Study Kit Paper 2 Contents and Sample Materials

Contents of The Brochure
1. Contents Details of Study Kit Paper - II
(I) Contents of Comprehension & English Language Comprehension
· Contents of Comprehension
· Contents of English Language Comprehension
(II) Contents of Interpersonal & Communication Skills, Decision Making & Problem Solving
· Contents of Interpersonal & Communication Skills
· Contents of Decision Making & Problem Solving
(III) Contents of General Mental Ability, Logical Reasoning& Analytical Ability
· Contents of General Mental Ability
· Contents of Logical Reasoning & Analytical Ability
(IV) Contents of Basic Numeracy
· Contents of Basic Numeracy
(V) Contents of Data Interpretation & Data Sufficiency
· Contents of Data Interpretation & Data Sufficiency
2. Sample Study Materials of Our Study Kit
(I) Comprehension
· Comprehension
(II) English Language & Comprehension
· Sentence
· Sentence Correction
· English Language Comprehension
(III) Interpersonal & Communication Skills
· Interpersonal & Communication Skills
(IV) Decision Making and Problem Solving
· Decision Making and Problem Solving
(V) General Mental Ability
· Blood Relations
· Sitting Arrangements
· Arithmetical Reasoning
· Number, Ranking and Time Sequence Test

(VI) Logical Reasoning & Analytical Ability
· Syllogism
· Statement and Courses of Action
· Assertion and Reason
· Situation Reaction Tests
· Statement and Assumptions
(VII) Basic Numarcy
· Number System
· Simplification
· Average
· Percentage
· Time and Distance
· Probability
· Set Theory, Venn Diagrams, Functions & Relations
· Sequences & Series
(VIII) Data Interpretation & Data Sufficiency
· Introduction to Data Interpretation
· Pie Charts
· Mix Diagrams
· Data Sufficiency

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Contents of Study Kit Paper -2
Contents of Comprehension & English Language Comprehension
Contents of Comprehension
i Comprehension
Contents of English Language Comprehension
i Narration (Direct and Indirect)
i Sentence
i Common Error
i One-Word Substitution
i Synonyms and Antonyms
i Idioms and Phrases
i Sentence Correction
i Sentence Arrangement
i Analogy
i Foreign Words and Phrases
i English Language Comprehension
Contents of Interpersonal & Communication Skills, Decision
Making & Problem Solving
Contents of Interpersonal & Communication Skills
i Interpersonal & Communication Skills
Contents of Decision Making & Problem Solving
i Decision Making and Problem Solving
Contents of General Mental Ability, Logical Reasoning &
Analytical Ability
Contents of General Mental Ability
i Analogy
i Classification
i Series
i Coding-Decoding

i Blood Relations
i Direction Sense Test
i Logical Venn Diagrams
i Alphabet Test
i Sitting Arrangements
i Mathematical Operations
i Arithmetical Reasoning
i Inserting the Missing Character
i Number, Ranking and Time Sequence Test
i Eligibility Test
Contents of Logical Reasoning & Analytical Ability
i Syllogism
i Statement and Arguments
i Statement and Assumptions
i Statement and and Courses of Action
i Statement and Conclusions
i Deriving Conclusion
i Assertion and Reason
i Punch lines
i Situation Reaction Tests
i Cause and Effect
i Analytical Reasoning
Contents of Basic Numeracy
Contents of Basic Numeracy
i Number System
i Fractions
i Indices and Surds
i Square Root & Cube Root
i Simplification
i HCF & LCM
i Orders of Magnitude
i Unitary Method
i Average
i Percentage
i Profit and Loss
i Ratio & Proportion
i Partnership
i Alligation or Mixure
i Time and Work
i Time and Distance
i Simple Interest
i Compound Interest

i Area of Plane Figures
i Volume and Surface Area of Solid Figures
i Clocks and Calendar
i Concepts of Geometry
i Coordinate Geometry
i Combinatorics
i Probability
i Basic Algebra
i Set Theory, Venn Diagrams, Functions & Relations
i Statistics
i Sequences & Series
Contents of Data Interpretation & Data Sufficiency
Contents of Data Interpretation & Data Sufficiency
i Introduction to Data Interpretation
i Approaches to Data Interpretation
i Table Chart
i Pie Charts
i Bar Charts
i Line Graphs
i Mix Diagrams
i Data Sufficiency

SAMPLE
MATERIAL
OF
OUR
STUDY KIT
PAPER 2
Basic Numeracy

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A Civil Servant should be well-versed in basics of Number System. In the Civil Services Aptitude Test Paper
2, in Basic Numeracy, certainly there will be asked some questions based on types of, and operations on numbers.
In Indian system, numbers are expressed by means of symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, called digits. Here, 0
is called insignificant digit whereas 1, 2, 3, 4, 5, 6, 7, 8, 9 are called significant digits. We can express a number in
two ways.
Notation: Representing a number in figures is known as notation as 350.
Numeration: Representing a number in words is known as numeration as Five hundred and forty five .
Place Value (Indian)
Crore Lakh Thousand Unit
Ten Crore Crore Ten Lakhs Lakh Ten Thousands Thousand Hundred Tens One
100000000 10000000 1000000 100000 10000 1000 100 10 1
108
107
106
105
104
103
102
101
100
Place Value (International)
Million Thousand Unit
Hundred Ten Millions One Million Hundred Ten Thousand Hundred Tens One
Millions Thousands Thousands
100000000 10000000 1000000 100000 10000 1000 100 10 1
108
102
106
105
104
103
102
101
100
Face Value and Place Value of a Digit
Face Value: It is the value of the digit itself eg, in 3452, face value of 4 is four , face value of 2 is two .
Place Value: It is the face value of the digit multiplied by the place value at which it is situated eg, in 2586,
place value of 5 is 5 102
= 500.
Number Categories
Natural Numbers (N): If N is the set of natural numbers, then we write N = {1, 2, 3, 4, 5, 6, }
The smallest natural number is 1.
Whole Numbers (W): If W is the set of whole numbers, then we write W = {0, 1, 2, 3, 4, 5, }
The smallest whole number is 0.
Number System
CHAPTER 1

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Integers (I): If I is the set of integers, then we write I = { 3, 2, 1, 0, 1, 2, 3, }
Rational Numbers: Any number which can be expressed in the form of p/q, where p and q are both integers
and q # 0 are called rational numbers.
eg,
3 7
, ,5, 2
2 9
-
- ¼
There exists infinite number of rational numbers between any two rational numbers.
Irrational Numbers Non-recurring and non-terminating decimals are called irrational numbers. These
numbers cannot be expressed in the form of
p
q
.
eg, 3, 5, 29,¼
Real Numbers: Real number includes both rational and irrational numbers.
Basic Rules on Natural Numbers
1. One digit numbers are from 1 to 9. There are 9 one digit numbers. ie, 9 100.
2. Two digit numbers are from 10 to 99. There, are 90 two digit numbers. ie, 9 10.
3. Three digit numbers are from 100 to 199. There are 900 three digit numbers ie, 9 102.
In general the number of n digit numbers are 9 10(n 1)
4. Sum of the first n, natural numbers ie, 1 + 2 + 3 + 4 + + n =
( )1
2
n n +
5. Sum of the squares of the first n natural numbers ie. 12
+ 23
+ 32
+ 42
+ + n2
=
( )( )1 2 1
6
n n n+ +
6. Sum of the cubes of the first n natural ie, 12
+ 23
+ 32
+ + n3
=
( )2
1
2
n né ù+
ê ú
ê ú
ë û
Example: What is the value of 51 + 52 + 53 + + 100 ?
Solution. 51 + 52 + 33 + ... + 100 = (1 + 3 + + 100) (1 + 2 + 3 + ... + 50)
=
100 101 50 51
2 2
´ ´
- = 5050 1275 = 3775
Different Types of Numbers
Even Numbers: Numbers which are exactly divisible by 2 are called even numbers.
eg, 4, 2, 0, 2, 4
Sum of first n even numbers = n (n + 1)
Odd Numbers: Numbers which are not exactly divisible by 2 are called odd numbers.
eg, 5, 3, 1, 0, 1, 3, 5
Sum of first n odd numbers = n2
Prime Numbers: Numbers which are divisible by one and itself only are called prime numbers.
eg, 2, 3, 5, 7, 11
2 is the only even prime number.

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1 is not a prime number because it has two equal factors.
Every prime number greater than 3 can be written in the form of (6K + 1) or (6K 1) where K is an integer.
There are 15 prime numbers between 1 and 50 and l0 prime numbers between 50 and 100.
Relative Prime Numbers: Two numbers are said to be relatively prime if they do not have any common
factor other than 1.
eg, (3, 5), (4, 7), (11, 15), (15, 4)
Twin Primes: Two prime numbers which differ by 2 are called twin primes.
eg, (3, 5), (5, 7), (11, 13),
Composite Numbers Numbers which are not prime arc called composite numbers
eg, 4, 6, 9, 15,
1 is neither prime nor composite.
Perfect Number: A number is said to be a perfect number, if the sum of all its factors excluding itself is
equal to the number itself. eg, Factors of 6 are 1, 2, 3 and 6.
Sum of factors excluding 6 = 1 + 2 + 3 = 6.
6 is a perfect number.
Other examples of perfect numbers are 28, 496, 8128 etc.
Rules for Divisibility
Divisibility by 2: A number is divisible by 2 when the digit at ones place is 0, 2, 4, 6 or 8.
eg, 3582, 460, 28, 352, ....
Divisibility by 3: A number is divisible by 3 when sum of all digits of a number is a multiple of 3.
eg, 453 = 4 + 5 + 3 = 12.
12 is divisible by 3 so, 453 is also divisible by 3.
Divisibility by 4: A number is divisible by 4, if the number formed with its lasttwo digits is divisible by 4. eg,
if we take the number 45024, the last two digits form 24. Since, the number 24 is divisible by 4, the number 45024
is also divisible by 4.
Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
eg, 10, 25, 60
Divisibility by 6: A number is divisible by 6, if it is divisible both by 2 and 3.
eg, 48, 24, 108
Divisibility by 7: A number is divisible by 7 when the difference between twice the digit at ones place and
the number formed by other digits is either zero or a multiple of 7.
eg, 658
65 2 8 = 65 16 = 49
As 49 is divisible by 7 the number 658 is also divisible by 7.
Divisibility by 8: A number is divisible by 8, if the number formed by the last 3 digits of the number is
divisible by 8. eg, if we take the number 57832, the last three digits form 832. Since, the number 832 is divisible
by 8, the number 57832 is also divisible by 8..
Divisibility by 9: A number is divisible by 9, if the sum of all the digits of a number is a multiple of 9.
eg, 684 = 6 + 8 + 4 = 18.
18 is divisible by 9 so, 684 is also divisible by 9.
Divisibility by 10: A number is divisible by 10, if its last digit is 0. eg, 20, 180, 350, .
Divisibility by 11: When the difference between the sum of its digits in odd places and in even places is
either 0 or a multiple of 11.

