this case; 5 = 21 min past 4

Ex.3 At what time between 4 and 5 o’clock will the hand of clock be at right angle?

Sol. At 4o’clock there are 20 min. spaces between hour and minute hands. To be at right angle, they should be 15 min

spaces apart. So, there are two cases:

Case I:

When the minute hand is 15 min spaces behind the hour hand

To be in this position, the min hand should have to gain 20 – 15 = 5 min spaces.

Now, we know that 55 min spaces are gained in 60 min.

Therefore 5 min spaces are gained in

Therefore they are at right angle at past 4

Case II:

When the minute hand is 15 min spaces ahead of the hour hand

To be in this position, the min hand should have to gain 20 + 15 = 35 min spaces.

Now, we know that 55 min spaces are gained in 60 min

Therefore 5 min spaces are gained in

Therefore they are at right angle at past 4

Alternate Method:

As the hands of the clock are at right angle therefore

Also time is between 4 and 5 o’clock, no of hours = 4

Required time =

= = 38 or 5 min

Therefore they are at right angle 38 or 5 min past 4

Another Formula:

Between x and (x + 1) o’clock the two hands are right angle at (5X min past X

In this case; they will be at right angle at (5 min past 4 = 38 or 5 min past 4

Concept of slow and fast clocks

Too Fast and Too Slow: Back to Table of Contents



If a watch indicates 9.20 when the correct time is 9.10, it is said to be 10 minutes too fast. And if it indicates 9.00

when the correct time is 9.10, it is said to be 10 minute too slow.

Ex.4 Two clocks are set at 1 p.m. Fast clock gains 1 min for every hour. Find the time when the fast clock shows

6 p.m.

Sol. For every 60 min of true clock, the fast clock will show 61 min.

For 61 minutes of fast clock, true time = 60 minutes

For 300 minutes (5hrs) of fast clock; true time = =

Actual time in the true clock =5

Overall gain/loss

After every 65 min = min the two hands will coincide. If the hands of a clock coincide every ‘X’ min, then Gain/Loss per

day by a watch, is given by [If answer is (+) then there will be gain and if (-) then there will be loss.]

CALENDAR

There are 12 months in 1 year.

Jan –31 (days), Feb –28; 29 (for a leap year), Mar –31, Apr – 30, May –31, June – 30; July – 31; Aug – 31; Sept –30;

Oct – 31; Nov –30; Dec – 31

Total 365 (ordinary year); 366 days (for a leap year)

To find the number of weeks in an ordinary year:

= 52 weeks + 1 odd day (remainder)

To find the number of weeks in a leap year:

= 52 weeks + 2 odd day (remainder)

Leap year: It is so called as it comes after a leap of 3 years from the previous leap year.

Method to find whether a given year is a leap year or an ordinary year

Every year which is not a century year (i.e. which is not a multiple of 100) is a leap year if and only if it is completely

divisible by 4.

Every century year is a leap year if and only if it is completely divisible by 400 or is an integral multiple of 400 (i.e. the

remainder ought to be 0).

e.g. 2000 is a leap year.

1900 is not.

1996 is a leap year, 1998 is not

Odd day’s concept Back to Table of Contents



To find the number of odd days in a century

A century, i.e. 100 year has 76 ordinary year and 24 leap year

= [(76 × 52) weeks + 76 days] – [(24 × 52) weeks + 24 × 2 days]

= 5200 weeks + 124 days

= 5200 weeks + 17 weeks + 5 odd days

= 5217 weeks + 5 odd days

Therefore, 100 years contain 5 odd days.

Now,

(i) 200 years contain 5 × 2 = 10, i.e., 3 odd days.

(ii) 300 years contain 5 × 3 = 15 i.e., 1odd day.

(iii) 400 years contain 5 × 4 + 1 = 21, i.e., no odd day.

Similarly, 800, 1200 years etc. contain no odd day.

Note:

(i) 5 × 2 = 10 days = 1 week + 3 days i.e., 3 odd days

(ii) 5 × 3 = 15 days = 2 weeks + 1 day i.e. 1 odd day.

(iii) 400th year is a leap year therefore one additional day is added.

Odd days and their numeral values

When we have to calculate the number of days on any particular Extra Days and their numeral Values

0 → Sunday

1 → Monday

2 → Tuesday

3 → Wednesday

4 → Thursday

5 → Friday

6 → Saturday

Ex.5 Father of Nation Mahatma Gandhi died on 30th January 1948. What was the day on which he died?

Sol. Up to 1600 AD we have 0 odd days; up to 1900 AD we have 1 odd day. Now for in 47 years we have 11 leap years and

36 normal years.

Odd days from 1901 to 1947 = 11 x 2 +36 x1 = 22 + 36 =58 odd days

= 8 weeks + 2 odd days

Total odd days up to 31st December 1947 = 1 + 2 = 3 odd days

30 days of January contain only 4 weeks + 2 odd days Back to Table of Contents



So 30th January 1948 has total 5 odd days

Day on 30th January 1948 = Friday.

Ex.6 How does the number of odd days help us in finding the day of a week? (Please take care of this point) When a

specific day is given:

Suppose a question like:

Jan 1, 1992 was Wednesday. What day of the week will it be on Jan 1, 1993? If you recall, 1992 being a leap year it

has 2 odd days. So, the above said day will be two days beyond Wednesday, i.e., it will be Friday.

When no specific day is given:

Here, we count days according to number of odd days.

Sunday for 0 odd day, Monday for 1 odd day and so on. (i.e. from 0 to 6; 6 being Saturday)

Suppose someone asks you to find the day of the week on 12th January 1979.

12th Jan, 1979 means 1978 year + 12 days

Now, 1600 years have 0 odd day.

300 years have 1 odd day

78 years have 59 ordinary year + 19 leap years = 6 odd days.

Total no. of odd days = 0 + 1 + 6 + 12 = 19 or 5 odd days. So, the day was “Friday”.

Points not to be ignored:

1. 400th year is a leap year or a century multiple of 400 is a leap year, rests are not.

2. 100 years has 5 odd days

200 years has 3 odd days

300 years has 1 odd day

400 years has 0 odd days. And so on




Geometry (2D & 3D)/ Coordinate Geometry


1. A fly is 1 ft below the ceiling right across a wall length is 30m at equal distance from both the ends. There is a spider 1 ft

above floor right across the long wall equidistant from both the ends. The width and height of the room are 12m and 12m.

What is the minimum distance to be travelled by the spider to catch the fly?


Geometry (2D & 3D)/ Coordinate Geometry

Properties of lines

Intersecting Lines and Angles: If two lines intersect at a point, then opposite angles are called vertical angles and have the

same measure.

Perpendicular Lines: An angle that has a measure of 90o is a right angle.

Parallel Lines: If two lines that are in the same plane do not intersect, the two lines are parallel.

Parallel lines cut by a transverse: If two parallel lines L1 and L2 are cut by a third line called the transverse.

Polygon

A closed plane figure made up of several line segments that are joined together is called a polygon. The sides do not cross

each other. Exactly two sides meet at every vertex.

Types of Polygons:

Regular: all angles are equal and all sides are the same length. Regular polygons are both equiangular and equilateral.

Equiangular: all angles are equal.

Equilateral: all sides are the same length.

Triangles

A triangle is a polygon of three sides.

A triangle with three sides of different lengths is called a scalene triangle. An isosceles triangle has two equal sides.

The third side is called the base. The angles that are opposite to the equal sides are also equal. An equilateral triangle has

three equal sides. In this type of triangle, the angles are also equal, so it can also be called an equiangular triangle. Each angle




of an equilateral triangle must measure 60o, since the sum of the interior angles of any triangle must equal to 180o.

Obtuse angled triangle: is a triangle in which one angle is always greater than 90o

Acute angled triangle: In which all angles are less than 90o

Right Angled Triangle: A triangle whose one angle is 90o is called a right (angled) Triangle.

Properties of a Triangle

1. Sum of the three angles is 180o.

2. An exterior angle is equal to the sum of the interior opposite angles.

3. The sum of the two sides is always greater than the third side.

4. The difference between any two sides is always less than the third side.

5. The side opposite to the greatest angle is the greatest side and the side opposite to the smallest angle will be the shortest

side.

Centroid:

(a) The point of intersection of the medians of a triangle. (Median is the line joining the vertex to the mid-point of the

opposite side. The medians will bisect the area of the triangle.)

(b) The centroid divides each median from the vertex in the ratio 2 : 1.

Orthocentre:

The point of intersection of altitudes. (Altitude is a perpendicular drawn from a vertex of a triangle to the opposite side.)

Circumcentre:

The circumcentre of a triangle is the centre of the circle passing through the vertices of a triangle. It is also the point of

intersection of perpendicular bisectors of the sides of the triangle.

If a, b, c, are the sides of the triangle, Δ is the area, then abc = 4R Δ where R is the radius of the circum-circle.

Incentre:

The point of intersection of the internal bisectors of the angles of a triangle

Congruent triangles

Two triangles ABC and DEF are said to the congruent, if they are equal in all respects (equal in shape and size).

The tests for congruency

(a) SAS Test: Two sides and the included angle of the first triangle are respectively equal to the two sides and included angle

of the second triangle.

(b) SSS Test: Three sides of one are respectively equal to the three sides of the other triangle.

(c) AAS Test: Two angles and one side of one triangle are respectively equal to the two angles and one side of the other

triangle.




(d) RHS Test: The hypotenuse and one side of a right-angled triangle are respectively equal to the hypotenuse and one side of

another right-angled triangle.

Similar Triangles

Two figures are said to be similar, if they have the same shape but not the same size. If two triangles are similar, the

corresponding angles are equal and the corresponding sides are proportional.

Test for similarity of triangles

(a) AAA Similarity Test: Three angles of one triangle are respectively equal to the three angles of the other triangle.

(b) SAS Similarity Test: Two sides of one are proportional to the two sides of the other and the included angles are equal.

Properties of similar triangles: If two triangles are similar, the following properties are true:

(a) The ratio of the medians is equal to the ratio of the corresponding sides.

(b) The ratio of the altitudes is equal to the ratio of the corresponding sides.

(c) The ratio of the circumradii is equal to the ratio of the corresponding sides.

(d) The ratio of in radii is equal to the ratio of the corresponding sides.

(e) The ratio of the internal bisectors is equal to the ratio of the corresponding sides.

(f) Ratio of areas is equal to the ratio of squares of corresponding sides.

Pythagoras Theorem

The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides i.e. in a right

angled triangle ABC, right angled at B, AC2 = AB2 + BC2

Pythagorean triplets are sets of 3 integers which can be three sides of a right-angled triangle.

Examples of Pythagorean triplets are (3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41) etc.

Quadrilaterals

A polygon with 4 sides, is a quadrilateral

1. In a quadrilateral, sum of the four angles is equal to 360°.

2. The area of the quadrilateral = ½ × one diagonal x sum of the perpendicular to it from vertices.

Cyclic Quadrilateral:

If a quadrilateral is inscribed in a circle, it is said to be cyclic quadrilateral.

1. In a cyclic quadrilateral, opposite angles are supplementary.

2. In a cyclic quadrilateral, if any one side is extended, the exterior angle so formed is equal to the interior opposite angle.

Circles

If O is a fixed point in a given plane, the set of points in the plane which are at equal distances from O is a circle.

Properties of the circle Back to Table of Contents



1. Angles inscribed in the same arc of a circle are equal.

2. If two chords of a circle are equal, their corresponding arcs have equal measure.

3. A diameter perpendicular to a chord bisects the chord.

4. Equal chords of a circle are equidistant from the centre.

5. When two circles touch, their centres and their point of contact are collinear.

6. If the two circles touch externally, the distance between their centres is equal to sum of their radii.

7. If the two circles touch internally, the distance between the centres is equal to difference of their radii.

8. Angle at the centre made by an arc is equal to twice the angle made by the arc at any point on the remaining part of the

circumference.

9. The angle inscribed in a semicircle is 90o.

10. Angles in the alternate segments are equal.

11. If two chords AB and CD intersect externally at P, then PA × PB = PC × PD

12. If PAB is a secant and PT is a tangent, then PT2 = PA × PB

13. If chords AB and CD intersect internally, then PA × PB = PC × PD

14. The length of the direct common tangent (PQ)

CO-ORDINATE GEOMETRY

Rectangular axes (general)

(i) Distance formula: If A (x1, y1) and B (x2, y2) be two points, then in particular, of a point

P(x, y) form O(0, 0) is

(ii) Section formula: The point which divides the join of two distinct points A (x1, y1) and B (x2, y2) in the ratio m1: m2 Internally,

has the co-ordinates m1 ≠ 0, m2 ≠ 0, m1 + m2 ≠ 0 and externally, is

In particular, the mid-point of the segment joining A (x1 y1) and B (x2, y2) has the coordinates

Straight Line

Slope of a line

Slope of a non-vertical line L is the tangent of the angle θ, which either of half ray of

the line L makes with the positive direction of x-axis. In particular,

(a) Slope of a line parallel of x-axis is zero.

(b) Slope of a line parallel to y-axis is not defined.

(c) Slope of a line equally inclined to both the axis is − 1 or 1.

(d) Slope of a line making equal intercepts on the axis is − 1 or 1. Back to Table of Contents



(e) Slope of the line through the points A (x1, y1) and B (x2, y2) is

(f) Slope of the line ax + by + c = 0, b ≠ 0, is – a / b.

(g) Slopes of two parallel (non-vertical) lines are equal then m1 = m2

(h) If m1 and m2 be the slopes of two perpendicular lines (which are oblique), then m1m2 = – 1.

Equation of a Line

An equation of the form ax + by + c = 0 is called the general equation of a straight line, where x and y are variable and a, b, c

are constants.

Equation of a line parallel to X axis or Y - axis

(i) Equation of any line parallel to x-axis is y = b, b being the directed distance of the line from the x-axis. In particular equation

of x-axis is y = 0

(ii) Equation of any line parallel to y-axis is x = a, a being the directed distance of the line from the y-axis. In particular equation

of y-axis is x = 0.

One point form

Equation of a line (non-vertical) through the point (x1, y1) and having slope m is y – y1 = m (x – x1).

Two- points form

Equation of a line (non-vertical) through the points (x1, y1) and (x2, y2) is

Slope-intercept form

Equation of a line (non-vertical) with slope m and cutting off an intercept c from the y-axis is y = m x + c.

Intercept form

Equation of a line (non-vertical) with slope m and cutting off intercepts a and b from the x-axis and y-axis respectively is

Counting Techniques and Probability


1. On rolling six dice, it is found that

Three of dice show the same number

Only one die shows 6

Not more than three dice show 4 or more.




I.What is the maximum possible total numbers on the faces, if the three dice having same number show 2?

A) 14 B) 22 C) 11 D) 9

II. What is the maximum possible total numbers on the faces, if four of the dice show less than 4?

A) 28 B) 31 C) 17 D) 22

III. What is the maximum possible total numbers on the faces, if three dice are faulty and have only 5 on all faces?

A) 30 B) 25 C) 34 D) Not possible

IV. If only one die shows one, what is the maximum number of dice with number greater than 4?

A) 3 B) 1 C) 2 D) cannot be determined

V. What is the maximum number that can be on the face of the three dice which shows the same number?

A) 2 B) 4 C) 3 D) 5

2. In a party, there are certain number of men and women. Every man dances with three women and every woman dances

with three men. In addition, two men have two women in common. Find the number people attended the party. Ans: 8

3. In a Knockout tournament, 51 teams participated. Moreover, every team will be thrown out of the tournament, when they

lose twice. Find the number of matches to be held to decide the winner. Ans: 101 matches

4. Two people are playing with a pair of dice. Instead of numbers, the dice have different colors on their sides. The game has

following condition to select the winner.

First person wins the game, if the same color appears on both the dice; else, the second one will win the game.

The odds of their winning are equal.

If the first die has 5 red sides and 1 blue side, then the second die should have___________ Ans: 3 Red, 3 Blue

5. A cardboard of 34 * 14 cm has to be attached to a wooden box and 35 pins are to be used on the each side of the

cardboard. Find the total number of pins used. Ans: 210

6. Last year my cousin came to my house during the summer vacation and we played a game. The game has the condition that

the looser has to give one chocolate to the person who wins the game. At the end of the vacation (i.e. the day my cousin was

leaving, he counted number of games each). Finally, he gave me a total of 8 chocolates. Moreover, he knew the he won 12

games in all. Find the number of games that we played. Ans: 20

7. In a World football competition, there are 100 nations competing for a world-cup. The board decided to make it as a

knockout series. Find the number of matches to be played by each team before deciding the world champion. Ans: 99.

8. Find the number of four digit number formed by using the digits 1, 2, 3 and 4, which are divisible by 4. (Repetitions allowed)


Counting Techniques and Probability Back to Table of Contents



Fundamental Principle of Counting

Sum Rule

Suppose a work A can occur in m ways and B can occur in n ways and both cannot occur simultaneously. Then A or B (at least

one of term) can occur in (m + n) ways. This rule is also applicable for two or more exclusive events.

Product Rule

Suppose there are two works A and B Let A can occur in m ways and B in n ways. Suppose that the ways for A and B are not

related in the sense that B can occur in n ways regardless the outcome of A. Then both A and B occur in m ×n ways.

Event A (can occur in) m ways

Event B (can occur in) n ways

The at least one (A or B) m + n ways (Sum Rule)

A and B (both simultaneously) m n ways (Product Rule)

Ex. Consider a room with 2 entry doors (E1, E2) & three exit doors (X1, X2,X3).

E1 X1

X2

E2 X3

Number of ways in which a person can enter and exit the room

If he enters through E1, then he can exit through X1, X2, X3, so there are three ways. Similarly for E2. So totally there

are six ways.

Event A (entering the room) → 2 ways (E1, E2)

Event B (exit) → 3 ways (X1, X2, X3)

According to the product rule:

Enter and exit (Event A and B) → 2 × 3 = 6 ways.

Number of ways a person can enter or exit.

In case of ‘OR’, add the no. of ways of two events.

Enter or Exit → 2 + 3 = 5 ways.




Another Example

Let there be two questions A and B which can be solved by two methods and 3 methods respectively, Then A or B can be

solved in 2 + 3 = 5 ways and both A and B can be solved in 2 x 3 = 6 ways.

Problem based on Sum/Product

These two rules are bases of Permutation & Combination and can be directly used for following of questions:

Ex. How many numbers are between 100 and 1000 such that

(i) All the digits are distinct

A number between 100 and 1000 has three digits. So we have to form all possible 3 digit numbers with distinct digits.

We cannot have 0 at the hundred’s place.

So, the hundred’s place can be filled with any of the 9 digits 1, 2, 3….., 9. So, there are 9 ways of filling the hundred’s

place. Now, 9 digits are left including 0. So, ten’s place can be filled with any of the remaining 9 digit in 9 ways. Now,

the unit’s place can be filled by any one of the remaining 8 digits. So, there are 8 ways of filling the unit’s place. Total

numbers are equal to

(ii) If repetition is allowed (We can use same digit again),

Then, the hundred’s place can be filled with any of the 9 digits 1, 2, 3….., 9. So, there are 9 ways of filling the

hundred’s place. Ten’s place can be filled with any of the 10 digits in 10 ways. Now, the unit’s place can be filled in

any of the 10 digits in 10 ways. Total numbers is equal to

(iii) Even numbers with distinct digits

What most of student will do?

• Unit’s place can be filled in 5 ways. (0, 2, 4, 6, 8)

• Hundred digit can be filled with 8 digits (excluding 0 & the number that is already taken at unit place). Now we are left with

8 digits, so there are 8 ways to fill the ten’s place. Total number = 8 × 8 × 5 = 320. Back to Table of Contents



But this is incorrect.

Unit’s place can be filled in 5 ways (0, 2, 4, 6, 8). This is fine.

• Hundred’s place can be filled in 8 ways. This is wrong

If unit’s digit is 0, then hundred’s place can be filled in 9 ways. If unit’s digit is a number other than 0 then hundred’s place can

be filled in 8 ways. So you have to divide into two cases.

CASE I: If unit’s digit is 0.then

CASE II: Unit digit is digit other than 0.

Total numbers = 72 + 256 = 328.

(iv) All the even numbers.

Then

(v) Numbers are divisible by 5 if all the digits are distinct

Again two cases: -

Case I: If unit digit is 0.

Case II: If unit digit is 5.




Total numbers = 72 + 64 = 136.

All these questions can be done with permutation formula.

Permutation and Combination

Most important here is how to distinguish between permutation & combination. (i.e. where to apply permutation / where

combination)

Ex. Suppose, there are three quantities A, B, C. The different orders of arrangements of these three quantities by

taking two at a time are

AB, AC, BC

BA, CA, CB

So there are 6 different arrangements.

But In combination AB & BA means same thing. Similarly AC & CA and BC & CB, therefore there are 3 groups only AB,

AC, BC

Permutation and Combination Formulae

Permutation of n things taking r at a time =

In above example n = 3, r = 2,

Combination of n things taking r at a time =

In above example, number of groups =

Problem based on Permutation

Ex. Consider the word: O R D I N A T E

Find the number of different words that can be formed with the letters of above word

(i) Taking all at a time

Sol. 8 objects taking all at a time can be arranged in 8P8 i.e. 8! ways. OR By fundamental principle of counting

(ii) Taking all at a time starting with ‘O’ Back to Table of Contents



First place must be been filled with ‘O’. We are left with 7 objects that to be arranged in 7 places. This can be done in

7P7 or 7! ways.

(iii) Starting with ‘O’ & ending with E.

2 objects fixed and other six can be arranged in 6P6 i.e. 6! ways.

(iv) If O and E occupy extreme positions

Two cases possible: - O at first & E at the last or E at first & O at the last .So there are two ways to fill extreme places.

Middle places can be filled in |6 ways. Total ways = 2 × 6! ways

(v) A & D are always together.

Considering A & D as one object, now we have AD as one object & six other (O, R, I, N, T, E) i.e. total of seven objects

to arrange. This can be done in 7P7 ways.

A and D themselves can be arranged in 2! (two) ways i.e. AD or DA.

Total words in which A & D are always together = 2! × 7P7 = 2 × 7!

(vi) A & D are never together.

Use negation approach i.e. Total no of ways minus no. of ways when A and D are always together.

Total no. of words = |8

No. of words in which A & D are together = 2! × 7! = 2 × 7!

Required ways = 8! – 2 7!

(vii) A, D and T are always together

Considering A, D & T as one object, now we have (ADT) as one object & five others (O, R, I, N, T, E) i.e. total six objects

to arrange. This can be done in 6P6 ways.

A, D and T themselves can be arranged in |3 ways i.e. AD or DA.

Total words in which A, D and T are always together = 3! × 6P6

(viii) Vowels occupy odd places ORDINATE contains 8 letters: 1, 2, 3, 4, 5, 6, 7, 8 It has 4 odd places, 4 vowels.

⇒ Number of arrangements of the vowels 4 ! Also number of arranging consonants is 4!

⇒ Number of words = 4! x 4! = (4 x 3 x 2 x 1)2 = 576.

Ex. Arrangement of 4 men & 5 women in a line

(i) All women are always together.

Consider 5 women as one object.




Number of objects: 5

Number of ways in which 5 objects can be arrangement = 5!

Women themselves can be arrangement is 5! ways.

Required arrangements = 5! × 5!

(ii) All men are always together

Consider all men as one object.

Total objects – 6 can be arranged in |6 ways & 4 men can be arranged in 4! ways. Total no. of arrangements when all

men are together is 6! × 4!

(iii) All men/women are never together.

Use negation approach i.e. (Total ways – when they are together)

(iv) No two women are together.

In this case arrange the men first, which can be done in 4! ways.

X M1 X M2 X M3 X M4 X

In between men there are 5 places (Mark as ‘X’) & there are 5women.5 Women can be arranged at 5 places in |5

ways. Total number of ways = 4! × 5!

(v) No two men are together

Then arrange the women first which can be done in 5! ways.

X W1 X W2 X W3 X W4 X W5 X

In between women there are 6 places & 4 men.

Men can be arranged in 6P4.

Total number of ways = 5! × 6P4.

Permutation with repetitions

The number of permutations of n different things taking r at a time when each thing may be repeated any number of times in

any permutation is given by (n x n x n x n x n……………..r times) i.e. nr ways. Back to Table of Contents



PROBABILITY

“In a world as crazy as this one, it ought to be easy to find

something that happens solely by chance. It isn't”

Ex. A spinner has 4 equal sectors colored yellow, blue, green, and red. What are the chances of landing on blue after

spinning the spinner?

Sol. The chances of landing on blue are 1 in 4, or one fourth. This problem asked us to find the probability that the spinner

will land on blue. Let's look at some definitions and examples relating to the problem above.

Definition

Example

An experiment is a situation involving chance or

probability that leads to results called outcomes.

The experiment is spinning the spinner.

An outcome is the result of a single trial of an

experiment.

The possible outcomes are landing on

yellow, blue, green, or red.

An event is one or more outcomes of an

experiment.

The event being measured is landing on

blue.

Probability is the measure of how likely an event

is.

The probability of landing on blue is one

fourth.

In order to measure probabilities, mathematicians have devised the following formula:

Probability of an event occurring =




Ex. A spinner has 4 equal sectors colored yellow, blue, green, and red. After spinning the spinner, what is the

probability of landing on each color?

Outcomes:

The possible outcomes of this experiment are yellow, blue, green, and red.

Probabilities:

P (yellow) =

P (blue) =

Similarly P (green) = P (red) =

Ex. A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even

number (2, 4 or 6)? an odd number (1, 3 or 5)?

Sol.

Outcomes:

The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:

P (1) =

P (2) =

Similarly P (3) = P (4) = P (5) = P (6) =

This experiment illustrates the difference between an outcome and an event. A single outcome of this experiment is

rolling a 1, or a 2, or a 3, or a 4, or a 5, or a 6. Rolling an even number (2, 4 or 6) is an event, and rolling an odd

number (1, 3 or 5) is also an event.

In experiment 1 the probability of each outcome is always the same. The probability of landing on each color of the spinner is

always one fourth. In Experiment 2, the probability of rolling each number on the die is always one sixth. In each of these

experiments, the outcomes are equally likely to occur. Let's look at an experiment in which the outcomes are not equally

likely.

Ex. A glass jar contains 6 red, 5 green, 8 blue, and 3 yellow marbles. If a single marble is chosen at random from the jar,

what is the probability that it is red? green? blue? yellow? Back to Table of Contents



Sol.

Outcomes:

The possible outcomes of this experiment are red, green, blue and yellow.

Probabilities:

P (red) =

P (green) =

Similarly P (blue) = and P (yellow) =

The outcomes in this experiment are not equally likely to occur. You are much more likely to choose a blue marble

than any other color marble. You are least likely to choose a yellow marble.

Complement of an event A

The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A.

The complement of event A is represented by (read as A bar ( ))

Ex. A spinner has 4 equal sectors colored yellow, blue, green, and red.What is the probability of landing on a sector

that is not red after spinning this spinner?

Sample Space:

{yellow, blue, green, red}

Probability:

P (not red) = 1 –P (red) = 1 -

Ex. A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a

card that is not a king?

Probability:

P (not king) = 1 – P (king) =

Impossible event and certain event

An impossible event has no chance of occurring. If event A is impossible, then P (A) = 0.

A certain event is certain to occur. If event A is certain, then P (A) = 1




Ex. A single card is chosen at random from a standard deck of 52 playing cards. What is the probability that the card

chosen is a joker card?

Probability:

It is impossible to choose a joker card since a standard deck of cards does not have any jokers. This is an impossible

event.

P (joker) =

Mutually Exclusive Events

Two events are mutually exclusive if one happens, the other cannot happen and vice versa. In other words, the events have no

simultaneous occurrence. For example,

1. In rolling a die

A: The event that the number is odd

B: The event that the number is even

C: The event that the number is multiple of 2.

In the above case events A and B are mutually exclusive but the events B and C are not mutually exclusive or disjoint since

they may have common outcomes.

With the help of Venn diagram:

Sum Rule

If E and F two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given

by: P (E or F) or P (E U B) = P (E) + P (F)

If the event are not mutually exclusive, then P (E U F) = P (E) + P (F) – P (E and F together)

Note: Compare this with n (A ∪ B) = n (A) + n (B) – n (A ∩ B) of set theory

Similarly P (neither E nor F) = 1 – P (E or F).

Independent Events & Dependent Events

(A) Two events are independent if the occurrence of one has no effect on the occurrence of the other.

For example,

1. On rolling a die and tossing a coin together Back to Table of Contents



E: - The event that number 6 turn up.

F: - The event that head turn up.

Ex. A dresser drawer contains one pair of socks of each of the following colors: blue, brown, red, white and black. Each

pair is folded together in matching sets. You reach into the sock drawer and choose a pair of socks without looking.

The first pair you pull out is red -the wrong color. You replace this pair and choose another pair. What is the

probability that you will get the red pair of socks twice?

There are a couple of things to note about this problem. Choosing two pairs of socks is a compound event. Since the

first pair was replaced, choosing a red pair on the first try has no effect on the probability of choosing a red pair on

the second try. Therefore, these events are independent.

Independent Event and Dependent Event

Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.

Two events, A and B, are dependent if the fact that A occurs affect the probability of B occurring.

Some other examples of independent events are:

Tossing a coin and landing on heads, and rolling a die and getting a 5.

Choosing a card from a deck of cards and getting a three, replacing it, and choosing a second card and getting an ace.

Rolling a die and getting a 4, and then rolling a second die and getting a 1.

To find the probability of two independent events that occur in sequence, find the probability of each event occurring

separately and then multiply the answers. This multiplication rule is defined symbolically below.

When two events, A and B, are independent, the probability of both occurring is: P (A and B) = P (A) · P (B).

Logical Reasoning


Odd man out (with figures)

Coding & Decoding

Directions

1. Four prisoners Mr. East, Mr. West, Mr. South, Mr. North escape from a prison, head towards different directions.

The following information of their escape was supplied:

The escape routes were North Road, South Road, East Road and West Road. Back to Table of Contents



None of the prisoners took the road, which was in their name.

Mr. East did not take the South Road

Mr. West did not the South Road.

Mr. East did not take the West Road

What road did each of the prisoners take to make their escape?

Ans:

Mr. East - North Road

Mr. West - East Road

Mr. North - South Road

Mr. South - West Road.

Logical Deduction

Picture Series

Ranking

1. If Ram's rank from the beginning is 10th in a classroom of students and ranks 12th from the last. How many students are

there in the classroom? Ans: 21

2. If Mr. Rajesh stands 13th from beginning in the class of 31 students. What is his rank from the end?

3. In a class rank list, Ram ranks 12th from the beginning and Shyam ranks 11th from the end. If Ram and Shyam exchange their

places, then Shyam's rank from end is 15. What is the rank of Ram from the beginning now?

Logical Puzzles/Logical Games

1. Four persons A, B, C and D are playing cards. Each person has one card, laid down on the table, which has two different

colors on either side. The colors visible on the table are Red, Green, Red and Blue. The four persons made the following

comments about color on the reverse side:

A: Yellow or Green

B: Neither Blue nor Green

C: Blue or Yellow

D: Blue or Yellow

Out of the four persons two always lie. Find out the colors on each card.

2. Four Friends A, B, C and D are living in a hostel. One day A’s uncle came and gave him a box of Chocolates. They decided to

divide equally among them. A divided the Chocolates into four equal parts and he ate his part. Then B came to room, divided

the Chocolates into four equal parts and he ate his part. Followed that C came to room, divided the Chocolates into four equal

parts and he ate his part. Finally, D came to room, divided the Chocolates into four equal parts and he ate 3 chocolates as his

share. Find the number of Chocolates did each one eat Back to Table of Contents



3. Amit, Avinash, Govind, Pramod and Ashok are five students of Delhi University. They belong to five different states, namely

UP, Tamil Nadu, Maharashtra, Haryana and Gujarat not in the same order. They all cleared I P S exam and posted in different

states. However, none of them is posted in their home state. Each person likes a different folk dance, i.e., Tippani, Gagor,

Kavadi, Dahikala and Jhora.

One who lives in Maharashtra does not like Gagor and one who is posted in UP likes neither Gagor nor Dahikala.

One who is posted in Tamil Nadu likes Jhora, while Govind likes Kavadi.

Amit’s home state is Haryana, but he is posted in Gujarat.

Avinash and Govind are not associated with UP.