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eg, 30426
3 + 4 + 6 = 13
0 + 2 = 2
13 2 = 11
As the difference is a multiple of 11 the number 30426 is also divisible by 11.
‘Smart’ Facts
• If p and q are co-primes and both are factors of a number K, then their product p x q will also be a factor of r. eg,
Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 prime factors of 24 are 2 and 3, which are co-prime also. Product of
2 × 3 = 6, 6 is also a factor of 24.
• If ‘p’ divides ‘q’ and ‘r’, then p’ also divides their sum or difference. eg, 4 divides 12 and 20. Sum of 12 and 20 is 32
which is divisible by 4. Difference of 20 and 12 is 8 which is divisible by 4.
• If a number is divisible by another number, then it must be divisible by each of the factors of that number. 48 is
divisible by 12. Factors of 12 are 1, 2, 3, 4, 6, 12. So, 48 is divisible by 2, 3, 4 and 6 also.
Division on Numbers
In a sum of division, we have four quantities.
They are (i) Dividend, (ii) Divisor, (iii) Quotient and (iv) Remainder. These quantities are connected by a
relation.
(a) Dividend = Divisor Quotient + Remainder.
(b) Divisor = (Dividend Remainder) Quotient.
(c) Quotient = (Dividend Remainder) Divisor.
Example 2: In a sum of division, the quotient is 110, the remainder is 250, the divisor is equal to the sum of
the quotient and remainder. What is the dividend ?
Solution. Divisor = (110 + 250) = 360
Dividend = (360 110) + 250 = 39850
Hence, the dividend is 39850.
Example 3: Find the number of numbers upto 600 which are divisible by 14.
Solution. Divide 600 by 13, the quotient obtained is 46. Thus, there are 46 numbers less than 600 which are
divisible by 14.
Factors and Multiples
Factor: A number which divides a given number exactly is called a factor of the given number,
eg, 24 = 1 24, 2 12, 3 8, 4 6
Thus, 1, 2, 3, 4, 6, 8, 12 and 24 are factors of 24.
1 is a factor of every number
A number is a factor of itself
The smallest factor of a given number is 1 and the greatest factor is the number itself.
If a number is divided by any of its factors, the remainder is always zero.
Every factor of a number is either less than or at the most equal to the given number.
Number of factors of a number are finite.
Number of Factors of a Number: If N is a composite number such that N = am
bn
c ... where a, b, c ... are
prime factors of N and m, n, o ... are positive integers, then the number of factors of N is given by the expression
(m + 1) (n + 1) (o + 1)

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1. Evaluate:
( )
9 3 5 5 4 10
3 5 2 4 2
- -
- -
(a) 9/10 (b) 8/17
(c) 16/19 (d) 4/7
2. The sum of three consecutive natural numbers
each divisible by 3 is 72. What is the largest
among them?
(a) 25 (b) 26
(c) 27 (d) 30
3. 55% of a number is more than one-third of that
number by 52. What is two-fifth of that number?
(a) 96 (b) 240
(c) 144 (d) 142
4. The digits of a two-digit number are in the ratio
of 2 : 3 and the number obtained by interchanging
the digits is bigger than the original number by
27. What is the original number?
(a) 63 (b) 48
(c) 96 (d) 69
5. What least number would be subtracted from
427398 so that the remaining number is divisible
by 15?
(a) 13 (b) 3
(c) 16 (d) 11
ANSWER
1. (c) 2. (c) 3. (a) 4. (d) 5. (b)
EXPLANATIONS
1.
( )
- - ¸
- - ´ ¸
9 3 5 5 4 10
3 5 2 4 2
=
( ) ( )
( )
´ - ´ ¸
- ´ - ¸
9 2 5 4 10
3 5 8 2 =
-
- -
18 2
15 4
= -
16
19
2. 3x + (3x + 3) + (3x + 6) = 72
Þ 9x + 9 = 72 = 9x = 72 9
or x =
63
9
= 7
The largest of them = 27.
3. Let the number be x.
55
100
x× =
1
52
3
x +
Þ
13
60
x× = 52 Þ x = 240

2
5
x =
2
240
5
= 96
4. Let the number be 10x y
x : y = 2 : 3
(10y + x) (10x + y) = 27 ...(i)
Þ 9y 9x = 27 Þ y x = 3 Þ y = x + 3
Putting this value of y, in (i)
3
x
x +
=
2
3
Þ x = 6 y = 9
Hence the number is 69.
5. Apply the divisibility tears of 3 and 5.
Example 4: Find the number of factors that 224 has.
Solution. 224 = 25
71
Hence, 224 has (5 + 1) (1 + 1) = 6 2 = 12 factors.
Multiple: A multiple of a number is a number obtained by multiplying it by a natural number eg,
Multiples of 5 are 5, 10, 15, 20
Multiples of 12 are 12, 24, 36, 48
Every number is a multiple of 1.
The smallest multiple of a number is the number itself.
We cannot find the greatest multiple of a number.
Number of multiples of a number are infinite.
EXERCISE

Percentage
Per cent means per hundred . It is given by % symbol. Here x% means x per hundred or .
100
x
Thus, any
percentage can be converted into an equivalent fraction by dividing it by 100.
eg 20% =
20 1
;
100 5
= 150% =
150 3
100 2
=
Also, any fraction or decimal can be converted into its equivalent percentage by multiplying with 100.
eg
1 1
100
5 5
= ´ = 20%;
3 3
100
2 2
= ´ = 150%.
Important Formulae
1. Percentage increase =
Increase
100
Original value
´
2. Percentage decrease =
Decrease
100
Original value
´
3. If the price of the commodity increases by r% then the reduction in consumption so as not to increase the
expenditure is 100 %
100
r
r
é ù
´ê ú+ë û
4. If the price of the commodity decreases by r% then the reduction in consumption so as not to increase the
expenditure is 100 %
100
r
r
é ù
´ê ú-ë û
5. If A s income is r% more than B s income then B s income is less than A s income by 100 %
100
r
r
é ù
´ê ú+ë û
.
6. If A s income is r% less than B s income then B s income is more than A s income by 100 %
100
r
r
é ù
´ê ú-ë û
.
Percentage
CHAPTER 10

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7. Let the population of a town be P and it increases at the rate of r% per annum, then
(a) Population after n years = 1
100
n
r
P
æ ö
+ç ÷
è ø
(b) Population n years ago = 1
100
n
P
ræ ö
+ç ÷
è ø
8. Let the present value of the machine be P and if it depreciates at the rate of r% per annum.
(a) Value of machine after n years = 1
100
n
r
P
æ ö
-ç ÷
è ø
(b) Value of machine n years ago = 1
100
n
n
r
P
æ ö
-ç ÷
è ø
Example 1: Express 3/2 as rate per cent.
Solution.
3
2
=
3
100 %
2
æ ö
´ç ÷
è ø
= 150%
Example 2: Find 25% of 1000.
Solution. 25% of 1000 =
25
1000
100
´ = 250
Example 3. What per cent of 6 is 144?
Solution. Required percentage =
144
100 %
6
æ ö
´ç ÷
è ø
= 2400%
Example 4: What per cent of 2.5 kg is 15 g?
Solution. Required percentage =
15
100 %
2.5 1000
æ ö
´ç ÷´è ø
= 0.6%
Example 5. If the price of tea falls by 12%, by how much pr cent must a house holder increase its consumption,
so as not to decrease its expenditure on tea?
Solution. (Short cut method)
Increase % in consumption = 100 %
100
r
r
ì ü
´í ý
-î þ
=
12
100 %
100 12
ì ü
´í ý
-î þ
=
12
100 %
88
æ ö
´ç ÷
è ø
=
150
%
11
=
7
13 %
11
Example 6: The value of a machine depreciates at the rate of 10% per annum. If its present value is
` 162000, what was the value of the machine 2 year ago?
Solution. Value of machine 2 year ago = ` 2
162000
10
1
100
é ù
ê ú
ê ú
ê ú
æ öê ú-ç ÷ê úè øë û
= `
10 10
162000
9 9
æ ö
´ ´ç ÷
è ø
= ` 200000

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Example 7: Due to a reduction of 5% in prices of sugar, a man is able to buy 1 kg more for ` 95. Find the
original and reduced rate of sugar.
Solution. Let the original rate be ` x per kg.
Reduced rate = `
1
(100 5)
100
x- ´ = `
95
100
x

95 95
1
95
100
x x
- =
Þ
5
1
x
= Þ x = 5
Original rate = ` 5 per kg
Reduced rate = `
19 5
.
20 1
æ ö
´ç ÷
è ø
per kg = `
19
4
= 4.75 er kg
Example 8: If the price of 1 kg cornflakes is increased by 25%, the increase is ` 10. Find the new price of
cornflakes per kg.
Solution. Original price =
Difference in price
100
Difference in per cent
´ =
10
100
25
´ = 400
New price =
125
40
100
´ = ` 50
EXER CISE
1. The difference of two numbers is 20% of the larger
number. If the smaller number is 20, then the
larger number is:
(a) 25 (b) 46
(c) 27 (d) 82
2. When any number is divided by 12, then dividend
becomes 1/4th of the other number. By how much
percent first number is greater than the second
number?
(a) 165 (b) 200
(c) 300 (d) 400
3. If one number is 80% of the other and 4 times the
sum of their squares is 656, then the numbers
are:
(a) 6,8 (b) 8, 10
(c) 16, 20 (d) 10, 15
4. Two numbers A and B are such that the sum of
5% of A and 4% of B is two-third of the sum of 6%
of A and 8% of B. Find the ratio of A : B.
(a) 1 : 2 (b) 3 : 1
(c) 3 : 4 (d) 4 : 3
5. Three candidates contested an election and
received 1136, 7636 and 11628 votes respectively.
What percentage of the total votes did the winning
candidate get?
(a) 57% (b) 77% (c) 80% (d) 90%
ANSWERS
1. (a) 2. (b) 3. (b) 4. (d) 5. (a)
EXPLANATIONS
1. Let the larger number be x.
Then, x 20 =
20 1
20
100 5
x x xÛ - =
Û
4
5
x = 20 Û x =
5
20
4
ì ü
´í ý
î þ
= 25%
2. Let the numbers be x and y. Then,
3 .
12 4
x y
x y= Û =
Required percentage =
100 %
x y
y
æ ö-
´ç ÷
è ø
=
2
100 %
y
y
æ ö
´ç ÷
è ø
= 200%
3. Let one number = x. Then, other number = 80%

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of x =
4
5

2
2 4
4
5
x x
é ùæ ö
+ê úç ÷
è øê úë û
= 656
Û x =
2 216
25
x x+ = 164
Û
241
25
x = 164
Û x2 =
164 25
41
´æ ö
ç ÷
è ø
= 100
Û x = 100
So, the numbers are 10 and 8.
4. 5% of A + 4% of B =
2
3
(6% of A + 8% of B)
5 4 2 6 8
100 100 3 100 100
A B A B
æ ö
Û + = +ç ÷
è ø
1 1 1 4
20 25 25 75
A B A BÛ + = +
1 1 100 4
100 75 75 3
A
A B
B
Û + Û = =
5. Total number of votes polled
= (1136 + 7636 + 11628) = 20400.
Required percentage =
11628
100 %
20400
æ ö
´ç ÷
è ø
= 57%.

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Probability
Probability is used to indicate a possibility of an event to occur. It is often used synonymously with chance.
(i) In any experiment if the result of an experiment is unique or certain, then the experiment is said to be
deterministic in nature.
(ii) If the result of the experiment is not unique and can be one of the several possible outcomes then the
experiment is said to be probabilistic in nature.
Various Terms Used in Defining Probability
(i) Random Experiment: Whenever an experiment is conducted any number of times under identical
conditions and if the result is not certain and is any one of the several possible outcomes, the experiment is
called a trial or a random experiment, the outcomes are known as events.
eg, When a die is thrown is a trial, getting a number 1 or 2 or 3 or 4 or 5 or 6 is an event.
(ii) Equally Likely Events: Events are said to be equally likely when there is no reason to expect any one of
them rather than any one of the others.
eg, When a die is thrown any number 1 or 2 or 3 or 4 or 5 or 6 may occur. In this trial, the six events are
equally likely.
(iii) Exhaustive Events: All the possible events in any trial are known as exhaustive events. eg, When a die
is thrown, there are six exhaustive events.
(iv) Mutually Exclusive Events: If the occurrence of any one of the events in a trial prevents the occurrence
of any one of the others, then the events are said to be mutually exclusive events. eg, When a die is thrown
the event of getting faces numbered 1 to 6 are mutually exclusive.
Classical Definition of Probability
If in a random experiment, there are n mutually exclusive and equally likely elementary events in which n
elementary events are favourable to a particular event E, then the probability of the event E is defined as P (E)
P(E) =
Favourabel Events
Total number of Events
=
( )
( )
n E m
n S n
=
If the probability of occurrence of an event E is P(E) and the probability of non-occurrence is P( )E , then,
Probability
CHAPTER 25

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P(E) + P( )E = 1. Hence, P( )E = 1 ,
m n m
ie
n n
-
- = the sum of the probabilities of success and failure is 1.
Also, 0 £ P(E) £ 1 and 0 £ P( )E £ 1.
If P(E) = 1, the event E is called a certain event and if P(E) = 0, the event E is called an impossible event.
If E is an event, then the odds in favour of E are defined as P(E) : P(E) and the odds against E are defined
as P(E): P ( )E . Hence, the odds in favour of E are
( )
( ): : ,
n mm
m n m
n n
-
= - the odds against E are
( )
:
n m m
n n
-
= (n m):m
Addition Theorem on Probability
If El
and E2
are two events in a sample space S, then P (El
È E2
) = P (El
) + P (E2
) P (El
Ç E2
). If E1
and E2
are
mutually exclusive events (disjoint), then P(El
È E2
) = P (El
) + P (E2
) . (Q P(El
Ç E2
) = q)
Independent and Dependent Events
Two or more events are said to be independent if the happening or non-happening of any one does not
depend (or not affected) by the happening or non-happening of any other. Otherwise they are called dependent
events.
eg, Suppose a card is drawn from a pack of cards and replaced before a second card is drawn. The result of the
second drawn is independent of the first drawn. If the first card drawn is not replaced, then the second drawn is
dependent on the first drawn.
If El
and E2
are independent events, then
P(El
Ç E2
) = P(El
) P(E2
)
Simple Event
An event which cannot be further split is called a simple event. The set of all simple events in a trial is called
a sample space.
Compound Event
When two or more events occur in relation with each other, they are called compound events.
Conditional Event
If El
and E2
are events of a sample space S and if E2
occurs after the occurrence of El
, then the event of
occurrence of E2
after the event El
is called conditional event of E2
given El
. It is denoted by E2
/El
.
Conditional Probability
If El
and E2
are the events in a sample space S and P(El
) ¹ 0, then the probability of E2
after the event El
has
occurred is called conditional probability of E2
given El
. It is denoted by 2
1
E
P
E
æ ö
ç ÷
è ø
and we define,
2 2 2 1 2
1 1 1
( ) ( )
( ) ( )
E P E E n E E
P
E P E n E
æ ö Ç Ç
= =ç ÷
è ø

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Smart Facts
When a die is rolled six events occur. They are {1, 2, 3, 4, 5 and 6}
When two dice are rolled 36 events occur. They are [(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2),
(2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1),
(5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)]
When a coin is tossed 2 events occur. They are {H, T}
When two coins are tossed 4 events occur. They are {HH, HT, TH, T T}
When three coins are tossed 8 events occur. They are {HHH HHT, HTH, HT T, T HH, THT, T TH, T
T T}
In a pack of 52 cards there are 26 red cards and 26 black cards. The 26 red cards are divided into 13
heart cards and 13 diamond cards. The 26 black cards are divided into 13 club cards and 13 spade
card. Each of the colours, hearts, diamonds, clubs and spades is called a suit. In a suit, we have 13
cards (ie, A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3 and 2)
Example 1: In a toss of a coin, find the probability of getting a tail.
Solution. Here, S = [H, T] and E = [T] P(E) =
( )
( )
n E
n S
=
1
2
Example 2: Two unbiased coins are tossed, what is the probability of getting both heads.
Solution. Here, S = [HH, HT, TH, TT] and E = [HH]
P(E) =
( )
( )
n E
n S
=
1
4
Example 3: In a simultaneous throw of a pair of dice, find the probability of getting a total more than 9.
Solution. Here n(S) = 6 6 = 36
Let E = Event of getting a total more than 9 = [(4, 6), (5, 5), (5, 6), (6, 5), (6, 4), (6, 6)]
P(E) =
( )
( )
n E
n S
=
6
36
=
1
6
Example 4: In a simultaneous throw of a pair of dice, find the probability that the sum of numbers shown on
the two faces is divisible by 5 or 6.
Solution. Here, n(S) = 6 6 = 36
Let E = Event of getting a sum of numbers shown on the two faces divisible by 5 or 6.
= [(1, 4), (1, 5), (2, 3), (2, 4), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1), (5, 5), (6, 4), (6, 6)]
N(E) = 12
P(E) =
( )
( )
n E
n S
=
12
36
=
1
3
Example 5: A card is drawn from a well shuffled pack of cards. Find the probability that it is a (i) queen (ii)
a red card (iii) a space.
(i) Let E be event of drawing a queen card. Then, one queen card can be drawn from 4 queens in 4
C1
ways.
P(E) =
( )
( )
n E
n S
=
4
1
52
1
C
C
=
4 1
52 13
=
(ii) Let E be the event of drawing a spade card. Then, one spade card can be drawn from 13 spade cards in 13
C1
ways.

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P(E) =
( )
( )
n E
n S
=
26
1
52
1
C
C
=
13 1
52 4
=
(iii) Let E be the event of drawing a spade card. Then, one spade card can be drawn from 13 spade cards in
13
C1
ways.
P(E) =
( )
( )
n E
n S =
13
1
52
1
C
C
=
13 1
52 4
=
Example 6: If three cards are drawn simultaneously from a pack of well shuffled cards, then find the
probability of them being (i) all queens (ii) all red cards (iii) all spades.
Solution. The total number of ways of selecting three cards from 52 cards is 53
C3
ways.
(i) Let E be event of drawing the queen cards. Then, three queen cards can be drawn from 4 queen cards in
4
C3
ways.
P(E) =
( )
( )
n E
n S
=
4
3
52
3
C
C
=
4 1
22100 5525
=
(ii) Let E be event of drawing the red cards. Then, three red cards can be drawn from 26 red cards i 26
C3
ways.
P(E) =
( )
( )
n E
n S
=
26
3
52
3
C
C
=
2600 26
22100 221
=
(iii) Let E be event of drawing spade cards. Then, three spade cards can be drawn from 13 spade cards in
13C3 ways.
P(E) =
( )
( )
n E
n S
=
13
3
52
3
C
C
=
286
22100
=
143
11050
Example 7: A card is drawn at random from a normal pack of cards. What is the probability that it is either
a diamond or a king?
Solution. Out of 52 cards one card can be drawn in 52
C1
ways.
Let E1
be the event that the card drawn is a diamond, E2
be the event that the card drawn is king and (E1
Ç
E2
) be the event that the card drawn is both diamond the king and S be the sample space.
P(E1
) =
1( )
( )
n E
n S
=
13
1
52
1
13
52
C
C
=
P(E2
) = (
2( )
( )
n E
n S
=
4
1
52
1
4
52
C
C
=
P (E1
Ç E2
) =
1
1
52
1
1
52
C
C
=
P (E1
È E2
) = P(E1
) + P(E2
) P (E1
Ç E2
)=
13 4 1
52 52 52
+ - =
16 4
52 13
=
Example 8: A bag contains 4 red balls and 5 white balls. a ball is drawn at random. Find the probability that
it is a red ball or a white ball.