Ashok, Govind and Pramod are posted in home states of Pramod, Avinash and Govind respectively.

(i) One who likes Kavadi is posted in

a) Haryana b) Maharashtra c) Tamil Nadu d) None of these

(ii) One who is posted in UP, likes

a) Gagor b) Dahikala c) Tippani d) Kavadi

(iii) Avinash is posted in the home state of

a) Amit b) Govind c) Pramod d) None of these

(iv) Which of the following pairs is not associated with 4 states?

a) Ashok—Avinash b) Pramod---Avinash c) Govind---Amit d) Pramod—Govind

(v) Which of the following statement is correct?

a) Amit’s home state is Haryana and he is posted in Tamil Nadu

b) Govind likes Dahikala, a folk dance of Maharashtra.

c) Pramod’s home state is UP; he likes either Jhora or Tippani; and he is posted in Gujarat

d) None of these

4. Answer the question based on the following statements made by five friends.

L says, “All of my other 4 friends have money”

M says, “P said exactly one has money”

N says, “L said precisely two have money”

O says, “M said that 3 of others have money”

P says, “Land N said they have money”

In fact all are lying. Find who has money and does not have.




5. Three thieves A, B and C went for a robbery in different directions and they theft one Horse, one Mule and one Camel. The

police caught them. When they were interrogated, made the following statements:

A: B has stolen the Horse

B: I did not rob anything.

C: Both A and B are false and B has stolen the Mule.

The person who has stolen the Horse always tells the truth and the person who has stolen the Camel always tells the lie. Find

who has stolen which animal.

Ans: A- Camel B- Mule C- Horse

6. After World War II, there countries A, B and C, exchanged their weapons.

First A gave some tanks to B and C equal to the number they are having. Then B gave some tanks to A and C equal to the

number they are having. Finally, C gave some tanks to A and B equal to the number they are having. At the end each country

had 24 tanks. Find the initial number of tanks that each country had.

Ans: A-39 B-21 C-12

7. A, B, C, D and E are having their birthdays on consecutive days of the week not necessarily in the same order. A’s birthday

comes before G's as many days as B's birthday comes after E's. D is older than E by 2 days. G's birthday is on Wednesday.

Then find the day of each of their birthdays?

Ans: D – SUNDAY B – MONDAY E – TUESDAY G – WEDNESDAY A – THURSDAY

8. There are three friends A B C. Either A or B is oldest. Either C is oldest or A is youngest. Who is Youngest?

Ans: A - youngest B - oldest

9. There are six people related to a murder case Fillip, Walt, Mark, Joseph, Peter, and George. Among them, there is a police,

Murderer, Judge, Witness, Victim, and Hangman not in the same order. The murderer is hanged up for his crime.

Joseph knows both witness and murderer

Walt is the last one who saw Fillip alive.

The Policeman made sure that he arrested George at the murder site.

Peter did not meet Walt.

Find who played following role in the above drama

1. Murderer

2. Judge

3. Witness

4. Hangman

Ans:

1. Murderer – Walt. 2. Judge – Joseph. 3. Witness – George. 4. Hangman – Peter.




10. There are three friends; Fodder, Pepsi and Cereal often eat dinner out.

Each orders either coffee or tea after dinner.

If Fodder orders coffee, then Pepsi orders the drink that cereal orders

If Pepsi orders coffee, then Fodder orders the drink that Cereal does not order

If Cereal orders tea, then Fodder orders the drink that Pepsi orders

Which person/persons always order the same drink after dinner? Ans: Fodder

11. In a birthday party, there were four mothers and their children aged 1, 2, 3 and 4. From the clues below, find out whose

child is whose and their age.

It was Jane’s child’s birthday party.

Brian is not the oldest child.

Sarah had Anne just over a year ago.

Laura’s Child will have next birthday.

Daniel is older than Charlie.

Teresa’s child is the oldest.

Charlie is older than Laura’s child.

Jane – Charlie – 3

Laura – Brian – 2

Teresa – Daniel – 4

Sarah – Anne – 1

12. There are three guesses on the color of a Horse

1st says: it is not black

2nd says: it is brown or grey

3rd says: it is brown

At least one of them is wrong and one of them is true. What is the color of the Horse? Ans: Grey

13. An artist has exactly seven paintings: T, U, V, W, X, Y and Z, from which he must choose exactly five to show in an

exhibition. Any combination is acceptable provided it meets the following conditions:

If T is chosen, X cannot be chosen

If U is chosen, Y must also be chosen

If V is chosen, X must also be chosen

(i) Which one of the following is an acceptable combination of paintings?

A. T, U, V, X, Y B. T, U, V, Y, Z C. T, W, X, Y, Z D. U, V, W, Y, Z E. U, V, W, Z, Y

(ii) If the painting T is chosen to be among the paintings included in the exhibition which one of the following cannot be

chosen to be among the paintings included in the exhibition.




A. U B. V C. W D. Y E. Z

(iii) Which one of the following substitutions can the artist always make without violating restrictions affecting the

combination of paintings given that the painting mentioned first was not, and the painting mentioned second was, originally

going to be chosen?

A. T replaces V B. U replaces Y C. V replaces X D. W replaces Y E. Z replaces W

(iv) If the artist chooses painting V to be included among the paintings in the exhibition, which one of the following must be

true of that combination of paintings?

A. T is not chosen B. Y is not chosen C. U is chosen D. W is chosen E. Z is chosen

14. Some five friends, they made the following statements

A said the day before yesterday is either Friday\Saturday

B said no u r wrong, today is Saturday

C said no today is not Saturday nor Sunday or Monday

D said the tomorrow is Tuesday

E said the day after tomorrow is either Saturday\Sunday

All are liars except one.

Find out what is the day today.

15. Four persons A, B, C, D were there. All were of different weights. All four gave a statement. Among the four statements

only the person who is lightest in weight gave a true statement.

A Says: B is heavier than D.

B Says: A is heavier than C.

C Says: I am heavier than D.

D Says: C is heavier than B.

Find the order their weights.

Ans: A, C, D, B

16. Andy, Brian, Cedric, Dave are Architects, Barber, Caseworker and Dentist but not in the order.

Architect will have the letter ‘r’ in his name.

At least one of the persons should have coincidence in the first letter of their name and their occupation but not all

Barber and Dentist share their name by only one letter.

Find the occupation of each person.

Ans:

Andy – Caseworker

Brain – Barber

Cedric – Architect

Dave – Dentist Back to Table of Contents



Cubes

1. There is a cube (we cut into 27 pieces with six cuts) with all faces painted red.

Find the number of cubes with three sides painted red. Ans: 8

Find the number of cubes with two sides painted red. Ans: 12

Find the number of cubes with one side painted red. Ans: 6

Find the number of cubes with no side painted red. Ans: 1

2. A cube of 10 cm side was painted with a black border of width 1 cm on all its sides. The rest of the part on the surface of the

cube was painted blue. Then the cube was cut into 1000 parts.

Find the number of cubes were uncolored.

Arrangements (Linear/Circular)

1. In a 10-storey building, there is floor, which is above the floor below the floor, below the floor above the floor, below the

floor above the fifth. Ans: 6th floor.

2. Seven members sat around a table for a 3 - day conference. The member's names were Abhishek, Amol, Ankur, Anurag,

Bhuwan, Vasu and Vikram.

Vikram chaired all the meetings.

On the first evening members sat around the table alphabetically.

On the following two nights, Vikram arranged the seating so that he could have Abhishek as near to him as possible and

absent minded Vasu as far away as he could.

On no evening did any person sit next to a person, who had previously been his neighbor.

How did Vikram manage to seat everybody to the best advantage on the second and third evenings?

Ans:

Second evening: Vikram, Ankur, Abhishek, Amol, Vasu, Anurag and Bhuwan

Third evening: Vikram, Anurag, Abhishek, Vasu, Bhuwan, Ankur, Amol

3. Ram, Shyam and Guna are friends. Ram is a widower and lives alone and his sister takes care of him. Shyam is a bachelor

and his niece cooks his food and looks after his house. Guna is married to Gita and lives in large house in the same town. Gita

gives the idea that all of them could stay together in the house and share monthly expenses equally.

First month, each person contributed Rs.25.

At the end of the month, it was found that Rs.92 was the expense so that the remaining amount was distributed equally

among everyone.

The distribution was such that everyone received a whole number of Rupees.

How much did each person receive?

Ans. Rs.2 (Hint: Ram's sister, Shyam's niece and Guna's wife are the same person)




4. I was sitting in a merry-go-round. of the people in front of me, added to of those behind me, gave the total number

of people in the merry-go-round. How many people were there?

5. There are five men; Anil, Dick, Jack, Tom are fishing. They all had some number of fishes.

Tom had more than the Dick

Anil and Dick had same as Tom and Jack

Anil and Jack had less than Tom and Dick

Write their names in the descending order (based on numbers fishes that they had).

6. There were four boys playing a game in which the person who is the strongest would win. The four boys were Tom, Hank,

Bill and Joe. Hank could pull Bill and Joe with some effort. Hank and Bill together could just hold Joe and Tom (i.e., neither

could pull each other). However, when Hank and Joe interchanged their places, Bill and Joe could be easily pulled. Who is the

strongest? Ans: T > H > B > J

7. There are four persons named Jack, Brown, Smith and White. They are sitting in a train in different compartments

numbered as 1,2,3,4. Out of these four, two are bank robbers, one is a black mailer, and one is a forger.

Jack is not a bank robber.

One of the bank robbers sitting in 4th compartment.

White is sitting in the 3rd compartment.

Jack is sitting in the 2nd compartment.

Forger sitting in 1st compartment.

Smith had worked with one of the bank robber not with the other.

Find the name of the forger.

8. A, B, C, D and E are sitting around a table on five chairs which are numbered as 1, 2, 3, 4, and 5.The highest aged person sits

on the highest numbered chair and the least aged person sits on the lowest numbered chair. Given that their ages are in AP.

Find the age of C.

D is the oldest with the age 52 and he is sitting on chair 5.

The Common difference in the AP is two and C is sitting on chair 3.

9. A, B, C, D, and E visit five different cities: P, Q, R, S, and T not necessarily in the same order.

A does not go to P; B does not visit Q; and so on.

A and C visit the cities Q and S respectively, while B does not visit City T

D will visit either R or T.

Find the city that D visits.

10. CAP, SIP, TAP, IPS, and ECS are the top five coaching institutes in India. Which one is the oldest?

CAP is older than SIP and TAP but not older than IPS.

ECS is older than SIP and TAP but not older than CAP. Back to Table of Contents



Number Series

Find the next number in the following series:

3, 6, 13, 26, 33, 66, _____

364, 361, 19, 16, 4, 1, ____

5, 20, 24, 6, 2, 8, ____ Ans: 12

Blood Relations

1. Given that A, B, C, D, E are five members of a family. Four of them make one true statement each as follows:

E is my mother in law

C is my son in law's brother

B is my father's brother

A is my brother's wife

Who made the statements?

Ans: E

|

A<-->B--C

|

D

2. In a family, there are several children. Each boy in this family has as many sisters as brothers but each girl has twice as many

brothers as sisters. How many brothers and sisters are there? Ans: Brothers-4 , Sisters-3

3. Five people A, B, C, D, and E are related to each other. Four of them make one true statement each as follows.

(i) B is my father's brother.

(ii) E is my mother-in-law.

(iii)C is my son-in-law's brother

(iv) A is my brother's wife.

Who made the statements?

Ans: (i) D (ii) B (iii) E (iv) C





Logical Reasoning

TYPES:

1. Prime number Series:

Ex.1 2, 3, 5, 7, 11, 13, …

(1) 15 (2) 17 (3) 18 (4) 19

Sol. The given series is prime number series. The next prime number is 17. Answer: (2)

Ex.2 2, 5, 11, 17, 23, …, 41.

(1) 29 (2) 31 (3) 37 (4) 39

Sol. The prime numbers are written alternately. Answer: (2)

2. Difference Series:

Ex.1 2, 5, 8, 11, 14, 17,…, 23.

(1) 19 (2) 21 (3) 20 (4) 18

Sol. The difference between the numbers is 3. (17 + 3 = 20). Answer: (3)

Ex.2 45, 38, 31, 24, 17,…, 3.

(1) 12 (2) 14 (3) 10 (4) 9

Sol. The difference between the numbers is 7. (17 – 7 = 10)

Answer: (3)

3. Multiplication Series:

Ex.1 2, 6, 18, 54, 162,…, 1458.

(1) 274 (2) 486 (3) 1236 (4) 1032

Sol. The numbers are multiplied by 3 to get next number. (162 × 3 = 486)

Answer: (2)

Ex.2 3, 12, 48, 192,…, 3072.

(1) 768 (2) 384 (3) 2376 (4) 1976

Sol. The numbers are multiplied by 4 to get the next number. (192 × 4 = 768)

Answer: (1)

4. Division Series: Back to Table of Contents



Ex.1 720, 120, 24,…, 2, 1

(1) 12 (2) 18 (3) 20 (4) 6

Sol. Answer: (4)

Ex.2 32, 48, 72, 108,…, 243.

(1) 130 (2) 162 (3) 192 (4) 201

Sol. Number × 3/2 = next number. 32 × =48, 48 = 72, 72 = 108, 108 = 162

Answer (2)

5. n2 Series:

Ex.1 1, 4, 9, 16, 25,…, 49

(1) 28 (2) 30 (3) 32 (4) 36

Sol. The series is 12, 22, 32, 42, 52…, The next number is 62 = 36.

Answer: (4)

Ex.2 0, 4, 16, 36, 64,…, 144.

(1) 100 (2) 84 (3) 96 (4) 120

Sol. The series is 02, 22, 42, 62 etc. The next number is 102 = 100. ]

Answer: (1)

6. n2 – 1 Series:

Ex.1 0, 3, 8, 15, 24, 35, 48,…

(1) 60 (2) 62 (3) 63 (4) 64

Sol. The series is 12 – 1, 22 – 1, 32 – 1 etc. The next number is 82 – 1 = 63.

Answer: (3)

Another Logic:

Difference between numbers is 3, 5, 7, 9, 11, 13 etc. The next number is (48 + 15 = 63).

7. n2 + 1 Series:

Ex.1 2, 5, 10, 17, 26, 37,…, 65.

(1) 50 (2) 48 (3) 49 (4) 51

Sol. The series is 12 + 1, 22 + 1, 32 + 1 etc. The next number is 72 + 1 = 50.

Answer: (1)

8. n2 + n Series (or) n2 – n Series:

Ex.1 2, 6, 12, 20,…, 42.

(1) 28 (2) 30 (3) 32 (4) 36 Back to Table of Contents



Sol. The series is 12 + 1, 22 + 2, 32 + 3, 42 + 4 etc. The next number = 52 + 5 = 30.

Answer: (2)

Another Logic:

The series is 1 × 2, 2 × 3, 3 × 4, 4 × 5. The next number is 5 × 6 = 30.

Another Logic:

The series is 22 – 2, 32 – 3, 42 – 4, 52 – 5. The next number is 62 – 6 = 30.

9. n3 Series:

Ex.1 1, 8, 27, 64, 125, 216,…

(1) 256 (2) 343 (3) 365 (4) 400

Sol. The series is 13, 23, 33 etc. The missing number is 73 = 343.

Answer: (2)

10. n3 + 1 Series:

Ex.1 2, 9, 28, 65, 126, 217, 344,…

(1) 513 (2) 500 (3) 428 (4) 600

Sol. The series is 13 + 1, 23 + 1, 33 + 1 etc. The missing number is 83 + 1 = 513.

Answer: (1)

LETTER SERIES

Introduction:

In these types of problems a series of the letters of alphabet will be given which follow a pattern or a sequence. The letter

series mainly consists of skipping of the letters.

To solve these types of problems, assign numbers 1 to 26 to the letters of the alphabet as shown below. In some cases it is

useful to assign the numbers in the reverse order.

1 2 3 4 5 6 7 8 9 10 11 12 13

A B C D E F G H I J K L M

Z Y X W V U T S R Q P O N

26 25 24 23 22 21 20 19 18 17 16 15 14

Here the table is showing both forward as well as reverse place value of any alphabet. A very important fact about the

position of any alphabet is that both the sum of forward position and reverse position for any alphabet is always constant and

equal to 27. Such as Sum of both positions of H is (8 + 19 = 27) or for W is (23 + 4 = 27).