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P(E1
) =
1( )
( )
n E
n S
=
4
1
9
1
C
C
=
4
9
P(E2
) =
2( )
( )
n E
n S
=
5
1
9
1
C
C
=
5
9
P(E1
) = 0
P(E1
È E2
) = P(E1
) + P(E2
) =
4 5
9 9
+ =
9
1
9
=
Example 9: A bag contains 4 red balls and 4 white balls. Two balls are drawn in succession from the bag
with replacement, what is the probability that the two balls are of different colours?
Solution. Let E1
and E2
denote the events of drawing balls of different colours in the first and the second
draw with replacement and S be the sample space.
P(E1
) =
1( )
( )
n E
n S
=
4
1
8
1
C
C
=
4 1
8 2
=
P(E2
) =
2( )
( )
n E
n S
=
4
1
8
1
C
C
=
4 1
8 2
=
P(E1
Ç E2
) = P(E1
) P(E2
) =
1 1 1
2 2 4
´ =
Example 10: A family has two children. What is the probability that both the children are boys given that at
least one of them is a boy?
Solution. Let b stand for boy and g for girl. The sample space of the experiement is
S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events:
E : both the children are boys
F : at least one of the child is a boy
Then, E = {(b, b)} and F = {(b, b), (g, b), (b, g)}
Now, E È F = {(b, b)}
Thus, P(F) =
3
and ( )
4
P E FÇ =
1
4
Therefore, P(E|F) =
( )
( )
P E F
P F
Ç
=
1
4
3
4
=
1
3
Example 11: In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430,
10% of the girls study in class XII. What is the probability that a students choosen randomly studies in Class XII
given that the choosen student is a girl?
Solution. Let E denote the event that a student choosen randomly studies in Class XII and F be the event
that the randomly choosen student is a girl. We have to find P(E|F).

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Now, P(F) =
430
1000
= 0.43 and P(E Ç F) =
43
0.043
1000
=
[Q no. of girls studying in XII 10% of 430 = 43]
Then, P(E|F) =
( )
( )
P E F
P F
Ç
=
0.043
0.43
= 0.1
Example 12: A person has undertaken a construction job. The probabilities are 0.65 that there will be
strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction
job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if
there is a strike. Determine that probability that the construction job will be completed on time.
Solution. Let a be the event that the construction job will be completed on time, and B be the event that
there will be a strike. We have to find P(A).
We have P(B) = 0.65, P(no strike) = P(B¢) = 1 P(B) = 1 0.65 = 0.35
P(A|B) = 0.32, P(A|B¢) = 0.80
Since events B and B¢ form a partition of the sample space S therefore, by theorem on total probability, we
have
P(A) = P(B) P(A|B) + P(B¢) P(A|B¢)
= 0.65 0.32 + 0.35 0.8 = 0.208 + 0.28 = 0.488
Thus, the probability that the construction job will be completed in time is 0.488.
EXERCISE
1. Which of the following cannot be the probability
of an event ?
(a) 1/4 (b) 20% (c) 1.2 (d) 0.3
2. If P(E) = 0.03, what is the probability of not E ?
(a) 0.90 (b) 0.97 (c) 0.07 (d) 0.70
3. A bag contains orange flavoured candies only. A
girl takes out one candy without looking into the
bag. What is the probability that she takes out
an orange flavoured candy ?
(a) 1 (b) 0
(c) 1/2 (d) 1/4
4. A bag contains orange flavoured candies only. A
girl takes out one candy at random from the bag.
What is the probability that she takes out a
strawberry flavoured candy ?
(a)
1
3
(b) 1 (c) 0 (d)
1
2
5. An unbiased die is thrown once. What is the
probability of getting a prime number ?
(a)
1
2
(b)
1
4
(c)
2
3
(d)
1
3
ANSWERS
1. (c) 2. (b) 3. (a) 4. (c) 5. (a)
EXPLANATIONS
1. 1.2 cannot be the probability of an event because
0 £ P (E) £ 1
2. Probability of not E
P (not E) = ( )p E = 1 P(E) = 1 0.03 = 0.97
3. Probability that the girl takes out an orange
flavoured candy is 1 because the bag contains
orange flavoured candies only.
4. Probability that she takes out a strawberry
flavoured candy is 0 because the bag contains only
orange flavoured candies.
5. Here, n(S) = {1, 2, 3, 4, 5, 6} and E = Event of
getting a prime number = {2, 3, 5)
P(E) =
( )
( )
n E
n S =
3
6
=
1
2

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Sets
A set is a collection of well defined objects.
The objects of the sets are called elements.
(i) Sets are usually denoted by capital letters A, B, C,..., X, Y, Z.
(ii) The elements of the sets are denoted by small letters like a, b, c,..., x, y, z etc.
Representation of Sets
Sets are usually described into two ways.
(i) Tabular form or roster form, in this form, all the elements of the set are separated by commas and enclosed
between the bracket { }.
For example
(a) The set of vowels of English Alphabet as
A = {a, e, i, o, u)
(b) The set of numbers on a clock face is written as
B = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(ii) Set builder from: We define a set by stating properties which its elements must satisfy. For example the
set of all even integers. Then, we use the letters usually x, and we write
A = {x | x is an even integer}
This is to be read as A is a set of numbers x such that x is an even integer. The vertical line | to be read as
such that some times we use x in place of vertical line.
A = {x : x is an even integer}
eg, C = {1,w, w2} = {x | x3 1 = 0}
If an object x is an element of a set A, we write x Î A which is read as x belong to A and if an object x is not
a member of A we write x Ï A and read as x does not belong to A .
Some Important Terms
(i) Empty or Null set The set which contains no elements is called the empty set or the null set. The empty
set is written as f.
Thus, f = { } as there is no element in the empty set.
For example; the set of odd numbers divisible by 2 is the null set.
SetTheory,
Venn Diagrams,
Functions & Relations
CHAPTER 27

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(ii) Singleton set A set containing only one element is called a singleton for example, {1}, {4} are singleton
sets.
(iii) Equality of sets. The sets A and B are equal if they have same members that is if every elements of A is
an element of B and every element of B is an element of A, then A = B
eg, if A = { l, 3,5,7} and B = {7, 3, 1, 5}, then A = B
If the two sets are not equal we write A ¹ B
Important Formulae
1. A set does not change if its elements, are repeated.
2. A set does not change even if the order of its elements is different.
(iv) Finite and Infinite set. The set which contains a definite number of elements is called a finite set. The set
which contains an infinite number of elements is called an infinite set.
eg, (I) The set of days in a week.
eg, (II) The set of natural numbers.
(v) Disjoint set. Two sets A and B are said to be disjoint, if they do not have any element in common.
eg, A = { 1, 2, 3}, B = { 4, 5, 6} are disjoint sets.
(vi) Subset. If every element in set A is also an element of another set B. Then A is called a subset of B. Also B
is said to be super set of A.
Symbolically, we write
A Í B (ie, A contained in B)
B Í A (ie, B contains A)
More specifically A Í B if x A Þ x B
eg, (I) Let A = { 2, 4, 7}, B = { 1, 2, 3, 4, 7}
Then, A Î B since every element of A is in B.
eg, (II) A = {x | x a real number} and B = {x | x is an integer} Then, A Ê B
1. If there is at least one element of A which is not in B, then A is not a subset of B written as A Í B.
2. Every set is a subset of itself ie, A Í A.
3. If A Í B and B c Í A, then A = B.
(vii) The Null set f is a subset of every set A.
(viii) Proper Subset: A is a proper subset of B. if A Í B and A ¹ B and is written as A Ì B ie, if B contains at least
one element more than A, then A is a proper subset of B
(ix) Power set: Set of all the subsets of a set is called the power set
eg, A = {a, b, c} subsets of A are f, {a}, {b}, {c}, {a, b}, {b, c}, {c,a}, {a, b, c}
Hence, P(A) = [f, {a}, {b}, {c}, {a, b}, {b, c}, {c, a}, {a, b, c}]
If n is the number of elements of a set A, then the number of subset of A ie, the number of elements of P (A)
= 2n.
(x) Universal set: If all the sets under consideration are the subsets of a fixed set U, then U is called the
Universal set.
Union of sets
Union of two sets A and B is the set of all elements which belongs to A or B (or to both) and is written as
A È B (ie, A union B)
The same is defined in set builder form as
A È B = {x|x Î A or x Î B}

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If A = {1, 3, 5, 7, 9} and B = {2, 4, 5, 6, 9}
Then, A È B = { 1, 2, 3, 4, 5, 6, 7, 9}
1. From the definition of Union of sets A u B = B u A (Commutative Law)
If A is any set, then A È A = A and A È f = A
2. If A and B are any two sets, then A Í (A È B) and B Í (A È B)
If x Î A È B,then x Î A or x Î B and if x Ï A È B,then x Î A and x Ï B.
3. If A, B, C are three sets, then A È (B È C) = (A È B) È C
Intersection of Sets
If A and B are any two sets, then intersection of A and B is the set of all elements which are in A and also in
B. It is written as A Ç B and is read as A intersection B
If A = {2, 4, 6, 8} and B = {4, 5, 6, 9}
Then A Ç B = {4, 6}
1. From the definition of the intersection, it follows A Ç B = B Ç A (Commutative Law)
2. If A is any set, then A Ç A = A and A Ç f) = f
3. For any two sets A and B.
A Ç B = A and A Ç B Í B
4 If A and B have no elements in common ie, A and B are disjoint, then A Ç B = f
If x Î A Ç B = x Î A and x Î B
eg, (I) If A = { 2, 3, 6, 8, 9} and B = (1, 3, 5, 6, 7, 9}, then A Ç B = {3, 6, 9}
eg,(II) If A = {x1|< x < 4 } and B = {x|2 < x < 5}, then A Ç B = {x|2 < x < 4}
If A, B, C are three sets, then
(i) (A Ç B) Ç C = A Ç (B Ç C) Associative Law
(ii) A Ç (B Ç C) = (A Ç B) È (A Ç C) Distributive Law
Difference of Sets
The difference of two sets A and B is set of elements which belongs to A but do not belong to B. This is written
as A B
A B = {x| x Î A and x Î B}
1. Set A B subset of A ie, A B Í A
2. Set (A B) and B are disjoint ie, (A B) Ç B = f
3. A B = (A È B) (A Ç B)
Symmetric Difference of Sets
The symmetric difference of two sets A and B is (A B) È (B A) and is written as A D B
Thus, A D B = (A B) È (B A)
In the set builder form A D B = {x | x Î A or x Î B, but x Ï A Ç B}
Demorgan Laws
If A, B, C are three sets, then
(i) A (B È C) = (A B) Ç (A C)
(ii) A (B Ç C) = (A B) È (A C)

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Complement of a Set
Let A be a subset of universal set U, then the complement of A is denoted by AC
is defined by
AC
= {x Î U, x Ï A} x Î AC
Û x Ï A
eg, (I) If U = {1, 2, 3, 4, 5, 6} and A = { 1, 3, 5}, then AC
= {2, 4, 6}
eg, (II) U be the set of all letters in English alphabet and A is a set of all vowels, then AC
is the set of all
consonants.
1. (A È B)C = AC Ç BC 2. (A Ç B)C = Ac È BC
Venn Diagrams
A simple way of explaining the relation between sets is by a diagram which is called Venn diagram. In this a
set is generally represented by a circle and its elements by points in the circle.
CaseI: A Ì U and B Ì U and A Ç B ¹ f
A B
U
Here A and B are represented by a circle.
A B is the lined region
B A is dotted region and A Ç B is plane region.
Case II: A = {a, c, e}, B = {b, d}
f
g
a b
c
de
A B
A Ç B = f and A B = A and B A = B
Case III: When A Í B Í U
In adjoining figure, in Venn diagram
A
B
U
A È B = B, A Ç B = A and A B = f
Some results from the Venn diagram
(i) n(A È B) = n (a) + n(B) n(A Ç B)
(ii) n(A È B) = n (a) + n(B), when A Ç B = f
(iii) n(A B) + n (A Ç B) = n(A)
(iv) n(B A) + n (A Ç B) = n(A)
(v) n(A B) + n (A Ç B) + n (B A) = n (A Ç B)

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Example 1: If in a factory of 30 workers, 10 take tea but not coffee and 14 take tea. Then how many take
only coffee ?
Solution. Total number of workers = n (T È C) = 30
Number of workers who take tea n(T) = 14
n(T)
10 4 x
n
U = 30
( )c
Who take tea but not coffee = n(T C) = 10
Who drinks both coffee and tea = n (T) n (T C) = 14 10 = 4
Who takes only coffee = n (C T) = x
From the figure = x + 4 + 10 = 30 = x = 30 14 = 16
The worker who drinks only coffee = 16
Example 2: An elocution competition was held in English and Hindi. Out of 80 students, 45 took part in
English, 35 in Hindi, 15 in both English and Hindi, then for the number of students.
(a) Who took part in English but not in Hindi.
n(E) = 45
n(H) = 35
only E only H
(b) Who took part in Hindi but not in English.
(c) Who took part in either English or Hindi.
(d) Who took part in neither.
Solution. Suppose E is the set of students who took part in English, His the set of students who took part in
Hindi, then E n H gives the set of students who took part in both English and Hindi.
(a) The number of students who took part in English but not in Hindi
= n(E) = n(E Ç H) = 45 15 = 30
(b) The number of students who took part in Hindi but not in English
= n(H) n(E Ç H) = 35 15 = 20
(c) The number of students who took part either in English or in Hindi is
n(E È H) = n(E) + n(H) n(E Ç H) = 45 + 35 15 = 65
(d) The number of students who took part neither in English nor in Hindi
= n (S) n (T È H) = 80 65 = 15
Ordered Pair
If a, b be any two objects, then the pair (a, b) is called the ordered pair. The object a is called the first
coordinate (or first number) and b is called the second coordinate (or second number) of the ordered pair (a, b).
1. The ordered pair (a, b) ¹ (b, a)
Two ordered pairs (a, b) and (c, d) are said to be equal, if and only if a = c and b = d.

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Cartesian Product of Sets
If A and B be any two sets, then cartesian product of A and B is the set of all ordered pair (a, b), where a Î
and b Î B
Cartesian product of A and B is written as A B (ie, A cross B)
ie A B = {(a, b) | a Î A and b Î B}
eg, If A = {a, b, c} and B = {1, 2}, then
A B = {(a, 1), (a, 2),(b, 1),(b, 2),(c, 1),(c, 2)}
B A = {(1, a),(1, b),(1, c),(2, a),(2, b),(2, c)}
Thus, A B ¹ B A
A A = {(a, a), (a, b), (a, c),(b, a),(b, b),(b, c) (c, a),(c, b),(c, c)}
B B = {(1, 1), (1, 2),(2, 1),(2, 2))
1. A (B È C) = (A B) È (A C)
2. A (B Ç C) = (A B) È (A C)
Relations
A relation is a set of ordered pairs. If (x, y) is a member of a relation R, we write it as x R y (ie relation R to y).
eg, If R is the ordered pairs of positive integers where R = {(x, y)| x2
= y} The relation is y is a square of x and
the set is {(1, 1), (2, 4),(3, 9),(4, 16),...}
Types of Relations
(i) Reflexive: A relation R on a set A is said to be reflexive for every x Î A
(x, x) Î R
(ii) Symmetric Relation: A relation R on a set A is said to be symmetric if x R y Þ y R x
(x, y) Î R = (y,x) Î R
eg, Let A = {1, 2, 3} and R = {(1, 1), (2, 2),(1, 3),(3,1)}
Clearly, R is a symmetric relation.
(iii) Transitive Relation: A relation R in a set A is called transitive if x R y and y R z Þ x R z
eg, Let R be a relation in the real number defined by x less than y then
x < y and y < z = x < z
(iv) Equivalence Relation A relation which is reflexive, symmetric and transitive is a equivalence relation.
Functions
If each element of a set A is associated with exactly one element in the set B, then this association is called a
function from A to B.
The set A is called the domain and the set B is called the co-domain of the function.
Consider : A = {1, 2}, and B = (3, 4, 5, 6), then {(1, 4),(2, 5)) is a function
{(1, 4),(2, 5),(2, 6)} is not a function since element 2 in the set A have two images 5 and 6 in the set B
1. Each element of A must be associated with exactly one element in the set B.
2. All the elements of the set B need not have the association.
3. The set of elements of B which are associated with the elements of the set A is called the range of the
function.
4. The range is the subset of the co-domain.

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Types of Functions
(i) One-one Function (injection): A function f : A ® B is said to be a one-one function elements of A have
different images in B ie,
f(x) = zx, x Î {1, 2, 3}
f = {(1, 2),(2, 4),(3, 6)}
(ii) Many-one Function: A function f : A ® B is said to be a many-one function if two are n of A have the
same images in B.
(iii) Onto Function: A function f : A ® B is called an onto function if every element of B is an image of some
elements of A ie, if co-domain = range.
eg, Let A = {a, b, c, d} and B = {1, 2, 3}
f = {(a, 3),(b, 2),(c, 2),(d, 1))
(iv) Into Function: A function f: A B is called an into function if co-domain ¹ range.
Example 3: A is set of prime numbers less than 20, write A in Roster form.
Solution. Prime numbers Less than 20 are 2, 3, 5, 7, 11; 13, 17, 19 set A in Roster form.
A = {2, 3, 5, 7, 11, 13, 17, 19}
Example 4: Let A = {4, 5, 6, 7} and B = {6, 4, 7, 5}, then
Solution. {4, 5, 6, 7} = {6, 4, 7, 5}, since each of the elements 4 , 5, 6, 7 belongs to Band each of the elements
6, 4, 7, 5 belongs to A, then A = B.
The set does not change if its elements are rearranged.
Example 5: A = {x2
= 16, x is odd}, then
Solution. A is a empty set.
x 2
= 16 Þ x = + 4 or x = 4, but x is not odd
A does not contain any element, A = f
Example 6: Rewrite the following statements using set notations.
(a) x does not belongs to A (b) A is not a subset of B
(c) H does not include D (d) d is a member of E.
Solution.
(a) x Ï A (b) A Ë B
(c) H Ê D (d) d Î E
Example 7: Let A = {a, b, c}; ie, A contains the elements a, b, c, state whether each of the four statements is
correct or incorrect tell why.
(a) a Î A (b) a Í A
(c) {a} Î A (d) {a} Í A
Solution.
(a) a Î A, correct.
(b) Incorrect. The symbol Í must connect two set it indicates that one set a subset of other. Therefore,
a Í A is incorrect since a is a member of A, not a subset.
(c) Incorrect. The symbol a connects an objects to a set. It indicates that object is a member of the set.
Therefore, {a} Î A is incorrect since {a} is a subset of A.
(d) Correct.
Example 8: If S be the universal set of English alphabet and let A = {a, b, c}, then complement of A is
Solution. AC
= {d, e, f x, y, z}
Example 9: If A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, find A B, B A and A D B.
Solution. A B = {1, 3},