We can also remember the relative positions of these alphabets by just remembering the word EJOTY.

Letters E J O T Y

Position 5th 10th 15th 20th 25th Back to Table of Contents



Just remember the word EJOTY and its values i.e. 5, 10, 15, 20, 25

e.g. If you are asked to complete the series F, K, P, U, __

Then from EJOTY, you know that values of F = 6, K = 11, P = 16, U = 21 i.e. difference is 5, so the answer should be 21 + 5 = 26

i.e. Z

Various types of letter series are given below.

TYPE – 1

One Letter Series:

Ex.1 A, C, E, G, I, …

(1) J (2) K (3) L (4) M

Sol. The series is (+ 2). i.e., A + 2 = C; C + 2 = E; E + 2 = G; G + 2 = I.

The missing letter is I + 2 = K.

Answer: (2)

Another Logic:

Skip one letter is I + 2 = K.

After I skip J to get K; the missing letter is K.

Note: "Skip" process saves time.

Ex.2 A, B, D, G, K, ...…

(1) P (2) N (3) O (4) L

Sol. The series is + 1, + 2, + 3 etc.

The missing letter is (K + 5) = P.

Answer: (1)

Skip Process:

First no letter is skipped, then 1, 2, 3 etc. letters are skipped to get next letter. Skip 4 letters after 'K' to get P.

Ex.3 B, E, H, K, N,…

(1) P (2) O (3) Q (4) R

Sol. The series is + 3. The missing letter is N + 3 = Q.

Answer: (3)

Another Logic:

Skip two letters to get the next letter. Skip Q, P after N to get Q. The missing letter is Q.

Ex.4 B, D, G, I, L, N,…

(1) N (2) O (3) P (4) Q

Sol. The series is alternately + 2 and + 3. The missing letter is N + 3 = Q. Back to Table of Contents



Answer: (4)

Another Logic:

Skip one and two letters alternately to get the next letter. Skip two letters O, P after N to get Q.

Ex.5 B, C, E, G, K,…

(1) M (2) N (3) O (4) P

Sol. If numbers are assigned, the series becomes prime number series.

The next prime number is 13 and the corresponding letter is M.

Answer: (1)

Ex.6 A, E, I, O,…

(1) Q (2) R (3) U (4) S

Sol. The series is a series of Vowels.

Answer: (3)

Ex.7 A, D, I, P,…

(1) U (2) V (3) X (4) Y

Sol. If numbers are assigned, the series becomes square series.

The next number is 52 = 25 and the corresponding letter is Y.

Answer: (4)

Ex.8 D, F, H, I, J, L,…

(1) M (2) N (3) O (4) P

Sol. If numbers are assigned, the series becomes composite number series.

The next composite number is 14 and the corresponding letter is N.

Answer: (2)

Ex.9 A, Z, B, Y, C, X, D,…

(1) U (2) V (3) W (4) X

Sol. The sequence consists of two series A, B, C, D etc., and Z, Y, X, W etc.

Answer: (3)

TYPE – 2

Two Letter Series:

The first letters of the series follow one logic and the second letters follow another logic. Also, the first two letters, the next

two letters and so on follow a logic.

Ex.1 AM, BN, CO, DP, EQ,…

(1) FG (2) FR (3) GR (4) ER Back to Table of Contents



Sol. The first letters are A, B, C, D, E, F and the second letters are M, N, O, P, Q and R.

Answer: (2)

Ex.2 AB, DE, GH, JK, MN,…

(1) OP (2) NO (3) PQ (4) RS

Sol. After every set of letters one letter is skipped. Skip O to get next two letters PQ.

Answer: (3)

Ex.3 AA, CE, EI, GO,…

(1) IU (2) IQ (3) IR (4) IT

Sol. The first letters follow a sequence of A, C, E, G, I. (+ 2 series) and the second letters are vowels.

Answer: (1)

TYPE – 3

Three Letter Series:

This sequence consists of 3 letters in each term. The first letters follow one logic, the second letters follow another logic and

the third letters follow some other logic, (or the same logic in all the three cases)

Ex.1 ABD, CDF, EFH, GHJ,…

(1) IJK (2) IJL (3) HIJ (4) HIK

Sol. The first letters follow a sequence of A, C, E, G, I etc.

The second letters follow a sequence of B, D, F, H, J etc.

And the third letters form a sequence of D, F, H, J, L etc.

Answer: (2)

Ex.2 CKZ, DLY, EMX, FNW,…

(1) GOV (2) GOU (3) GNU (4) GNV

Sol. The first letters form a series of C, D, E, F, G etc.

The second letters form a series of K, L, M, N, O etc, and the third letters form a series of Z, Y, X, W, V etc.

Answer: (1)

Ex.3 MAB, NEC, OIE, POG,…

(1) QPH (2) QUH (3) QUI (4) QUK

Sol. The first letters form a series of M, N, O, P, Q etc. The second letters form Vowels; the third letters form prime

number series (if numbers are assigned to letters).

Answer: (4)

Ex.4 ABC, CBA, DEF, FED, GHI, ..…

(1) JKL (2) IHG (3) DFE (4) IJK Back to Table of Contents



Sol. The second term is the reverse order of first term.

In addition to the above types a number of other types can also be identified.

Answer: (2)

CODING / DECODING

Introduction:

For conveying secret messages from one place to another, especially in Defence Services, coding is used. The codes

are based on various principles/patterns such that the message can be easily be deciphered at the other end. Now-a-days, in

certain competitive examinations, such questions are given to judge the candidates’ intelligence and mental ability. They are

required to encode and decode words and sentences after observing the pattern and principles involved. These questions can

be broadly classified into 5 main categories, as follows:

(i) Coding with Letters of Alphabets

(ii) Coding with Numerical Digits (Numbers)

(iii) Mixed Coding (Both Alphabetical and Numerical)

(iv) Coding with Arbitrary Signs / Symbols

(v) Miscellaneous Type

TYPE – 1

Coding with Letters of Alphabet:

In these questions, the letters of the alphabets are exclusively used. These letters do not stand for themselves but are allotted

some artificial values based on some logical patterns/analogies. By applying those principles or observing the pattern

involved, the candidates are required to decode a coded word or encode a word. These can be further classified into the

following categories :

Simple Analogical Letter Coding:

These are also called arbitrary codes. There are 2 definite principles/pattern involved. Codes are based on the analogy of one

example from which different codes are to be formed.

Ex.1 If NETWORK is coded as O P C T R S Q, how is CROPS written in that code; is written in actual code?

Sol. N = O C = T

E = P R = O

T = C then O = N

W = T P = E

O = R S = R

R = S

K = Q Back to Table of Contents



Hence CROPS can be coded as TONER.

Ex.2 The code ‘TABLESTESF’ stands for the word ‘BELONGINGS’ how will you code the following :

(1) LONG (2) ON (3) GIN (4) SONG (5) NO

(6) SING (7) SINGS (8) GONE (9) IS (10) GO

Sol. The coding is done as follows:

(1) BLES (2) LE (3) STE (4) FLES (5) EL

(6) FTES (7) FTESF (8) SLEA (9) TF (10) SL

Ex.3 If INLAND is coded as BSTRSI, make codes of the following letters.

(1) IN (2) LAND (3) INN (4) AND (5) AN (6) LAID

Sol. The coding is done as follows:

(1) BS (2) TRSI (3) BSS (4) RSI (5) RS (6) TRBI

Ex.4 If EWFGHONTISO stands for OBSERVATION, code the following letters.

(1) RATION (2) RATE (3) SEAT (4) NOT (5) NOTE (6) BEST

Sol. The coding is as follows:

(1) HNTISO (2) HNTG (3) FGNT (4) OST (5) OSTG (6) WGFT

Ex.5 If PROMISED is coded as RMNIOSTD, decode the following codes.

(1) RNST (2) MNIT (3) DOI (4) RMNST (5) SOM (6) INMT

Sol. The decoding is as follows:

(1) POSE (2) ROME (3) DIM (4) PROSE (5) SIR (6) MORE

Ex.6 Column A contains certain words numbered from (1) to (6). Column B goes with the codes for column A, but with

different order. You have to match the words of column A with their respective coded word in column B. The pattern

of coding used here is BLADES = CMBEFT.

Column (A) Column (B)

(1) BASE (1) CBE

(2) BALE (2) CBTF

(3) SALE (3) CFE

(4) SAD (4) CBMF

(5) BAD (5) TBE

(6) BED (6) TBMF

Sol. A (1) B (2), A (2) B (4), A (3) B (6), A (4) B (5), A (5) B (1), A (6) B (3).




Letter Coding on Specific Pattern:

In such questions, letters of alphabets are no doubt allotted artificial values but based on certain specific pattern/principles.

The candidates are required first to observe the specific pattern involved and then proceed with coding or decoding; as the

case may be.

Ex.1 If POSTED is coded as DETSOP, how will be word SPEED be coded?

Sol. A careful observation of the above example will reveal that letters of the first word have been reversed

Ex.2 If GREET is coded as FQDDS, decode the following codes:

(1) KDS (2) SNQD (3) CNBI (4) CDDO (5) ONS (6) ANRR

Sol. Here, each letter is allotted the value of its preceding letter in the sequence; the pattern of coding used here is B = A,

C = B. Based on this pattern, the answers to the above questions will be follows:

(1) LET (2) TORE (3) DOCK (4) DEEP (5) POT (6) BOSS

Ex.3 If A = E, how will you code the following words.

(1) BLACK (2) ACT (3) BAT (4) CADRE (5) LOOT (6) FOOL

Sol. (1) FPEGO (2) EGX (3) FEX (4) GEHVI (5) PSSX (6) JSSP

Ex.4 If “CAT” is coded as “DEBCUV”, how will you code “RACE”.

Sol. The pattern of coding is such that each letter has been allotted value of 2 letters following the sequence, i.e. A = BC, B

= CD, C = DE, etc.

Hence, the word RACE will be coded as “STBCDEFG”

Based on the above principle, try to code the following.

(1) FATHER (2) DATED (3) LATE (4) FAKES (5) MAIN (6) PLANE

Sol. (1) G H B C U V F G S T (2) E F B C U V F G E F (3) M N B C U V F G

(4) G H B C L M F G TU (5) N O B C J K O P (6) Q R M N B C O P F G

Ex.5 Column (A) contains coded words and column (B) contains equivalent decoded words given in a different serial order.

Match the words of the column (A) with column (B) and indicate the first and last letters of the coded word in column

A from the answer choices.

The pattern of coding is Q = P, S = R, U = T, etc.

Column (1) Column (B)

(1) USJN (1) WORK

(2) CPOF (2) SHORT

(3) MPPU (3) FEET

(4) GFFU (4) LOOT

(5) TIPSU (5) BONE Back to Table of Contents



(6) XPSL (6) TRIM

Sol. A (1) B (6), A (2) B (5), A (3) B (4), A (4) B (3), A (5) B (2), A (6) B (1).

Ex.6 If “EGHJKMKM” is the code for “FILL”, how will you code the following :

(1) QSDFRTSU (2) SUDFKMKM (3) EGDFDFKM

(4) CENPDFRT (5) KMNPRTSU (6) ACDFCERT

Sol. The pattern of coding is such that the sequence follows the letters in between, each pair of letters in the code.

Pattern is AC = B, BD = C, CE = D, etc.

(1) REST (2) TELL (3) FEEL (4) DOES (5) LOST (6) BEDS

TYPE – 2

Coding with Numerical Digits:

The pattern of coding with numerical digits is similar to that of coding with alphabets except the use of numerical

digits with the assignment of some artificial values. The values are allotted based on some specific pattern which has to be

discerned by the candidate in order to solve the problem in the quickest possible time.

If TRAIN is coded as 23456, how will you code TIN and RAIN?

The answer will be 256 for TIN and 3456 for RAIN. T = 2, R = 3, A = 4, I = 5, and N = 6. These values have been allotted

arbitrary; based on logical relationship; the candidates will be able to solve the problem.

Analogical Coding with Numerical Digits:

Analogical coding with numerical digits involves the method of coding where the letters of alphabets are allotted numerical

values and the pattern of coding is based on the analogy of the example given in the question. There are no set of principles or

patterns involved. Candidates are required to study the examples given before getting started with the exercise.

Ex.1 If SELDOM is coded as “1 2 4 3 6 5”, how will you code the following words?

(A) DOES (B) SOLE (C) LED (D) DOLE (E) LODE (F) ODE

Choices:

(A) (1) 3 6 2 1 (2) 6 2 3 1 (3) 1 6 3 2 (4) 6 2 1 3

(B) (1) 1 4 6 2 (2) 1 6 4 2 (3) 1 4 2 6 (4) 1 6 2 4

(C) (1) 4 3 2 (2) 3 2 4 (3) 4 2 3 (4) 4 2 6

(D) (1) 3 6 4 1 (2) 3 6 4 2 (3) 3 6 2 4 (4) 3 6 4 1

(E) (1) 4 6 2 3 (2) 4 6 3 2 (3) 6 3 2 4 (4) 4 3 6 2

(F) (1) 6 2 3 (2) 2 3 6 (3) 6 3 2 (4) 6 3 4

Sol. If SELDOM stand for code 124365 which means S = 1, E = 2, L = 4, D = 3, O = 6, and M = 5. Based on this analogy, the

correct answer will be,

(A) (1) (B) (2) (C) (3) (D) (2) (E) (2) (F) (3) Back to Table of Contents



Ex.2 If “1 3 4 8 2 6 7 5 9” is the code for “O B S E R V A N T” how will you code the following words?

(A) SERVANT (B) SOBER (C) BENT (D) OVATE (E) ORATE (F) NOTES

Choices:

(A) (1) 4 8 2 6 7 5 9 (2) 4 8 2 6 7 6 0 (3) 4 8 2 6 7 5 0 (4) 4 2 8 6 7 5 9

(B) (1) 4 1 3 8 2 (2) 4 1 3 8 1 (3) 4 3 1 8 2 (4) 4 1 3 2 8

(C) (1) 3 8 9 5 (2) 3 8 5 9 (3) 3 5 8 9 (4) 9 8 3 5

(D) (1) 1 7 6 9 8 (2) 1 7 6 8 9 (3) 1 6 7 9 8 (4) 9 8 7 6 1

(E) (1) 1 7 2 9 8 (2) 1 2 7 9 8 (3) 1 2 7 8 9 (4) 8 9 2 7 1

(F) (1) 9 1 5 8 4 (2) 5 1 9 8 4 (3) 5 9 1 8 4 (4) 8 4 9 1 5

Sol. In the example, you’ll observe O = 1, B = 3, S = 4, E = 8 R = 2, V = 6, A = 7, N = 5 and T = 9.

Based on this analogy the correct answers will be :

(A) (1) B) (1) (C) (2) (D) (3) (E) (2) (F) (2)

Ex.3 The code 6 7 4 5 3 2 7, stands for B E C A U S E. Decode the following codes

(1) 4 5 3 2 7 (2) 6 5 2 7 (3) 4 5 2 7 (4) 6 7 7 (5) 4 7 5 2 7 (6) 3 2 7

Sol. (1) CAUSE (2) BASE (3) CASE (4) BEE (5) CEASE (6) USE

Ex.4 If 4 0 6 5 4 2 5 7 is the code for S T A N D I N G, decode the following codes.

(1) 4 0 6 7 (2) 6 5 4 (3) 4 0 6 2 5 (4) 42 5 2 5 7 (5) 4 6 5 4 (6) 4 0 2 5 7

Sol. (1) STAG (2) AND (3) STAIN (4) DINING (5) SAND (6) STING

Coding with Specific Pattern:

This is the pattern of coding which exhibits the natural correlation of Arabic numbers with alphabetic letters. For instance,

alphabets A to Z are assigned the numeric codes from 1 to 26 where each letter gets the assignment in the pattern as follow A

= 1, B = 2, C = 3, etc.

The sequence is classified as follows :

Forward sequence (e.g. A = 1, B = 2, etc.)

Backward sequence (e.g. Z = 1, Y = 2, A = 26, etc.)

Random Sequence (e.g. A = 2, B = 3 or A = 4, B = 6, C = 8 or any other pattern following a particular sequence).