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(A B) contains the elements of A but not the elements of B. Similarly, B A = 16,81
(B A) contains the elements of B but not the elements of A.
A D B = (A B) È (B A) = {1, 3, 6, 8}
Example 10: If S = {1, 2, 3, 4, 5}, A= {1, 2, 4}, B= {2, 4, 5}
Find
(a) A È B (b) A Ç B (c) BC (d) B A
(e) AC È B (f) A È BC (g) AC Ç BC (h) BC AC
(i) (A Ç B)C (j) (A Ç B)C
Solution.
(a) A Ç B = {1, 2, 4, 5}
(b) A Ç B = {2, 4}
(c) The complement of B consists of letters which are in S but not in B, therefore BC = {1, 3}
(d) B A consisted of elements in B which are not in A ie, B A = {5}
(e) AC = {3, 5} and B = {2, 4, 5}, therefore, AC È B = {2, 3, 4, 5}
(f) A = {1, 2,4) and BC
= {1, 3}, therefore, A È BC
= {1, 2,3,4}
(g) AC = {3, 5}, and BC = {1, 3} ; therefore, AC Ç BC = {3}
(h) BC = {1, 3}, and AC = (3,5); therefore, BC AC = {1}
(i) A Ç B = {2, 4}, therefore, (A Ç B)C = {1, 3,5}
(j) A È B = {1, 2, 4, 5}; therefore, (A Ç B)C = {3}
Example 11: A = { 1, 2, 3} and B = {a, b}, then find A B and B A
Solution. A B = {{1, a}, {1, b}, {2, a}, {2, b), {3, a}, {3, b}} and B A = {{a,1}, {a, 2}, {a, 3}, {b, l}, {b, 2}, {b, 3}}
A B ¹ B A since the ordered pair (1, a) ¹ (a, 1)
Example 12: If the set A contains 4 elements and set B contains 3 elements, then A B contains
Solution. The set A B contains 12 elements.
EXER CISE
1. The set in set builder form of vowels of English
alphabet is
(a) {a, e, i, o, u}
(b) A = {x | x is a vowel in English alphabet)
(c) f
(d) {a, b, c, d, e, f, g}
2. If A = {5} which of the following statement is
correct?
(a) A = 5 (b) 5 Ì A
(c) {5} Î A (d) 5 Î A
3. If A = {a, (b, c), d)}, which of the following is a
subset of A.
(a) {a, b} (b) {b, c}
(c) {c, d} (d) {a, d}
4. If A Ç B = f nd A and B are two sets, then
(a) A Ì B (b) B Ì A
(c) A ¹ B (d) A and B are disjoint
5. If n(A Ç B) = 13, n(A) = 20, n(B) = 44, then n(A È
B) = ?
(a) 27 (b) 13 (c) 75 (d) 51
ANSWERS
1. (b) 2. (d) 3. (d) 4. (d) 5. (d)
EXPLANATIONS
1. Set builder form of vowels of English alphabet is
A = {x | x is a vowel in English alphabet}
2. The set A contains an element 5, therefore 5 Î A
3. Clearly, {a, d} is a subset of {a, (b, c), d}
4. If A and B are disjoint sets, then A and B have no
common elements A Ç B = f
5. n (A È B)= n (A) + n (B) n (A Ç B)
= 20 + 44 13 = 51

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Arithmetic Progression (AP)
An arithmetic progression is a sequence in which terms increase or decrease by a constant number called the
common difference.
(i) The sequence 2, 6, 10, 14, 18, 22 is an arithmetic progression whose first term is 2 and common difference
4.
(ii) The sequence
5 7
2, ,3, ,4
2 2
is an arithmetic progression whose first term is 2 and common
difference .
An arithmetic progression is represented by a,(a + d), (a + 2d), (a + 3d) a + (n 1)d
Here, a = first term
d = common difference
n = number of terms in the progression
The general term of an arithmetic progression is given by Tn = a + (n - 1) d.
The sum of n terms of an arithmetic progression is given by S, =
2
n
[2a + (n 1) d] or Sn = 2 [a + l]
where l is the last term of arithmetic progression.
If three numbers are in arithmetic progression, the middle number is called the arithmetic mean of the
other two terms.
If a, b, c are in arithmetic progression, then b =
2
a c+
where b is the arithmetic mean.
Similarly, if n terms al, a2, a3 an are in AP, then the arithmetic mean of these n terms is given by
AM =
1 2 3
.na a a a
n
+ + +¼+
If the same quantity is added or multiplied to each term of an AP, then the resulting series is also an AP.
If three terms are in AP, then they can be taken as (a d), a, (a + d).
If four terms are in AP, then they can be taken as (a 3d), (a d), (a + d), (a + 3d).
If five terms are in AP, then they can be taken as (a 2d), (a d), a, (a + d), (a + 2d).
Sequences & Series
CHAPTER 29

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Geometric Progression (GP)
A geometric progression is a sequence in which terms increase or decrease by a constant ratio called the
common ratio.
(i) The sequence 1, 3, 9, 27, 81 is a geometric progression whose first term is 1 and common ratio 3.
(ii) The sequence
1 1 1 1
1, , , , ,
3 9 27 81
¼ is a geometric progression whose first term is 1 and common ratio
1
.
3
A geometric progression is represented by a, ar, ar2
arn 1
.
Here, a = first term
r = common ratio
n = number of terms in the progression.
The general term of a geometric progression is given by Tn = an 1
The sum to n terms of a geometric progression is given by
( )1
,
1
n
n
a r
S
r
-
=
-
when r < 1
( )1
,
1
n
a r
r
-
-
when r > 1
If three numbers are in geometric progression, the middle number is called the geometric mean of the other
two terms.
If a, b, c are in geometric progression, then b ac= where b is the geometric mean.
Similarly, if n terms a1
, a2
, a3
, a4
, an
are in geometric progression, then the geometric mean of 1 these n
terms is given by GM = ( )
1
1 2 3 .nna a a a´ ´ ´¼´
For a decreasing geometric progression the sum to infinite number of terms is
,
1
a
S
r
¥ =
-
where a = first term and | r | < 1.
If every term of a GP is multiplied by a fixed real number, then the resulting series is also a GP.
If every term of a GP is raised to the same power, then the resulting series is also a GP.
The reciprocals of the terms of a GP is also a GP.
If three numbers are in GP, then they can be taken as ,
a
r
a, ar.
If four numbers are in GP, then they can be taken as
3
3
, , , .
a a
ar ar
rr
If five numbers are in GP, then they can be taken as
2
2
, , , , .
a a
a ar ar
rr
Harmonic Progression (HP)
If the reciprocals of the terms of a series form an arithmetic progression, then the series is called a harmonic
progression.
(i) The sequence
4 3 12
, , ,
3 2 7
¼ is a harmonic progression as
3 2 7
, ,
4 3 12
is in arithmetic progression.
If a, b, c are in harmonic progression, then b =
2ac
a c+
where b is the harmonic mean.

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Sum of Natural Series
The sum of the first n natural numbers =
( )1
2
n n +
The sum of the square of the first n natural numbers =
( )( )1 2 1
6
n n n+ +
The sum of the cubes of the first n natural numbers =
( )22
1
4
n n +
The sum of first n even numbers = n(n + 1)
The sum of first n odd numbers = n2
Example 1: Find the nth term and the fifteenth term of the arithmetic progression 3, 9, 15, 21
Solution. In the given AP we have a = 3, d = (9 3) = 6
Tn
= a + (n 1)d = 3 + (n 1)6 = 6n 3
T15
= (6 15 3) = 87
Example 2: Find the 10th term of the AP 13, 8, 3, 2,
Solution. In the given AP, we have a = 13, d = (8 13) = 5
Tn
= a + (n 1)d = 13 + (n 1)( 5) = 18 5n
T10
= 18 5 (10) = 32
Example 3: The first term of an AP is -1 and the common difference is -3, the 12th term is
Solution. T1
= a = 1, d = 3
Tn
= a + (n 1)d = 1 + (n 1)( 3) = 2 3n
T12
= 2 3 12 = 34
Example 4: Which term of the AP 10, 8, 6, 4 is 28?
Solution. We have, a = 10,d = (8 10) = 2, Tn
= 28
Tn
= a + (n 1)d 28 = 10 + (n 1)( 2) = n = 20
Example 5: The 8th term of an AP is 17 and the 19th term is 39. Find the 20th term.
Solution. T8
= a + 7d =17 ...(i)
T19
= a + 18d = 39 ... (ii)
On subtracting Eq. (i) from Eq. (ii), we get 11d = 22 d = 2
Putting d = 2 in Eq. (i), we get a + 7(2) = 17 = a = (17 14) = 3
First term = 3, Common difference = 2
T20
= a + 19d = 3 + 19(2) = 41
Example 6 Find the sum of the first 20 terms of the AP, 2, 1, 4, 7,
Solution. Here, a = 5, d = ( 1 2) = 3 and n = 20
Sn
= ( )2 1
2
n
a n dé ù+ -ë û Þ ( )( )20
20
20 5 20 1 3 470
2
S é ù= ´ + - - = -ë û
Example 7: Find the sum of the series 5, 10, 15, 20, 125.
Solution. Here, a = 5, Tn
= 125, d = (10 5) = 5
Tn
= a + (n 1)d 125 = 5 + (n 1) 5 n = 25
Sn
= [ ] [ ]
25
5 125 1625
2 2
n
a l+ = + =

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Example 8: Find three numbers in AP whose sum is 36 and product is 1620.
Solution. Let the numbers be (a d), a, (a + d). Then, (a d) + a + (a + d) = 36 Þ 3a = 36 Þ a = 12
(a d) a (a + d) = 1620 (12 d) l2 (12 + d) = 1620
(144 d2
) 135 Þ d2
9 Þ d = 3
Numbers are 9, 12, 15 or 15, 12, 9.
Example 9: Find the nth term and 8th term of the GP 3, 6, 12, 24, 48,
Solution. In the given GP, we have a = 3, r =
6
2
3
=
-
Tn
= arn 1
= ( 3)( 2)n 1
T8
= ( 3)( 2)8 1
= ( 3)( 2)7
= 384
Example 10: The nth
term of GP is 3/2n
. Find the ratio of 5th to 10th term.
Solution. In the given GP, we have
3
2
n n
T =
5
5 10 5 105 10 5 10 5
3 3 3 3 1
, , : : 1 : 2 :1 32 :1
2 2 2 2 2
T T T T= = = = = =
Example 11: Determine the 9th term of GP whose 8th term is 192 and common ratio 2.
Solution. In the given GP, we have r = 2, T8
= ar7
= 192
Þ a(2)7
= 192 Þ a =
192 3
128 2
= T9
= ar8
= ( ) ( )8 73
2 3 2 384
2
= =
Example 12: The first term of a GP is 50 and the 4th term is 1350. Determine the 6th term.
Solution. Let a be first term and r be the common ratio.
Then, a = 50 ...(i)
T4
= ar3
= 1350 ...(ii)
On dividing Eq. (ii) by Eq. (i), we get r3
= 27 Þ r = 3
Example 13: Find the sum to infinity for the GP
1 1 1
, , ,
4 16 64
- - ¼
Solution. In the given GP, we have a =
1 1 /16 1
, ,
4 1/ 4 4
r
-
- = = - ¼
1 1
1 4 14 4
51 4 5 51
1
44
a
S
r¥
- -
= = = = - ´ = -
- é ùæ ö
- -ç ÷ê ú
è øë û
Example 14: In a certain colony of cancevous cells, each cell divides into two every minute. How many cells
will be produced from a single cell, if the rate of division continues for 12 min?
Solution. Total number of cells = 2 + (22
+ 23
+24
+...+212
)
= 21
+ 22
+ 23
+....+ 212
=
( )12
2 2 1
8190
2 1
-
=
-

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1. If five times the fifth term of an AP is equal to
seven times, the seventh term of the AP, then
what is the twelfth term?
(a) 1 (b) 0
(c) 1 (d) 2
2. Three terms are in AP such that their sum is 18.
The sum of the first two terms is six more than
the sum of the last two terms. Find the last term.
(a) 6 (b) 9
(c) 3 (d) 2
3. Determine k, so that (k + 2), (4k 6) and (3k 2)
are three consecutive terms of an AP.
(a) 3 (b) 2
(c) 4 (d) 6
4. In an AP, the first term is 2 and the sum of the
first five terms is one-fourth the sum of the next
five terms. Find the second term.
(a) 4 (b) 10
(c) 16 (d) 12
5. The sum of four terms in an AP is 64. The product
of the extreme terms is 220. Find the first and
fourth term.
(a) 14, 28 (b) 10, 22
(c) 28, 14 (d) 6, 30
ANSWERS
1. (b) 2. (c) 3. (a) 4. (a) 5. (b)
EXPLANATIONS
1. T5
= a + 4d, t7
= a + 6d
5 (a + 4d) = 7 (a + 6d) Þ 5a + 20d = 7a + 42d
Þ a = 11d
T12
= a + 11d = 11d + 11d = 0
The twelfth term is 0.
2. Let the three terms be a d, a, a + d.
(a d) + a + (a + d) = 18
EXER CISE
Þ 3a = 18
Þ a = 6
Also, [(a d) + a] [a + (a + d)] = 6 Þ d = 3
So, the three terms are 9, 6, 3 respectively, last
term = 3.
3. Since, (k + 2), (4k 6) and (3k 2) are in AP.
(4k 6) (k + 2) = (3k 2) (4k 6)
Þ 4k 6 k 2 = 3k 2 4k + 6 Þ k = 3
4. Given, a = 2, Sn
=
2
n
[2a + (n 1) d]
S5
=
5
2
[2 2 + (5 1) d] = 10 (d + 1) ...(i)
S10
=
10
2
[2 2 + (10 1) d] = 5 (9d + 4) ...(ii)
Sum of the sixth to tenth term is
S10
S5
= (35d + 10) ...(iii)
Now, given S5
=
1
4
(S10
S5
)
Þ 10(d + 1) =
1
4
(35d + 10)
Þ d = 6
T2
= a + d = 2 + ( 6) = 4
5. Let the four terms be
(a 3d), (a d), (a + d), (a + 3d)
(a 3d) + (a d) + (a + d) + (a + 3d) = 64
4a = 64 Þ a = 16 ...(i)
Also, (a 3d) (a + 3d) = 220
Þ a2
9d2
= 220
Þ (16)2
9d2
= 220 Þ d = 2
First term = a 3d = (16 3 2) = 10
Fourth term = a + 3d = (16 + 3 2) = 22

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Relation between Time, Speed and Distance
Distance covered, time and speed are related by
Time =
Distance
Speed
...(i)
Speed =
Distance
Time
...(ii)
Distance = Speed Time ...(iii)
Distance is measured in metres, kilometres and miles.
Time in hours, minutes and seconds.
Speed in km/h, miles/h and m/s.
1. To convert speed of an object from km/h to m/s multiply the speed by
5
18
.
2. To convert speed of an object from m/s to km/h, multiply the speed by
18
5
.
Average Speed
It is the ratio of total distance covered to total time of journey.
Average speed =
Total distance covered
Total time of journey
General Rules for Solving Time & Distance Problems
Rule 1
If a certain distance is covered with a speed of x km/h and another equal distance with a speed of y km/h,
then the average speed for the whole journey is the harmonic mean of the two speeds.
Average speed =
2
km/h
1 1
x y
æ ö
ç ÷
ç ÷
ç ÷+ç ÷
è ø
=
2
km/h
xy
x y
æ ö
ç ÷
+è ø
Time and Distance
CHAPTER 16

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Rule 2
If three equal distances are covered by three different speeds x, y and z km/h, then average speed for the
whole journey is given by
Average speed =
3
km/h
1 1 1
x y z
æ ö
ç ÷
ç ÷
ç ÷+ +ç ÷
è ø
=
3
km/h
xy yz zx
æ ö
ç ÷+ +è ø
Rule 3
If a certain distance is covered with a speed of x km/h and another distance with a speed of y km/h but time
interval for both journeys being same, then average speed for the whole journey is given by
Average Speed = km/h
2
x y+æ ö
ç ÷
è ø
Rule 4
If a certain distance is covered with a speed of x, y and z km/h, but time inverval for the three journey being
equal, then average speed is given by
Average speed = km/h
3
x y z+ +æ ö
ç ÷
è ø
Rule 5
If the ratio of speeds A and B is x : y, then the ratio of times taken by them to cover the same distance is
1 1
: .
x y
Relative Speed
(i) If two bodies are moving in the same direction at x km/h and y km/h, where (x > y), then their relative speed
is given by (x y) km/h.
(ii) If two bodies are moving in opposite direction at x km/h and y km/h, then the their relative speed is given
by (x + y) km/h.
General Rules for Solving Train Problems
Rule 1 Train Vs Stationary Object of no Length
Time taken by a train of length l metre to pass a stationary object such as a pole, standing man or a building
is equal to the time taken by the train to cover l metre.
Speed of the train =
Length of the train
Time taken to cross the stationary object
Rule 2 Train Vs Stationary Object of Certain Length
Time taken by a train of length l metre to pass a stationary object of length a metre such as another
standing train, bridge or railway platform is equal to the time taken by the train to cover (l + a) metre.
Speed of the train =
Length of the train +Lengthof the stationary object
Time taken to cross the stationary object

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Rule 3 Train Vs Moving Object of no Length
Time taken by the train of length l metre to pass a man moving is equal to the time taken by the train to
cover l metre
(i) When the train and man move in the same direction with speeds of x m/s and y m/s. Then,
(x y) =
Length of the train
Time taken to cross each other
(ii) When the train and man move in opposite directions with speeds of x m/s and y m/s. Then,
(x + y) =
Length of the train
Rule 4 Train Vs Moving Object of Certain Length
Time taken by the train of length l metre to pass a moving object of length a metre such as another moving
train is equal to the time taken by the train to cover (l + a) metre.
(i) When the two trains move in the same direction with speeds of x m/s and y m/s, (x > y), then
(x y) =
Length of the train + Length of train two
(ii) When the two tains move in opposite directions with speeds of x m/s and y m/s. Then,
(x + y) =
Length of the train one + Length of train two
Rule 5 Two Moving Train
If two trains start at the same time from points A and B towards each other and after crossing they take a and
b second in reaching B and A respectively. Then, (A s speed) : (B s speed) = :b a .
General Rules for Solving Boats and Streams Problem
Downstream Motion
When an object is moving against (opposite) direction in which the water in the stream is flowing, then the
bject is said to be moving upstream.
Upstream Motion
When an object is moving against (opposite) direction in which the water in the stream is flowing, then the
object is said to be moving upstream.
Motion in Still Water
When an object is moving in water where there is no motion in water, the object can move in any direction
with a uniform speed, then the object is said to be moving in still water.
Rule 1 Downstream and Upstream Speed
Let the speed of the boat in still water = x km/h and speed of the stream be y km/h, then
Speed of the boat with stream downstream speed = (x + y) km/h
Speed of the boat against stream = upstream speed = (x y) km/h
As, when the boat is moving downstream, the speed of the water aids the speed of the boat and when the boat
is moving upstream, the speed of the water reduces the speed of the boat.

at +91 8800734161, 011- 65023618
Rule 2 Speed of Boat in Still Water & Speed of Stream
If the downstream speed of boat is a km/h and the upstream speed of boat is b km/h, then
Speed of boat in still water =
1
( )km/h
2
a b+
Speed of stream =
1
( )km/h
2
a b-
General Rules for Solving Circular Tracks
Rule 1
When two people are running around a Circular Track starting at the same point and at the same time, then
whenever the two people meet the person moving with a greater speed covers one round more than the person
moving with lesser speed.
Rule 2
When two people with speeds of x km/h and y km/h start at the same time and from the same point in the
same direction around a circular track of circumference c km, then
The time taken to meet for the first time anywhere on the track = h
c
x y-
The time taken to meet for the first time at the starting point = LCM of ,
c c
h
x y
æ ö
ç ÷
è ø
Rule 3
When two people with speeds of x km/h and y km/h respectively start at the same time and from the same
point but in opposite direction around a circular track of circumference c km, then
The time taken to meet for the first time anywhere on the track = h
c
x y+
The time taken to meet for the first time at the starting point = LCM of , h
c c
x y
æ ö
ç ÷
è ø
Example 1: Convert 90 km/h into m/s.
Solution. 90 km/h =
5
90 m/s
15
æ ö
´ç ÷
è ø
= 25 m/s
Example 2: Convert 10 m/s into km/h.
Solution. 10 m/s =
18
10 m/s
5
æ ö
´ç ÷
è ø
= 36 km
Example 3: A man can cover a certain distance in 1 h 30 min by covering one-third of the distance at 6 km/
h and the rest at 15 km/h. Find the total distance.
Solution. Let the total distance be x km. Then,
2
3 3
6 15
x x
+ =
3
2
Þ
2
18 45
x x
+ =
3
2
Þ
9
90
x
=
3
2
Þ
10
x
=
3
2

at +91 8800734161, 011- 65023618
Þ x =
(3 10)
2
´
= 15
Total distance = 15 km
Example 4: An aeroplane started one hour later than the scheduled departure from a place 1200 km away
from its destination. To reach the distination on time, the pilot had to increase its speed by 200 km/h. What was
the normal speed of the aeroplane?
Solution. Let the time taken by the aeroplane in second case be x hour. Then,
1200
x
=
1200
200
1x
+
+
Þ
6
x
=
6
1
1x
+
+
Þ 6x + 6 = 6x + x2
+ x Þ x2
+ x 6 = 0
Þ (x + 3) (x 2) = x Þ x = 2h (Q = 3 is not possible)
Time taken in second case = 2 h
So, Speed =
1200
2
= 600 km/h
Hence, normal speed = 600 200 = 400 km/h
Example 5: The current of a stream runs at 1 km/h. A motor boat goes 35 km upstream and back again at
the starting point in 12 h. What is the speed of motor boat in still water?
Solution. Let speed of boat in still water be x km/h. Then,
35 35
1 1x x
+
- +
= 12 Þ 35 (x + 1 + x 1) = 12 (x2
1) Þ 6x2
35x 6 = 0
Þ (x 6) (6x + 1) = 0 Þ x = 6 (Q x = 1/6 is not possible)
So, speed of boat in still water = 6 km/h
Example 6: A train of length 100 m crosses a man who is coming to the train from opposite direction, in 6 s.
What is the speed of train?
Solution. Let speed of train = x km/h
Then, speed of tain relative to man = (x + 5) km/h
=
5
( 5) m/s
18
x + ´