Forward Sequence:

Ex. If ‘PACE’ is code as 16-1-3-5, how will you code the following :

(1) ACTED (2) BAIL (3) RACE (4) FRAME (5) GLAD (6) GAIN

Sol. (1) 1-3-20-5-4 (2) 2-1-9-12 (3) 18-1-3-5 (4) 6-18-1-13-5 (5) 7-12-1-4

(6) 7-1-9-14

Backward Sequence: Back to Table of Contents



Ex. If GREAT is coded as 20-9-22-26-7, how will you code the following words :

(1) FATE (2) DATE (3) MATE (4) RATE (5) GATE

Sol. (1) 21-26-7-22 (2) 23-26-7-22 (3) 14-26-7-22 (4) 9-26-7-22 (5) 20-26-7-22

Random sequence:

The sequence will not follow a specific pattern of assignment as in other cases but will surely show a pattern at a

strict analysis. The pattern can be established by various ways but in every case a set principle/pattern is involved which has to

be discovered by careful examination of the example given in the question.

Ex.1 If FRANCE is coded 9-21-4-6-8, code the following words after discerning the principle/pattern involved in this

example :

(1) INDIA (2) CANADA (3) GERMANY (4) NEPAL (5) PERU (6) KENYA

Sol. The pattern of assignment is read as given in the following table.

A B C D E F G H I J K L M N O P

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Q R S T U V W X Y Z

20 21 22 23 24 25 26 1 2 3

Based on above pattern, the answers will be follows :

(1) 12-17-7-12-4 (2) 6-4-17-4-7-4 (3) 10-8-21-16-4-17-2 (4) 17-8-19-4-15

(5) 19-8-21-24 (6) 14-8-17-2-4

Ex.2 If BREAD is coded as “2-18-0-0-4”, how will you code the following?

(1) COOL (2) COME (3) BROOM (4) GROOM (5) SHEETAL (6) CREAM

A strict analysis of the question reads that the vowels ‘E’ & ‘A’ are assigned the code ‘0’. The rest of the letters

follow the regular sequence of numerical assignment, i.e. B = 2, C = 3, etc. Based on the above pattern, the

answers are follows: Remaining letters of alphabet will follow the same order, i.e. B = 2, C = 3, D = 4 etc.

Sol. (1) 3-0-0-12 (2) 3-0-13-0 (3) 2-18-0-0-13

(4) 7-18-0-0-13 (5) 19-8-0-0-20-0-12 (6) 3-18-0-0-13

Ex.3 If 6 – 12 – 1 – 19 – 8 = FLASH and 6 – 15 – 15 – 12 – 9 – 19 – 8 = FOOLISH, find the sum with all the letters put

together.

(A) LATE (B) MAKE (C) ICED (D) ACT (E) FACT (F) LAND

Choices:

(A) (1) 38 (2) 59 (3) 56 (4) 58

(B) (1) 32 (2) 30 (3) 34 (4) 36

(C) (1) 22 (2) 23 (3) 21 (4) 24 Back to Table of Contents



(D) (1) 23 (2) 22 (3) 20 (4) 24

(E) (1) 41 (2) 51 (3) 21 (4) 30

(F) (1) 42 (2) 29 (3) 31 (4) 30

Sol.

(A) LATE = 12 + 1 + 20 + 5 = 38, hence the answer is (1).

(B) MAKE = 13 + 1 + 11 + 5 = 30 hence the answer is (2).

(C) ICED = 9 + 3 + 5 + 4 = 21 hence the answer is (3).

(D) ACT = 1 + 3 + 20 = 24 hence the answer is (4).

(E) FACT = 6 + 1 + 3 + 20 = 30 hence the answer is (4).

(F) LAND = 12 + 1 + 14 + 4 = 31 hence the answer is (3).

TYPE – 3

Mixed Coding (Letters + Digits):

Mixed coding takes the pattern of coding with both the letters of alphabets and numerical assignment. The candidates are

required to study the analogy given in question.

Ex.1 If “A – 3 – T – 5 – D” stands for ACTED and “D1T5D” stands for “DATED”, how will you code the following

(1) FADED (2) LOCATE (3) BAILED (4) FAILED (5) PRESS

(6) DREAM

Ans. (1) F1D5D (2) L15C1T5 (3) B 1 I 12E4 (4) F1I12E4 (5) P 18E 19S

(6) D 18 E 1 M

Ex.2 Decode the following:

(1) R 9 L 5 D (2) A 3 I 4 (3) 5A20I14G (4) B 1 I 12 (5) K 9L 12

(6) B 1 I 12E4

Ans. (1) RILED (2) ACID (3) EATING (4) BAIL (5) KILL

(6) BAILED

Ex.3 Decode the following:

(1) F 1 D 5 D (2) A 9 D 9 N 7 (3) R 5 Q 21 I 18 E 4

(4) D 5 A 12 I 14 G (5) O 2 S 5 R 22 E (6) A 3 I 4

Ans. (1) F A D E D (2) A I D I N G (3) R E Q U I R E D

(4) D E A L I N G (5) O B S E R V E (6) A C I D

TYPE – 4

Miscellaneous Types

Decoding the Rule Applied: Back to Table of Contents



This part of coding test required a careful examination of rules followed to code a certain word. Only after the analysis of the

pattern applied in coding, you can decode them.

Example: Study the five different ways of coding marked (1), (2), (3), (4) & (5). A specific rule has been applied to codify each

of them. Can you find out the rule of coding applied in the question that follows;

(1) N C E F R A

(2) F A C R N E

F R A N C E (3) E C N A R F

(4) A C E F N R

(5) F E R C A N

WORD CODE

1. C A N A D A C N D A A A

2. K E N Y A K A E Y N

3. N A T I O N S S N O I T A N

4. V A N D A N A V N A A A D N

5. V A R D H M A N N A M H D R A V

6. V A R I O U S A I O R U S V

7. C A R E E R E E R C A R

8. P O P U L A T I O N P N O O P I U T L A

9. M E D I C I N E M D C N E I I E

10. A P T I T U D E A D E I P T T U

Answers:

1. (2) 2. (5) 3. (3) 4. (2) 5. (3) 6. (4) 7. (1) 8. (5) 9. (2) 10. (4)

Contrasting and Marking Comparisons:

A set of words are given in column I and codes have been formed in column II. Here in such questions some

alphabets/letter are underline in column I and the corresponding codes in column II has been jumbled up thus making the

question more difficult to correspond. To find the formula to decode these type of question some logical rule/principle is

found by comparing or making contracts in all the questions. An example has been given below:

Example

In the following question the capital letters in column I are codified in small letters in column II. The small letters are

not arranged in the same order on the capital letters. Study the column (I) and (II) together and determine the small letters for

the corresponding underlined capital letter in column (I).

Column (I) Column (II) Back to Table of Contents



1. D I G I T w b z b m

2. T I G E R m b z x k

3. F E V E R x k y o x

4. G I T A R m t z b k

5. L I V E R b e x o k

Keys: 1. w 2. m 3. y 4. z 5. e

Explanation

If we compare question (1) & (2) we find that there are 3 alphabets (T, I, G) common and there corresponding small

letters will be (m, z, b) though not in the same order. This leaves us with (D and R) with small alphabets (w and k). Therefore,

we have now, Either ‘w or k’ is D’s code

Now, if we taken (2) and (3), we find that ‘w’ is not present is column II of either (2) or (3) and D is not there in column

II of either (2) & (3) the or conclude that D = w and therefore R = k.

Now, carrying on with this finding, we see in question (3) and (5) there are two common elements in column I, V, E & R.

Since E comes twice in (3), therefore code for E = x which leads to V = 0 and F = y in question is (1), I comes twice, this leads to

I = b. So we are left with ‘T’ and ‘G’, which are either ‘z’ or ‘m’.

Now, we cannot conclude anything more from these clues, but can fit in above observation to see what relation

capital letters have with small letters.

Therefore, G = z and T = m

Mathematical/Algebraic Operations

The code is always the sum of letters with the assignment of numbers put in the regular order. The order reads either

in a forward sequence or a backward sequence. Consider the table given below.

Ex.1 If DOLLY is 68, then how much will be SEEMA?

(1) 65 (2) 86 (3) 43 (4) 33 Back to Table of Contents



Sol. The coding is the sum of forward sequence of alphabets

DOLLY 4 + 15 + 12 + 12 + 25 = 68

SEEMA 19 + 5 + 5 + 13 +1 = 43, hence the answer is (3).

Ex.2 If NEERAJ is 109, then how much will be SHEETAL?

(1) 119 (2) 98 (3) 125 (4) 100

Sol. The coding is the sum of backward sequence of alphabets:

NEERAJ 13 + 22 + 22 + 9 + 26 + 15 = 109

SHEETAL 8 + 19 + 22 + 22 + 7 + 26 + 15 = 119, hence the answer is (1).

BLOOD RELATIONS

Introduction:

In these tests the success of a candidate depends upon the knowledge of the blood relations, some of which are summarized

below to help solve these tests.

Mother's or father's son: Brother

Mother's or father's daughter: Sister

Mother's or father's brother: Uncle

Mother's or father's sister: Aunt

Mother's or father's father: Grandfather

Mother's or father's mother: Grandmother

Brother's son: Nephew

Brother's daughter: Niece

Uncle or aunt's son or daughter: Cousin

Sister's husband: Brother – in – law

Brother's wife: Sister in – law

Son's wife: Daughter – in – law

Daughter's husband: Son – in – law

Husband's or wife's sister: Sister – in – law

Husband's or wife's brother: Brother – in – law

EXAMPLE:

Directions for questions 1 – 3: Read the following information to answer the given questions.

(i) A, B, C, D, E and F are six family members.

(ii) There is one Doctor, one Lawyer, one Engineer, one Pilot, one Student and one Housewife.

(iii) There are two married couples in the family. Back to Table of Contents



(iv) F, who is a lawyer, is father of A.

(v) B is a pilot and is mother of C.

(vi) D is grandmother of C and is a Housewife

(vii) E is father of F and is a Doctor.

(viii) C is brother of A.

1. Which of the following statements is definitely true?

(1) C is the brother of the Student (2) F is the father of the Engineer

(3) A is the Engineer (4) E is the father of the Pilot

2. How many female members are there in the family?

(1) Two only (2) Three only (3) Three or Four (4) Two or Three

3. How is A related to D?

(1) Grandson (2) Granddaughter (3) Either granddaughter or grandson (4) grandmother

Solutions 1 – 3:

1 Either A or C is Engineer, F is father of both A and C. Answer: (2)

2. Two or three; B and D (sex of A is not known). Answer: (4)

3. Either granddaughter or grandson. Answer: (3)

LOGICAL DIAGRAMS

Introduction:

There are three types of logical diagram tests generally asked by the examiner. These are explained below.

TYPE – 1

In Type 1, questions are based on the concept of class. A class is a group or collection of objects, all having something in

common. For example, a class of females will include all daughters and nieces in a group. There are three possible

relationships between any two different classes.

1. Class Containing Classes:

All those females that fall into class of nieces are contained in class of females. The idea that one class may contain another is

the most Fundamental logical principle underlying the logical diagram questions.

Females

Nieces Back to Table of Contents



2. Class Partially Containing Other Classes:

Consider two classes, doctors and females. Since not all doctors are females, no class of doctors can entirely contain the class

of females. The partial containment of one class by another class can be picturised in the following way:

Doctors Females

Now, in the above figure, the two-joined circles indicate that there are three classes.

1. Those who are lady doctors. (B)

2. Those who are doctors, but not females. (A)

3. Those who are females, but not doctors. (C)

3. Classes Independent of Each Other

The classes of all males and all females exclude each other, since no female can come into the class of males and also no male

can be included into the class of females. In actual logical diagram tests, you will be working with three circles rather than two,

with no new principles of relationships between classes. For example, let us take three different classes, Females, Doctors,

Girls Females

Now, each of these classes will stand in one of the three types of relationships to the other, i.e. three different two-circle

diagrams for females-girls, doctors-females, and doctors-girls. Instead of three different diagrams, you can represent all of

these relationships by intersecting the three circles as in the figure below.

Females Doctors

Girls


B C

A

Males



It will be helpful to familiarize yourself with the various patterns of three-circle relationships. Here are the seven most

common patterns. Though more than seven patterns are possible, these are the patterns based on which questions are

frequently asked.

Diagram I

1. Insects, Butterflies, Mosquitoes

2. Males, Fathers, Boys

Diagram II

1. Females, Mothers, Lady Doctors

2. People, Players, Asians

Diagram III

1. Males, Bachelors, Teachers

2. Females, Doctors, Girls

Diagram IV

1. Insects, Mosquitoes, Grapes

1. Mangoes, Fruits, Wheat

2. Men, Fathers, Girls

Diagram V

1. Hair, Nail, Teeth

2. Dogs, Cats, Birds

Diagram VI

Entrances, Gates, Barriers

1. Aunts, Congressman, Fathers

Diagram VII

Cousins, Sisters, Nieces

1. Teachers, Students, Professionals

Based on these principles, a variety of questions can be asked. Back to Table of Contents



TYPE – 2

In this type of questions, there are two sets of principles:

1. One to choose the figure that represents the logical relationship among the items (in the figure);

2. Another to find and measure the portion that represents a particular statement.

Example:

In the adjoining figure, represents citizens of Delhi, represents males, represents educated

and represents unemployed. Find out which of the numbers denote the following:

1. Educated female outside Delhi citizens who are also employed.

2. Uneducated unemployed males who are citizens of Delhi.

Answers:

(1) 2 (2) 7

TYPE – 3

In this type of logical diagram question, you’ll be able to see through reasoning yourself and deduce the right answer.

These are the “Logical Reasoning” questions falling under “Logical Diagrams” type questions; these reasoning questions

become easier to understand if we solve them by making use of methods/ tips of logical diagram. Here you’ll be provided with

a set of given statements which will be followed by another set of deductions / conclusions.

The conclusions are supposed to follow from the question statements, and the statements (or assumptions) are to be taken as

true even if they seem to be at a variance with commonly known facts or universal truths. The candidates are required to

practice such questions and find out by themselves whether answering/solving such type of questions is easier by using the

techniques of “Logical Reasoning” or “Logical Diagram” questions.

Example:

Statements: 1. Some Indians are Muslims. 2. Some Pakistanis are Muslims.

Conclusions:

(i) Every Muslim is either an Indian or a Pakistani.

(ii) Some Muslims are Indians as well as Pakistanis.

(iii) No Muslim is an Indian as well as a Pakistani. Back to Table of Contents



(iv) Some Muslims are neither Indians nor Pakistanis.

Now, you have to choose your answer from one of the following alternatives

(1) Only II follows (2) Only III follows (3) Either II or III follows (4) Only I, II and III follow

Answer: (3)

The following diagrams give a clear idea of what can be concluded from the above statements.

Indians Muslims Pakistanis Indians Muslims Pakistanis

Fig (i) Fig (ii)

Now, fig (i) infers that there are some Muslims who are Indians and are some who are Pakistanis. Also, there are

Muslims who are neither Indian nor Pakistani and fig (ii) denotes – there are some who are Pakistani, some who are both

Indian and Pakistani; which can not be possible so this figure is not a proper explanation of the statements.

So only figure (i) is true and statement 3 is true among all the statement. Answer: (3)

MATHEMATICAL OPERATIONS

These types of problems are common to be asked in good competitive examinations. In these some mathematical

operations are inter – changed among themselves such as if divide (÷) denotes multiplication (×), Greater to (>) denotes

(+)” etc are the type of statements given and on the basis of those statements we have to solve a given problem.

Let us understand this in much wider concept with the help of an example.

Ex. If ‘P’ denotes ‘divided by’; ‘Q’ denotes ‘added to’; ‘M’ denotes ‘subtracted from’; ‘B’ denotes ‘multiplied by’; then

1 8 B 1 2 P 4 M 8 Q 6 =?

(1) 108 (2) 46 (3) 17 (4) 52 (5) None of these

Sol. 1 8 B 1 2 P 4 M 8 Q 6

According to the given information put the signs assigned for each alphabet we get;

= 18 × 12 ÷ 4 – 8 + 6

Now applying the concept of BODMAS to solve the above expression we get

= 18 x 3 – 8 + 6

= 54 – 2 = 52 Answer: (4)

Ex: Some symbols have been given different meaning. Read them correctly carefully and find out the correct one out of

the four alternatives




SIGNS

> Stands for ÷

v Stands for ×

< Stands for +

^ Stands for –

+ Stands for =

– Stands for >

× Stands for <

(1) 5 v 4 < 2 + 10 v 2 < 2 (2) 8 v 4 – 2 + 5 > 7 ^ 6 (3) 8 v 6 – 3 + 4 > 7 ^ 6

(4) 9 v 3 – 1 + 6 > 8 ^ 9

Sol. 5 v 4 < 2 + 10 v 2 < 2 gives

5 x 4 + 2 = 10 x 2 + 2

22 = 22 Answer: (1)

EXAMPLE:

Directions for questions 1 – 5: In these questions, the symbols @, *, $, # and % are used with the following meanings as

illustrated below:

‘P @ Q’ means ‘P is neither greater than nor equal to Q’.

‘P # Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P % Q’ means ‘P is neither smaller than nor greater than Q’.

‘P $ Q’ means ‘P is neither smaller than nor equal to Q’.

Now in each of the questions given below, assuming the given statements to be true, find which of the two conclusions I and II

given below is/are definitely true?

Give your answer as –

(1) if only Conclusion I is true.

(2) if only Conclusion II is true.

(3) if only Conclusion I or II is true.

(4) if neither Conclusion I nor II is true.

(5) if both the Conclusion I and II are true.

1. Statements: T % B, M * B, J # B

Conclusions:

I. T & M II. T % J Back to Table of Contents



2. Statements: V # D, D * K, F $ K

Conclusions:

I. D @ F II. V @ F

3. Statements: W # D, D % M, M * F

Conclusions:

I. D @ F II. F % D

4 Statements: H * R, R $ N, N @, K

Conclusions:

I. H @ K II. K $ R

5 Statements: M * J, K $ J, K @ T

Conclusions:

I. T $ M II. J @ T

Solutions 1 – 5:

1. T = B, M ≤ Q, J ≥ B ⇒ T ≤ J, M ≤ Q Answer: (4)

2. V ≥ D, D ≤ K, F > K

⇒ D ≥ K < F ⇒ F ⇒ D ≥ F ⇒ D @ F Answer: (1)

3 W ≥ D, D = M, M ≤ F

⇒ D ≤ F ⇒ D L F or D = F

⇒ D ≥ F or D = F

⇒ D @ F or D = F Answer: (3)

4. Answer: (4)

5 M ≤ J, K > J, K < T

⇒ M ≤ J < K < T ⇒ M < T

⇒ T > M ⇒ T ≤ M ⇒ T $ M

Also J < T ⇒ J ≥ T ⇒ J @ T Answer: (5)

VISUAL REASONING

Instructions:

These are problems that are in the form of figures, drawings and designs. The problems may be in the form of series,

analogies, classification, cube turning, turning, mirror image, paper folding, paper cutting, completion of incomplete

pattern, figure perception, spotting the hidden designs or construction of square.




Analogies:

In these questions, there are two sets of figures viz. the problem figures and the answer figures. The problem figures

are presented in two units. The first unit contains a pair of related figures and the second unit contains one figure and a

question mark in place of the fourth figure. You have to establish a similar relationship between two figures and point out

which one of the answer figures should be in place of the question mark.

Problem Figures

Answer Figures

Consider the above problem figures. The second figure is related to the first figure in a certain way. That is the

elements in the second figure are double the elements in the first figure. The first figure has one square and the second has

two squares. The third and fourth figures should also have the same relationship as the first and second have. That means that

the fourth figure should have two circles.

Problem Figures

Answer Figures

Classification:

In classification the problem figures themselves are the answer figures. Out of the five given figures tour are similar in a

certain way. One figure is not like the other four. You have to identify the “odd man out”.


?

?



In the figures given below, of the five figures tour are straight lines whereas one is a circle. Thus the circle is the “odd

Man out”.

Series:

The four figures given at the left are the problem figures. The next five are the answer figures. The problem figures

make up a series. That means they change from left to right in a specific order. If the figures continue to change in the same

order what would the fifth figure be? In the example below, the line across the problem figures is falling down. Thus if

the line continues to fall its fifth position would be lying flat i.e. it will be horizontal. Therefore the answer is (4).

Figure perception:

In this type of problems, we have to count number of figures hidden in the given design

For example:

1. The number of squares in the given figure is

(1) 12 (2) 10 (3) 14 (4) 15

2. The number of rectangles excluding squares in the above figure is

(1) 12 (2) 13 (3) 14 (4) 17

3. The number of triangles in the figure is

(1) 54 (2) 48 (3) 69 (4) 70

Upon studying the figure one can easily state that the answer to the first question is (3), that to the second question is (4) and

to the third question is (3).

Cube turning:

In this type of problems we have to deal with different positions of the same cube.

For example:

The drawing on the left in each of the following figures represents a cube. There is a different design on each of the six

faces of the cube. Four other drawings of the cube are lettered (1), (2), (3) and (4). Point out which one of the four could

be the cube on the left turned to a different position. The cube on the left may have been turned over or around or over

and around. Back to Table of Contents



After studying all the choices, one can infer that the answer is (3).

Problems on Dice:

Sometimes we are given figures showing the same die in various positions. After looking at these figures, we have to find

the number opposite a given number on the die. The procedure for finding such a number will be clear from the example

given below.

Two positions of a block are given below. When one is at the top, which number will be at the bottom?

(1) 3 (2) 6 (3) 2 (4) 1 (5) 4

In both the figures 2 is at the top. To get the position of second figure, we have to rotate the dice in the first figure two times

in clockwise direction. After rotating the dice two times in the same direction, 6 comes in the place of 1. So 6 is on the side

opposite to the 1.

∴ Answer is (2).

Hidden Figure Test:

Hidden figure test is one more type of problem that one may encounter in visual reasoning. A simple figure is given. One

has to identify it in more complex figures.

For Example:

Find the simple figures hidden in this complex figure.

By inspection one can say that the figure (4) is hidden in the above figure.

Mirror Images:

In this type of problems the reflection of a design is seen in mirrors placed in different manners.

For Example:

A plane mirror is kept horizontally below the figure and then one kept on its side. Choose the correct image in the second

mirror.




In the given problem the image of the question figure in the mirror kept horizontally below the figure is

The image of this in the mirror kept at the side of the first mirror is

∴ Answer is (1).

LOGICAL GAMES

LOGICAL GAMES involve puzzles in which the relationships among the groups of objects, people, cities, activities etc. are

given. These puzzles may deal with such things as making a group, seating arrangement, scheduling the activities etc. After

reading and analyzing the statement, you’ll be asked to answer three to seven questions bout the relationships given, which

require you to accurately interpret the information given as well as draw logical inferences about relationships.

The analytical games can be categorized as follows:

1. Sequencing games

2. Grouping games

3. Matching games

4. Hybrid games

5. Mapping games

SEQUENCING GAMES

In these types of games you have to put the entities (persons, teaching, schedules etc.) in order. In a sequencing game, you

may be asked to arrange/schedule the entities from north to south, left to right, top to bottom, or Monday through Friday etc.

GROUPING GAMES

In grouping games, you may be asked to organize the entities into groups or teams etc. It can be a selection or distribution

problem e.g. selecting players or dividing the people into groups. In selection games you start with a large pool of entities and

you have to select a smaller group from these.

Tips for Sequencing and Grouping games

Use short hand language to write the rules

A and B both cannot be there AB

A is done before B A < B Back to Table of Contents



Two things are done between A and B A _ _ B or B _ _A

B is done two days/hours after A A_B

“Picturizing a problem is more important than making the diagram of the problem. Short handing and diagram forming are

only the tools to enhance your thinking and solve the question.”

TCY’s 5-step approach that can help you:

1. Get the overview of the problem

Establish the entities.

Note the action

2. Picturize the problem mentally (understanding).

Assemble the entities

Use a simple diagram

3. Consider individual rules.

Take time to understand the rules.

Short hand the rules (brief and clear)

4. Combine rules

Try deducing from the given set of rules

5. Answer:

Read the question carefully and try to pre-phrase the answer.

Use the elimination with the help of deductions you have made.

Note:

Don’t write the full name of cities, peoples etc., and the items should be designated simply by their first letter.

It’s unnecessary wastage of time.

Try to start the diagram with definite or concrete relationship/condition

Pay close attention to words like “could be”, “must be”, “may”, “not”, “except”, “necessarily”. Because answer to question

like “Which must be true” or “Which of the following may be true” will be different.

Don’t get confused with the one-way relationship. For example if A attends the seminar, then B also attends it. This means if A

is present, B should also be present. Do not interpret it as if B is present A should also be present.

Grouping games require us to answer the same basic questions: Who’s in and who’s out? Which group can include X, and who

else can or cannot reside in a group with X?

Tips to solve grouping game problems:

1. See what entities can, must or cannot be in what groups.

2. See what entities can, must or cannot be in the same group as other entities. Back to Table of Contents



3. Notice whether the game asks you to put ALL of the entities into groups or asks you to select SOME of the entities for a

smaller group.

4. Pay close attention to numbers: the number of entities in each group, the total number of entities available, the number of

entities already chosen.

5. For ambiguous entity names or to differentiate group names from entity names, use upper case and lower case letters.

MATCHING GAMES

In this type of problem some persons with some pet names or professions or states or cities or names of their wives etc. are

given but not in same order. You have to match the correct ones.

Method to solve these types of problems:

1. Draw a table with name of the person vertically and quality or other parameter horizontally

2. Read the statement. Put the cross mark (×) if quality or parameter is not applicable.

3. Put the tick mark (√) if some quality or parameter is applicable.

4. If in a row or column, a tick mark (√) appears, then put cross marks (×) in all the remaining boxes in that row or column.

5. If in a row or column, all the boxes except one have cross marks (×), then put tick mark (√) in that box.

HYBRID GAMES

It is a mixture of sequencing and grouping games. Mostly, these are considered to be the most difficult types of games. But

not every game is a hybrid, and not all hybrid games are difficult.

1. Don’t panic. Organization is the key to hybrid games.

2. As in other games there is no one ‘correct diagram’ for hybrid games.

3. Try making as many deductions as possible.

PROBLEMS

SEQUENCING GAMES

GAME 1

A professor plans his teaching schedule to deliver eight topics B,C, H, P, Q, R, V, and W. of his subject. The topic must be

taught one at a time in accordance with the following guidelines:

H must be the fourth topic and W must be the sixth topic.

Topic Q must be taught before topic H.

Topic B and topic V cannot be taught consecutively.

Topic C must immediately precede topic Q.

Exactly two topics must be taught between topic P and topic Q.

1. Topic P must come immediately between which of the following pairs of topics?

(1) Q and H (2) C and W (3) R and B (4) H and W Back to Table of Contents



2. What is the maximum number of topics that can be taught between the topics C and R?

(1) two (2) three (3) four (4) six

3. If Topic B is taught seventh, which of the following must be true?

(1) Topic C is taught second. (2) Topic V is taught third.

(3) Topic P is taught eighth. (4) Topic R is taught third.

4. Which of the following pairs of courses cannot be taught consecutively?

(1) Q and V (2) W and V (3) R and H (4) B and R

Solutions:

Since you’re asked to arrange eight topics in order, it makes senses to visualize the game by drawing eight slashes, and also

number the slashes like this:-

– – – – – – – –

1 2 3 4 5 6 7 8

Rule I is concrete rule .H is fourth and W is sixth. Build this right into your diagram

Rule II: Q < H (As H is at 4th position, Q must be first, second or third)

Rule III: (not together)

Rule IV: CQ ( C is immediately before Q. Rule I & II tell us Q must be first, second or third, so C must be first

or second and Q can’t be at first position)

Rule V: Exactly two topic between P and Q (Sequence can be P _ C Q or C Q _ _ P Now first not possible because Q must be

second or third, so only C Q _ _ P Sequence is possible.)

Q can only be taught second or third. If Q is taught second, C must be first and P must be fifth which is possible.

But if Q is taught third, C must be second and P must be sixth which is not possible as W must be taught sixth.

But that’s not all, Rule III said that B and V can’t taught consecutively .The only slot left are 3rd, 7th, and 8th.Since B and V must

be separated ,they can’t taught 7th,and 8th.Therefore ,either B or V must be taught third. By combining all the rules and

deductions now the problem can be visualized as follow:

B/V/R

C Q B/V H P W _ _

1 2 3 4 5 6 7 8

Answer to the problems:

1. Answer: (4)

2. Answer: (4)

C is definitely taught first and R can either be taught seventh or eighth.

Since you are looking for most topics between two, so R will be taught eighth Back to Table of Contents



(C Q B/V H P W V/B R )

3. If B is taught seventh then V must be taught third. Answer: (2)

4. All are possible except choice (3) i.e. R and H. Answer: (3)

GAME 2

Hardy’s world ride is composed of six dragon bogies, numbered 1 to 6. Six children must be put into the six bogies, one child

per bogie. The six children are Kailash, Mohair, Namarita, Onkar, Puneet, and Raman. Mohit must be in bogie 1 or 6.

Onkar and Puneet must be in adjacent bogies. Kailash must be closer than Raman to the front of the dragon bogies.

1. Kailash CANNOT be in which one of the following bogies?

(1) bogie 1 (2) bogie 2 (3) bogie 3 (4) bogie 6

2. If Onkar and Namarita are in adjacent bogies, and if Puneet is in bogie 6, Raman must be in which one of the following

bogies?

(1) bogie 1 (2) bogie 2 (3) bogie 3 (4) bogie 4

3. If Puneet is in bogie 1, which one of the following CANNOT be true?

(1) Mohit is in bogie 6 (2) Namarita is in bogie 3

(3) Onkar is in bogie 2 (4) Kailash is in bogie 5

4. Mohit must be in bogie 6 if which of the following children is in bogie 2?

(1) Namarita (2) Onkar (3) Puneet (4) Raman

Solutions:

Step –1: Establish the entities

There are Kailash, Mohit, Namarita, Onkar, Puneet, and Raman.These can be casted as K, M, N, O, P, R.

Step –2: Visualise the problem

This is a sequencing game which can be visualized with six slots:

1 2 3 4 5 6

– – – – – –

Step –3: Consider the individual rules

Abbreviate wherever possible, express simple rules in visually direct shorthand.

Rule- I: Mohit must be in bogie 1 or 6 can be abbreviated as

1 2 3 4 5 6

M – – – – –

or

– – – – – M

Rule – II : Onkar and Puneet must be in adjacent bogies. OP or PO Back to Table of Contents



Rule – III: Kailash must be closer than Raman to the front of the dragon bogies K < R

Step – 4: Answers to the questions

1. Kailash CANNOT be in which one of the following bogies?

Sol. As K < R. R follows K. So K cannot occur last i.e. bogie no – 6. Answer: (4)

2. If Onkar and Namarita are in adjacent bogies, and if Puneet is in bogie 6, Raman must be in which one of the following

bogies?

Sol. As P is on 6 no. So O must be at 5 (Rule no – II) and N must be at 4 (ON or NO given in the question)

1 2 3 4 5 6

M – – N O P

Also K < R so K must be in bogie no. 2 and R in bogie No. 3

Answer: (3)

3. If Puneet is in bogie 1, which one of the following CANNOT be true?

1 2 3 4 5 6

P O – – – M

Possible arrangements are P O K R N M or P O K N R M and P O N K R M

Let’s check the options

(1) Mohit is in bogie 6 ---- true

(2) Namarita is in bogie 3 ---- may be possible

(3) Onkar is in bogie 2 ---- true

(4) Kailash is in bogie 5 ---- not possible (As K < R)

(5) Raman is in bogie 4 ---- may be possible Answer: (4)

4. Mohit must be in bogie 6 if which of the following children is in bogie 2?

If R will be in bogie no 2 then K must be in bogie no 1 (Rule – III: K < R)

1 2 3 4 5 6

K R – – – M

Therefore M must be in bogie no 6 (Rule – I)

Answer: (4)

GAME 3

The coach of the Sports Club must choose two two-person Badminton teams for an upcoming tournament.

The players available are Chahail, Daman and Eshaan, who are experienced players; and Rajiv, Sahil, Tej, and Uday, who are

novices.

At least one experienced player must be in each team in the tournament. Back to Table of Contents



Daman and Sahil will be chosen only if the two are in different teams.

If either Chahail or Tej is chosen, the other must also be chosen.

Tej will not be chosen if Uday is chosen.

Eshaan will not be chosen if Rajiv is chosen.

1. Which one of the following must be true?

(1) Chahail and Rajiv cannot both be chosen. (2) Daman and Tej cannot both be chosen.

(3) Uday and Chahail cannot both be chosen (4) If Eshaan is chosen, Sahil cannot be chosen.

2. Which of the following is NOT an acceptable selection for the teams?

(1) Team 1: Daman and Rajiv; Team 2: Chahail and Tej

(2) Team 1: Chahail and Eshaan; Team 2: Tej and Rajiv

(3) Team 1: Daman and Chahail; Team 2: Eshaan and Tej

(4) Team 1: Eshaan and Sahil; Team 2: Daman and Uday

3. If Sahil is chosen and Tej is rejected for the tournament, which ones of the following must be the members of one of the

teams?

(1) Sahil and Daman (2) Sahil and Chahail

(3) Daman and Rajiv (4) Daman and Uday

4. If Uday is not chosen for the expedition, and Rajiv is chosen for team 1, which one of the following must be in team 2?

(1) Eshaan (2) Chahail (3) Sahil (4) Tej

Solutions:


Experienced players are Chahail, Daman, and Eshaan can be abbreviated as C, D & E

Novices are Rajiv, Sahil, Tej, and Uday i.e. R, S, T, U

Step –2: Visualise the problem

This is a grouping game which can be visualized with two two person teams:

T1 / T2

– – – –

Step –3: Consider the individual rule

Abbreviate where possible, express simple rules in visually direct shorthand

Rule – I: At least one experienced players must be in each team in the tournament.

Rule – II: Both Daman and Sahil will be chosen only if they are in different teams. D | S

Rule – III: If either Chahail or Tej is chosen, the other must also be chosen. CT or TC

Rule – IV: Tej will not be chosen if Uday is chosen. Back to Table of Contents



Rule – V: Eshaan will not be chosen if Rajiv is chosen.

Step – 4 Answers to the questions

1. Check the options:

(1) Chahail and Rajiv cannot both be chosen.

This is not true because we can choose C & R together. If we choose C then T will also be there. Fourth person will be D (Rule

no – I & Rule no – V)

T1 / T2

CT DR

(2) Daman and Tej cannot both be chosen.

This is also possible .SEE group made in option (1)

(3) Uday and Chahail cannot both be chosen.

This is not possible because if C is chosen, T should also be chosen (Rule-III). But U and T cannot be together (Rule IV). This

option must be true

(4) If Eshaan is chosen, Sahil cannot be chosen.

This is not true if we can make

T1 / T2 & T1 / T2

ES DU ES CT

(5) If Uday is chosen, Tej must also be chosen.

Not possible (Rule no IV)

Answer: (3)

2. Which of the following is NOT an acceptable selection for the teams?

All options except option (2) are possible (Rule V:

3. If Sahil is chosen and Tej is rejected for the tournament, which one of the following must be the members of one of the

teaming teams?

If T is rejected then C can’t be chosen (Rule V). But at least one-experienced players must be in each team in the tournament.

So D & E must be chosen. If E is chosen then R cann’t be chosen so the fourth person left is U. Possible teams are

T1 / T2

SU DR

Answer: (4)

4. If Uday is not chosen for the tournament, and Rajiv is chosen for team 1, which one of the following must be in team 2?

If R is chosen, then experienced player with R will be C and D (E can’t Rule V) As C and T are always together. So T must be

chosen. Possible teams are Back to Table of Contents



T1 / T2 & T1 / T2

RC DT RD CT

So T must be in team 2. Answer: (4)

MAPPING GAMES

Mapping games revolve around things like roads, messages relay thing with TO/FROM relationships

GAME 4

A telecommunication company has six satellite towers in cities: New Delhi, Orissa, Panipat, Quilon, Rajkot and Shimla

Because of an antiquated technology, signals can be directly sent only from:

New Delhi to Panipat

Panipat to New Delhi

New Delhi to Quilon

Quilon to Panipat

New Delhi to Shimla

Rajkot to New Delhi

Orissa to Rajkot

Shimla to New Delhi

A “relay” occurs when a tower receives a signal from another tower and sends it on to a third. A tower can relay a signal from

one tower to another in any combination allowed by the above conditions.

1. Which tower cannot receive signals from any other tower?

(1) New Delhi (2) Orissa (3) Panipat (4) Quilon

2. Which of the following would require exactly one relay?

(1) a signal sent from New Delhi to Shimla (2) a signal sent from Orissa to Quilon

(3) a signal sent from Quilon to Orissa (4) a signal sent from Rajkot to Quilon

3. A signal cannot possibly be sent from

(1) Shimla to Quilon (2) Rajkot to Panipat

(3) Panipat to Rajkot (4) Shimla to Panipat

4. If the telecommunication system at Panipat fails, so that Panipat may send but not receive signals, which of the following

would be IMPOSSIBLE?

(1)Sending a signal from Orissa to New Delhi (2) Sending a signal from Quilon to New Delhi

(3) Sending a signal from Orissa to Quilon (4) Sending a signal from Shimla to Quilon

5. Quilon would be able to send signals to all other cities either directly or by relay if which of the following capabilities were

added to the original list? Back to Table of Contents



(1) Sending signals from Orissa to Quilon (2) Sending signals from Rajkot to Orissa

(3) Sending signals from Quilon to Shimla (4) Sending signals from Panipat to Orissa

Solutions:

Step – 1: Establish the entities

There are six cities, which can be abbreviated as N, O, P, Q, R and S

Step – 2: Visualise the problem

This is a mapping game which can be visualized with a diagram.

Step – 3: Consider the individual rule

Rule No

8 1

S N P

5 2

6 3 4

O R Q

1. New Delhi to Panipat

2. Panipat to New Delhi

3. New Delhi to Quilon

4. Quilon to Panipat

5. New Delhi to Shimla

6. Rajkot to New Delhi

7. Orissa to Rajkot

8. Shimla to New Delhi

Solutions:


1. Which tower cannot receive signals from any other tower?

(2) Orissa (see the diagram)

2. Which of the following would require exactly one relay?

(1) a signal sent from New Delhi to Shimla ----- No relay

(2) a signal sent from Orissa to Quilon ----- 2 relays (O – R – N – Q)

(3) a signal sent from Quilon to Orissa ----- Not possible Back to Table of Contents



(4) a signal sent from Rajkot to Quilon ----- 1 relay (R – N – Q)

Answer: (4)

3. A signal cannot possibly be sent from

(1) Shimla to Quilon ---- Possible (S – N – Q)

(2) Rajkot to Panipat ---- Possible (R – N – P)

(3) Panipat to Rajkot ---- Not possible (S – N – Q)

(4) Shimla to Panipat ---- Possible (S – N – P)

Answer: (3)

4. If the telecommunication system at Panipat fails, so that Panipat may send but not receive signals, which of the following

would be IMPOSSIBLE?

Now diagram reduces to

S N P

O R Q

(1) Sending a signal from Orissa to New Delhi --- possible (O – R –N)

(2) Sending a signal from Quilon to New Delhi --- not possible

(3) Sending a signal from Orissa to Quilon --- possible (O – R –N –Q)

(4) Sending a signal from Shimla to Quilon --- possible (S – N –Q)

Answer: (2)

5. Quilon would be able to send signals to all other cities either directly or by relay if which of the following capabilities

were added to the original list?

S N P

O R Q




Q can send signals to P, N, S but not to R and O. So Q would be able to send signals to all other cities only when O would be

able to receive signals. (that is only in option (2) & option (4) ) But sending signals from R to O does not link O & Q. So Answer

is (4).

Another example of Mapping Game

Five villages linked by roads. The roads run directly between:

Village A and Village B

Village B and Village C

Village B and Village D

Village D and Village C

Village D and Village E

There are no other roads that provide access to any of the villages.

1. How many different ways are there to travel by road from village A to village E without going through any village twice?

(1) 1 (2) 2 (3) 3 (4) 4

Sol.

A B D E

C

We can see from the diagram that there are only two possible ways to travel from village A to village E that are:

A – B – D – E and A – B – C – D – E

Remember to:

Start your map with an entity frequently mentioned in the rules. This will form a hub at the center of your map.

Keep track of connections, not locations.

Now --- draw a map!

After drawing your map, think about its structure.

Which entities are centrally positioned, forming nodes or hubs?

Which are relatively cut off, forming dead ends?

GAME 5

Exactly four persons stand in a queue on the wait for their turn for collecting entrance ticket at ticket counter at a cinema

theater, numbered 1 through 4 from first to last. Two of the persons are males and other two are females. Two out of four are

doctors, one is lawyer and one is engineer. Exactly one of four is wearing a cap. The persons in a queue are standing according

to the following conditions. Back to Table of Contents



The person wearing a cap is either at No. 1 or No. 4. Doctor is at No. 2 position.

At least one male stands in line between the two females.

One of the doctors is wearing a cap.

1. Which one of the following must be true for a person at No. 3 position?

(1) She is a female. (2) He is a male.

(3) Person is a lawyer. (4) Person does not wear a cap.

2. If person at No. 4 position is a male who wears a cap, then all of the following must be true EXCEPT.

(1) No. 1 is a female. (2) No. 2 is a male.

(3) Engineer is at No. 3 position. (4) Doctor is at No. 4 position.

3. If the two males stand in line immediately adjacent to each other, then which one of the following must be false?

(1) A female wears a cap. (2) Engineer is a female.

(3) Person at No. 3 is a male. (4) Both doctors are male.

4. If the two doctors stand in line immediately adjacent to each other, and if person at No 2 is a male, then which one

of the following correctly describes person at No. 1?

(1) A female doctor wearing a cap. (2) A male doctor wearing a cap.

(3) A female engineer without cap. (4) A male engineer without cap.

Solutions:


M Stands for Male

F Stands for Female

D Stands for Doctor

L Stands for Lawyer

E Stands for Engineer

Step – 2: Visualise the problem

This is a hybrid game which can be visualized with a diagram

1 2 3 4

– – – –

Step – 3: Consider the individual rule

Two of the persons are males and other two are females.




Two out of four are doctors, one is lawyer, and one is engineer..

The person wearing a cap is either at No .1 or No. 4. Doctor is at No. 2 position.

_ D _ _

Cap or Cap

At least one male stands in line between the two females.

One of the doctors is wearing a cap.


1. Which one of the following must be true for a person at No.3 position?

(1) She is a female. ---- May be true

(2) He is a male. ---- May be true

(3) Person is a lawyer. ---- May be true

(4) Person does not wear a cap. ---- Must be true

(5) Person wears a cap. ---- False

Answer: (4)

2. If person at No.4 position is a male who wears a cap, then all of the following must be true EXCEPT.

F M F M

– D – D

Cap

(1) No. 1 is a female. ------ true

(2) No. 2 is a male. ------ true

(3) Engineer is at No. 3 position. ------ may be possible

(4) Doctor is at No.4 position. ------ true

(5) Exactly one person stands in the between the two doctors ------ true

Answer: (3) Back to Table of Contents



3. If the two male stand in line immediately adjacent to each other, then which one of the following must be false.

F M M F F M M F

D D – – or – D – D

Cap Cap

(1) A female wears a cap. ---- May be true

(2) Engineer is a female.. ---- May be true

(3) Person at No.3 is a male. ---- Definite true

(4) Both doctors are male. ---- False

(5) Lawyer is a female. ---- May be true

Answer: (4)

4. If the two doctors stand in line immediately adjacent to each other, and if person at No 2 is a male, then which one of the

following correctly describes person at No.1?

Answer: (1) A female doctor wearing a cap

IF – THEN (Conditional Statement)

Here’s an If-Then RULE. This is going to be very helpful for the Grouping and Hybrid Games section. “If A then B”: It means if

given A, then B must be true. It also means that if we have not given B then A must not be true. So this conditional statement

is equivalent to “If not B, then Not A” but we can’t tell “if not A then ….. “and “if B then …….”. Be careful while applying this

approach. For example: “If Amit attends the seminar then Ajay must attend it”. We can deduce from it that if Ajay does not

attend the seminar, then Amit must not attend it. But if Ajay attends the seminar then Amit may or may not attend it or if

Amit does not attend the seminar, then whether Ajit attends the seminar or not that we can’t tell.

Data Interpretation/Data Sufficiency


Option 1: if conclusion can be made by using either of the two statements.

Option 2: if conclusion can be made by using only one statement.

Option 3: if conclusion can be made by only by using both the statements. Back to Table of Contents



Option 4: if conclusion cannot be made by using both the statements

1. Total number of students in a class is 120. What is the total number of girls in the class?

a) The number of boys in a class is 90.

b) The number of girls is about 25%.

2. A shop sells 28 pairs of shoes and gives 20% discount on each pair. What is the cost of a pair of shoe?

1) The cost of a pair of shoe is Rs.89.

2) The total amount 60 pairs of shoe is 1250.

3. The cost of a book including postage charge and tax is Rs.100.What is the original price of the book.

1) The postage charge is 1.75 rupees.

2) The tax is 9%.

Data Interpretation

The following questions were asked in the actual written examination.

1. Industrial consumption of power doubled from 1980-81 to 1993-94, by how much percentage did the total power

consumption grow from 1980-81 to 1993-94.

1. 45% 2. 150% 3. 190% 4.Cant be determined

2. If in 1993-94, a total of 246.2 billion kW of power was consumed, how much power was consumed by the agricultural

sector?

1. 72,500 mn kW 2.73,860mn kW 3.738bn kW 4.75 bn kW

3.Per capita consumption of power in the country has increased from 100 kW in 1980-81 to 283 kW in 1993-94,whereas the

population increased from 64 crore to 87 crore, by how much % did the consumption of the Agricultural sector increase from

1980-81 to 1993-94(approx)?

1. 500% 2. 540% 3.600% 4. 700%

4. If a total of 357.8 bn kW of electricity was generated in1993-93 and 246.2 on kW was consumed, how many sectors

consumption was higher than the losses in distribution of electricity(losses in distribution =electricity generated-electricity

consumed)?

1. 0 2. 1 3. 2 4. 3

5. In how many sectors did the consumption in terms of % of total consumption has increased from 1980-81 to 1993-94?

1. 1 2. 2 3. 3 4. 4





Data Interpretation/Data Sufficiency

Data Interpretation

Data plays an important role in day to day life. If data is too large, it can be represented in precise form in a number of

ways. Once data is represented in precise form, the user of that data has to understand it properly. The process of

interpreting the data from its precise form is called Data Interpretation.

Now we will discuss different ways of representing data and we will see how we can extract the data from the given

representations.

Different ways of representing data:

1. Data Tables

2. Pie Charts

3. Two-Variable Graphs

4. Bar Charts

5. Venn Diagrams

6. Three-Variable Graphs

7. PERT Chart

8. Combination of 2 or more charts

Now, we shall study these methods in detail.

1. Data Table:

Here the entire data is represented in the form of a table. The data can be represented in a single table or in combination of

tables. To understand it better, look at the following example.

From the above table, we can find the following: Back to Table of Contents



1. Population of a particular city with respect to that in any other city for a given year.

2. Percentage change in the population of any city from one year to another.

3. The rate of growth of population of any city in any given year over the previous year.

4. The city, which has maximum percentage population growth in the given period.

5. For a given city, finding out the year in which the percentage increase in the population over the previous year was the

highest.

6. Rate of growth of the population of all the cities together in any given year over the previous year.

2. Pie Chart

In this, the total quantity is distributed over one complete circle. This circle is made into various parts for various

elements. Each part represents share of the corresponding element as portion of the total quantity. These parts can be

represented in terms of percentage or in terms of angle.


Look at the following Pie-chart representing crude oil transported through different modes over a specific period of time.

The above pie chart can also be represented as below

We can find the following from the above pie chart.

1. The oil that has been transported through any mode if the total transported amount is known.

2 The proportion of oil transported through any mode with respect to any other mode.

3. The total oil transported, if the oil transported through any particular mode is known.

3. Two-Variable Graphs

Here the data will be represented in the form of a graph. Generally it represents the change of one variable with respect to

the other variable. Look at the following graph. Back to Table of Contents



Car sales in India in different years (in 000’s)

From the above graph, we can calculate.

1. Percentage change in the sales of any brand in any year over the previous year.

2. Rate of growth of total sales of the cars (all the brands) in a given period.

3. Proportion of the sales of any brand with respect to those of any other brand in the given year. Back to Table of Contents

Bar Chart

Bar Chart is also one of the ways to represent data.

The data given in the above graph can also be represented in the form of bar chart as shown below.

Here also we can deduce all the parameters as we could do in the case of two-variable graph.

5. Venn Diagrams

If the information comes under more than one category, we represent such data in the form of a Venn diagram.

The following Venn diagram represents the number of people who speak different languages.




From the above Venn diagram, we can find

1. The number of people who can speak only English.

2. The number of people who can speak only Punjabi.

3. The number of people who can speak both Punjabi and Hindi.

4. The number of people who can speak all the three languages.

5. The number of people who can speak exactly one or two languages.

6. Three-Variable Graphs

Look at the following example to understand the concept. The graph represents percentage of GRE, GMAT and CAT

students in three institutes x, y, z.


The above diagram gives the percentage of students of each category (GRE, GMAT, CAT) in each of the institutes x, y, z.

Data Sufficiency

(a) Basic Data: The question has limited available data and could include geometrical figures, graphs and algebraic

statements.

e.g. (i) How old is Ravi?

(ii) Is x2 > x?

(iii) What is the area of the Δ ABC?

As a rule, the basic data is insufficient to arrive at a solution.

(b) Two Statements: To see whether we can arrive at a solution or not, there are two separate statements succeeding the

given question containing additional data which may be an aid to arrive at a possible solution. Normally, the alternatives given

are as follows:

(1) : Statement (I) alone is sufficient but statement (II) alone is not sufficient to answer the question asked.

(2) : Statement (II) alone is sufficient but statement (I) alone is not sufficient to answer the question asked.

(3) : Both statements (I) and (II) together are sufficient to answer the question asked but neither of the statement alone is

sufficient to answer the question asked. Back to Table of Contents



(4) : Each statement is sufficient by itself to answer the question (THOUGH THE ANSWER, IF COMPUTED, MAY BE DIFFERENT)

(5) : Statement (I) and (II) together are not sufficient to answer the question asked and additional data specific to the question

is needed.

In answering the questions, you are to indicate which of these five alternatives (1) to (5) applies to the given question. You

need not memorize these alternatives. They are always given before the questions.

Sometime you may be given four choices as follows:

Mark your answer as

(1) : If you can get the answer from any of the statement.

(2) : If you can the answer from either of the statement.

(3) : If you can get the answer by combining both the statements.

(4) : If neither of the statement is sufficient to answer the question.

The procedure to solve any given problem is as follows:

STEP (i) Read and comprehend the basic data. One important thing to remember is never assume any other information

except the basic rules and formulae.

STEP (ii) Take statement (I), combine the available data with the already existing information and check if you can arrive at a

solution.

DO NOT TRY SOLVING, JUST ENSURE THAT A SOLUTION CAN BE OBTAINED.

STEP (iii) Irrespective of whether a solution can be obtained from statement (I) alone, take statement (II) and check if a

solution can be arrived at. REMEMBER NOT TO USE THE DATA AVAILABLE IN STATEMENT (I).

STEP (iv) This step is optional. If a solution cannot be derived from either statement (I) or statement (II) individually, COMBINE

the data available from the two statements and check if a solution can be arrived at.

STEP (v) Select the right alternative.

Note: Always read the instructions given for DS questions before solving them.

Points not to be ignored

1. In DS questions our answer can be in “YES” or “NO”. It is not necessary that our answer should always be in “YES”. For

example let’s consider following example:-

Q. Is A son of B?

(A) C is A’s brother.

(B) B has two kids C & A, one son and one daughter.

Sol. (A) statement says C is a boy and A can be a boy or girl. Now when we combine statement (A) with (B), we can be sure

that C is a boy and A is a girl. So we can get the answer by combining both the statements. Please note that our answer is in

terms of “NO” i.e. A is not the son of B. But still we are able to answer the question being asked.




2. Don’t try to solve whole of the question. The moment you think that from this point onward question CAN BE SOLVED,

mark your answer.

3. If you can get the answer from one statement alone and by combining both the statements, then your answer will be from

one statement alone.

4. Always stick to your basics for these questions.

Let us understand answering a DS question with the help of a flow chart.

EXAMPLES:

1. What is the value of X, if X and Y are two distinct integers and their product is 30?

(A) X is an odd integer (B) X > Y

Sol. Statement (A) alone is not sufficient

As x can be 1, 3 or 5

Statement (B) alone is not sufficient

As x can be 10 or 6

Combing (A) and (B)

i.e. x is odd and x > y

x can be 10, 6, 30 So both statements together are not sufficient.

2. If X, Y, and Z are integers, Is X(Y+ Z) an odd integer?

(A) XYZ is an odd integer. (B) X + Y + Z is an odd integer." Back to Table of Contents



Sol. (A) xyz is odd

⇒ x, y, z are odd

⇒ x (y + z) is even ∴ Statement (a) alone is sufficient.

(B) x + y + z is on odd integer

⇒ Either x, y, z are odd or one of them is odd and other two are even

Hence x(y + z) can be either odd or even ⇒ Alone is not sufficient.

Miscellaneous


1. Given that A, B, C, D, E are different digits.

AB CD = EEE

(CD E)-AB = CC

Then find the digits. Ans: A, B, C, D, E = 3,7,1,2,4

2. 1 2 3 4

+ 3 4 5 5

----------

4 6 8 9

- 2 3 4 5

----------

2 3 4 4

+ 1 2 5 4

------------

3 6 9 8

Strike off any digit from each number in seven rows (need not be at same place) and combine the remaining digits and

perform the same operations with 3 digit numbers to get the same addition. After this strike off another digit from all and

combine the remaining digits and perform the same operations with 2 digit numbers to get the same addition. Similarly

perform the same process again with 1 digit numbers..

3. There is a safe with a 5-digit number. The 4th digit is 4 greater than second digit, while 3rd digit is 3 less than 2nd digit. The 1st

digit is thrice the last digit. There are 3 pairs with sum is 11. Find the number. Ans: 65292.




4. xxx) xxxxx (xxx

3xx

-------

xxx

x3x

------

xxx

3xx

------

Find the 5 digit number.

5. Replace each letter by a digit.

O N E

O N E

O N E

O N E

-------

T E N

------ Ans: 0 =1, N = 8 ,E = 2, T = 7.

6. Of all pets I have, except two all are rabbits

Of all pets I have, except two all are fish

Of all pets I have, except two all are cats

How many rabbits, fish and cats are there? Ans: r-1 c-1 f-1

7. There are nine cards with four colors (Red-1, Yellow-3, Blue-2 and Green-3) arranged in a 3*3 matrix.

One red card must be in first row or second row.

Two green cards should be in 3rd column.

Yellow cards must be in the 3 corners only.

Two blue cards must be in the 2nd row.

At least one green card in each row.

Solution:

Yellow Red Green

Blue Blue Green

8. How it is possible to place four points that are equidistance from each other?

9. You are given a cake; one of corners is broken. How will you cut the rest into two equal parts?




10. How will you measure height of building when you are at the top of the building? If you have stone with you.

Ans: Drop the stone and find the time taken for the stone to reach the ground then find the height using the formula

s = (s = height, a= initial velocity=0, g=9.8m/s, t = time taken)

In each question, a set of six statements are given, followed by four answer choices. Each of the answer choices has a

combination of three statements from the given set of six statements. you are required to identify the answer choices in

which the statements are logically related.

11. A. All cats are goats

B. All Goats are dogs.

C. No goats are cows

D. No goats are dogs

E. All Cows are dogs

F. All dogs are cows

1. FAB 2. ABE 3.AFB 4.ABF

12. A. Some lids are nibs

B. All hooks are lids.

C. All hooks are nibs

D. No lid is a nib

E. No lid is a hook

F. No nib is hook

1.EFD 2. BCA 3.DEA 4.CDA

13. A. All MBA’s are logical

B. Sudhir is rational.

C. Sudhir is a logical MBA

D. Sudhir is a man

E. Some men are MBA’s

F. All men are rational.

1. DEC 2. EAF 3. BCF 4.FDB

14. A. Competitive examinations are tough to pass.

B. Thre is heavy competition in any field.

C. No student can pass MAT

D. Very few students can pass MAT.

E. MAT is a competitive examination

F. MAT is tough to pass

1. AEF 2.ABC 3.DFB 4.CDE Back to Table of Contents



15. A. All Pens are knives

B. All knives are spoons

C. No knives are pens

D. No knives are spoons.

E. All pens are spoons.

F. All spoons are pens.

1. ABE 2. ABF 3. AFE 4. DBE

16. Find the values of each of the alphabets.

N O O N

S O O N

+ M O O N

----------

J U N E Ans: 9326

17. Using 8 8’s and using only ‘+’, arrange them to add up to thousand? Ans: 888+88+8+8+8=1000




Formula Booklet

1. Averages

Simple averages =

Weighted Average =

Geometric Mean =

Harmonic Mean =

For two numbers, Harmonic Mean =

2. Percentage Change

Change % = %

Total Successive Change% = %

3. Interest

Simple Interest =

Compound Interest =

Population after n years P’ =

4. Growth

Growth% = %

SAGR or AAGR = %

CAGR = %

[Here, S. A. G. R. = Simple Annual Growth Rate, A. A. G. R. = Average Annual Growth Rate and C. A. G. R. = Compound Annual

Growth Rate]

5. Profit and Loss

Profit = SP − CP

Loss = CP – SP

Percentage Profit = Back to Table of Contents



Percentage Loss =

6. False Weights

If an item is claimed to be sold at cost price, using false weights, then the overall percentage profit is given by

Percentage Profit =

7. Discount

Discount = Marked Price − Selling Price

Discount Percentage =

8. Buy x and Get y Free

If articles worth Rs. x are bought and articles worth Rs. y are obtained free along with x articles, then the discount is equal to y

and discount percentage is given by

Percentage discount =

9. Successive Discounts

When a discount of a% is followed by another discount of b%, then

Total discount =

10. Ratios

If a: b = c: d, then a: b = c: d = (a + c): (b + d) If 0<a < b, then for a positive quantity x,

If a > b>0, then for a positive quantity x,

11. Proportions

If a: b:: c:d or , then

Alternendo Law

Invertendo Law Back to Table of Contents



Componendo Law

Dividendo Law

Componendo and Dividendo Law

If , then

and p, q, r are real numbers, then

12. Successive Replacement

Where x is the original quantity, y is the quantity that is replaced and n is the number of times the replacement process is

carried out.

13. Alligation Rule

The ratio of the weights of the two items mixed will be inversely proportional to the deviation of attributes of these two items

from the average attribute of the resultant mixture.

Alligation Cross:

x1 x2

x

x2- x : x – x1

w1 w2

14. Time, Speed and Distance

Speed = Distance / Time

Important Conversion Factors:

1 km/hr = m/s and 1 m/s = km/hr




15. Average Speed

Average = =

A man travels first half of the distance at a speed , second half of the distance at a speed then, Average Speed [Average

speed is given by harmonic mean of two speeds] Savg =

If the time is constant, then average speed is given by arithmetic mean of two speeds: Savg =

16. Relative Speed

For Trains

Time =

For Boats and Streams

Sdownstream = Sboat + Sstream

Supstream = Sboat - Sstream

Sboat =

Sstream =

17. Time and Work/Pipes and Cisterns

Number of days to complete the work =

[This is our general formula to solve time & work problems. It is also known as Work Equivalence Method]

18. Application of H.C.F.

The greatest natural number that will divide x, y and z leaving remainders r1, r2 and r3, respectively, is the H.C.F. of (x − r1), (y

− r2) and (z − r3)

19. Application of L.C.M.

The smallest natural number that is divisible by x, y and z leaving the same remainder r in each case is the L.C.M. of (x, y and z)

+ r




20. H.C.F. and L.C.M. of Fractions

H.C.F of fractions =

L.C.M of fractions =

[Express all numbers as fractions in its simplest form]

21. Properties of Surds

22. Law of Indices

If a and b are non – zero rational numbers and m and n are rational numbers, the

If , then m = n

If and m 0, then a = b if m is odd and a = b if m is even

23. Laws of Logarithms




If

If

24. Binomial Theorem

If n is a natural number that is greater than or equal to 2, then according to the binomial theorem:

(x+a) n = ncoxna0 + nc1xn-1a1 + nc2xn-2a2 +………. +ncnx0an

Here,

25. Roots of Quadratic Equation

The two roots of the equation, ax2+bx+c =0 are given by:

26. Algebraic Formulae

(a + b) (a − b) = a2 − b2

(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a − b)3 = a3 − 3a2b + 3ab2 − b3

a3 + b3 = (a + b) (a2 − ab + b2)

a3 − b3 = (a − b) (a2 + ab + b2)

a3 + b3 +c3 – 3abc = (a+ b + c) (a2 + b2 + c2 – ab – bc – ac)

27. Arithmetic Progression




28. Geometric Progression

29. Harmonic Progression

30. Sum of Important Series

Sum of first n natural numbers

Sum of the squares of the first n natural numbers

Sum of the cubes of the first n natural numbers

=

31. Factorial

n! = 1 × 2 × 3 × … × (n − 1) × n

n! = n × (n − 1)!

32. Permutations

33. Combinations




Important Properties:

34. Partition Rule

Number of ways of distributing n identical things among r persons when each person may get any number of things

=

35. Probability

Probability of an event =

Odds in favour =

Odds against =

36. Pythagoras Theorem

For right triangle ABC

AC2 = AB2 +BC2




For acute triangle ABC

AC2 = AB2 + BC2 – 2 (BC) (BD)

For obtuse triangle ABC

AC2 = AB2 + BC2 + 2*BC*BD

37. Area of Triangle

When lengths of the sides are given

Area =

Where, semi perimeter (s) =




When lengths of the base and altitude are given

Area =

When lengths of two sides and the included angle are given

Area =

For Equilateral Triangle

Area =




For Isosceles Triangle

Area =

38. Apollonius Theorem

If AD is the median, then:

AB2 + AC2 = 2(AD2 + BD2)

39. Angle Bisector Theorem

If AD is the angle bisector for angle A, then: Back to Table of Contents



40. Area of Quadrilateral

For Cyclic Quadrilateral

Area =

Where, semi perimeter (s) =

If lengths of one diagonal and two offsets are given

Area =

If lengths of two diagonals and included angle are given

Area =




For Trapezium

Area =

For Parallelogram

Area = bh

For Rhombus

Area =




For Rectangle

Area = lb

For Square

Area = a2

41. Polygon

Number of Diagonals =

The sum of all the interior angles =

The sum of all exterior angles = 360

42. Area of Regular Hexagon



Area =

43. Circle

Circumference C =

Area (A) =

Length of Arc (l) = , where is in degrees.

Area of Sector = , where is in degrees.

Perimeter of Sector =




44. Ellipse

If semi-major axis (OD) = a and semi-minor axis (OA) = b,

Perimeter of the ellipse

Area of the ellipse

45. Trigonometric Ratios

For a right triangle, if P is the length of perpendicular, B is the length of base, H is the length of hypotenuse and is

the angle between base and hypotenuse,




46. Distance between Points

Distance between two points and is given by

AB =

47. Right Prism

Lateral Surface Area (L.S.A.) = Perimeter of base × height

Total Surface Area (T.S.A.) = L.S.A. + 2 × Area of base

Volume (V) = Area of base × height

48. Cuboid

S.A. = 2(lh + bh)

T.S.A. = 2(lh + bh + lb)

Volume (V) = lbh

Body Diagonal (d) =




49. Cube

L.S.A = 4a2

T.S.A = 6a2

Volume (V) =a3

Body Diagonal (d) =

50. Cylinder

Curved Surface Area (C.S.A.) = 2πrh

T.S.A. = 2πrh + 2πr2

Volume (V) = πr2h




51. Right Pyramid

L.S.A = Perimeter of Base Slant Height

T.S.A = L.S.A + Area of base

Volume (V) = Area of Base Height

52. Cone

C.S.A. = πrl

T.S.A. = πrl + πr2

Volume (V) =

Slant height (l) =




53. Frustum of a cone

Volume (V) =

54. Sphere

C.S.A. = 4πr2

T.S.A. = 4πr2

Volume (V) =

55. Hemisphere

C.S.A. = 2πr2

T.S.A. = 3πr2

Volume (V) = Back to Table of Contents



56. Spherical shell

T.S.A =

Volume (V) =

Documents

Analysis of INFOSYS Placement Papers