100
5
( 5)
18
x + ´ = 6 Þ
1800
5( 5)x + = 6
Þ x + 5 = 60 Þ x = 55 km/h
Example 7: Speed of three cars are in the ratio 2 : 3 : 4. What is the ratio of time taken by them in covering
the same distance.
Solution. Let speed of three cars be 2x, 3x and 4x km/h. If covered distance be d, then ratio of time taken by
them
= ; ;
2 3 4
d d d
x x x
=
1 1 1
: :
2 3 4
= 6 : 4 : 3
Example 8: A man can row 6 km/h in still waer. When the river is running at 4 km/h, it takes him 2 h 15 min
to row to a place and back. How far is the place?
Solution. Speed downstream = (6 + 4) km/h = 10 km/h

at +91 8800734161, 011- 65023618
Speed upstream = (6 4) km/h = 2 km/h
Let the required distance be x km.
Then,
10 2
x x
+ =
9
4
Þ x =
9 10
4 6
´æ ö
ç ÷´è ø
= 3.75 km
Example 9: Two men A and B start together from the same point to walk round a circular path 8 km long. A
walks 2 km and B walks 4 km an hour. When will they next meet at the starting point, if they walks in the same
direction?
Solution. Time to complete one revolution by A and B is
8
h
2
æ ö
ç ÷
è ø
and
8
h
4
æ ö
ç ÷
è ø
or 4 h and 2h.
The required time is the LCM of 4 and 2 which is 4 h.
Thus, they will next time at the starting point after 4 h.
EXER CISE
(c)
3
5
6
km (d)
7
15
8
km
5. A, B and C are on a trip by a car. A drives during
the first hour at an average speed of 50 km/hr. B
drives during the next 2 hours at an average
speed of 48 km/hr. C drives for the next 3 hours
at an average speed of 52 km/hr. They reached
their destination after exactly 6 hours. Their mean
speed was:
(a) 50 km/hr (b)
1
50
3
km/hr
(c) 51 km/hr (d) 52 km/hr
ANSWERS
1. (d) 2. (c) 3. (d) 4. (a) 5. (b)
EXPLANATIONS
1. Speed from A to B =
2
250
11
æ ö
´ç ÷è ø
mph =
500
11
æ ö
ç ÷è ø mph.
Speed from B to A =
2
250
9
æ ö
´ç ÷è ø
mph =
500
9
æ ö
ç ÷è ø
mph.
Average speed =
500 500
2
11 9
500 500
11 9
æ ö
´ ´
ç ÷
ç ÷
+ç ÷
è ø
mph
1. Mac travels from A to B a distance of 250 miles in
5 hours. He returns to A in 4 hours 30 minutes.
His average speed is:
(a) 42 mph (b) 49 mph
(c) 48 mph (d) 50 mph
2. A boy goes to his school from his house at a speed
of 3 km/hr and returns at a speed of 2 km/hr. If
he takes 5 hours in going and coming, the distance
between his house and school is:
(a) 8.5 km (b) 5.5 km
(c) 6 km (d) 9 km
3. The average speed of a train in the onward
journey is 25% more than that in the return
journey. The train halts for one hour on reaching
the destination. The total time taken for the
complete to and for journey is 17 hours, covering
a distance of 800 km. The speed of the train in
the onward journey is:
(a) 50 km/hr (b) 53 km/hr
(c) 52 km/hr (d) 56.25 km/hr
4. I started on my bicycle at 7 a.m. to reach a certain
place. After going a certain distance, my bicycle
went out of order. Consequently, I rested for 35
minutes and came back to my house walking all
the way. I reached my house at 1 p.m. If my
cycling speed is 10 kmph and my walking speed
is 1 kmph, then on my bicycle I covered a distance
of:
(a)
61
4
66
km (b)
4
13
3
km

at +91 8800734161, 011- 65023618
=
500000
4500 5500
æ ö
ç ÷è ø+
mph = 50 mph.
2. Average speed =
2 3 2
3 2
´ ´æ ö
ç ÷è ø+
km/hr =
12
5
km/hr.
Distance travelled =
12
5
5
æ ö
´ç ÷è ø km = 12 km.
Distance between house and school
=
12
2
æ ö
ç ÷è ø = 6 km.
3. Let the speed in return journey be x km/hr.
Then, speed in onward journey =
125
100
x
=
5
4
x
æ ö
ç ÷è ø
km/hr.
Average speed =
5
2
4
5
4
x x
x x
æ ö
´ ´
ç ÷
ç ÷
+ç ÷
è ø
km/hr
=
10
9
x
km/hr.

9
800
10x
æ ö
´ç ÷è ø = 16
Û x =
800 9
16 10
´æ ö
ç ÷è ø´ =
7200
160
= 45.
So, speed in onward journey
=
5
45
4
æ ö
´ç ÷è ø
km/hr.=
225
4
= 56.25 km/hr.
4. Time taken = 5 hrs 25 min =
65
12
hrs.
Let the required distance be x km.
Then,
10 1
x x
+ =
65
12
Û 11x=
650
12
Û x =
325
66
=
61
4
66
km.
5. Total distance traveled
= (50 1 + 48 2 + 52 3) km
= (50 + 96 + 156) km = 302 km.
Total time taken = 6 hrs.
Mean speed =
302
6
æ ö
ç ÷è ø km/hr
= 50
1
3
km/hr.

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SAMPLE
MATERIAL OF
OUR STUDY KIT
PAPER 2
Comprehension
&
English Language

at +91 8800734161, 011- 65023618
Directions (Q.1 11): Read the following passage
carefully and answer the questions given below
it. Certain words are printed in bold in the passage
to help you locate them while answering some of
the questions.
What is immediately needed today is the
establishement of a Wrold Government or an
International Federation of mankind. It is the utmost
necessity of the world today, and all those persons who
wish to see all human beings happy and prosperous
naturally feel it keenly.
Of course, at times we feel that many of the
problem of our political, social, linguistic and cultural
life would come to an end if there were one Govenment
all ovet the world. Travellers, businessmen, seekers of
knowledge and teachersof righteousness know very well
that great impediments and obstructions are faced by
them when they pass from one country to another,
exchange goods, get information, and make an efforts
to spread their good gospel among their fellow-men. In
the past, religious sects divided one set of people against
another, colour of skin or shape of the body set one
against the other.
But today when philosophical light has exploded
the darkness that was created by religious differences,
and when scientific knowledge has flasified the
superstitions, they have enabled human beings of all
religious views and of all races and colours to come in
frequent contact with one another . It is the governments
of various countries that keep poeple of one country
apart from, those of another. They create artificial
barriers, unnatural distinctions, unhealthy isolation,
unnecessary fears and dangers in the minds of common
men who by their nature want to live in friendship with
their fellow-men. But all these evils would cease to exist
if there were one Government all over the world.
1. What divides people of a country against another?
(a) Different religions
(b) Different language
(c) Different social and political systems of
different people
(d) Government of various countries
2. What is the urgent heed of the world today?
(a) The establishment of an international
economic order.
(b) The establishment of a world government.
(c) The creation of a cultural international social
order.
(d) The raising of an international spiritual army.
3. What will the world Government be expected to
do?
(a) it will arrange for interplanetary contacts
(b) it will end all wars for all time to come
(c) it will bring about a moral regeneration of
mankind
(d) it will kill the evil spirit in man
4. Choose the word which is SIMILAR in meaning
as the word "righteousness" as used in the
passage.
(a) rectitude (b) religiosity
(c) requirement (d) scrupulousness
EXERCISE 1
Comprehension

at +91 8800734161, 011- 65023618
5. Which of the following problems has not been
mentioned in the passage as likely to be solved
with the establishment of world Government?
(a) Social Problems (b) PoliticalProblems
(c) Cultural Problems (d) Economic Problems
6. Choose the word which is most OPPOSITE in
meaning of the word 'implediments' as used in
the passage.
(a) handicaps (b) furtherance
(c) providence (d) hindrances
7. The most appropriate title of the above passage
may be ...........
(a) The evils of the world order.
(b) The man can make his destiny.
(c) The need of world Government.
(d) The role of Religion in the Modern Times.
8. What was the factor, according to the passage,
that set one man against another?
(a) Material prosperity of certain people in the
midst of grinding poverty.
(b) Superior physical strength of some persons.
(c) Colour of skin or shape of the body.
(d) Some people being educated and other
illiterate.
9. The theory of racial superiority stands falsified
today by ........
(a) knowledge derived from scientific advances.
(b) the ascendancy of people who were here to fore
considered of inferior racial stock.
(c) the achievements of the so called backward
countries in every field of life.
(d) the precedence given to the physical powers
of different races.
10. In the part religious sects .............
(a) united the people with one another.
(b) Interfered in political affairs.
(c) did a good job by way of spreading message of
love and peace.
(d) divided one set of people from another.
Directions (Q. 12-18): Read the following passage
carefully and answer the question given below it.
Certain words are printed in bold to help you to
locate them while answering some of the
questions.
The window offered a view of the house opposite.
The two families did not speak to each other because of
a property dispute. One day, Ruchira's textbooks lay
untouched as the young girl's gaze was on the
happenings in the house opposite. There were two new
faces in the neighbouring household thatof an elderly
widow and a girls aged sixteen. Some times the elderly
lady would sit by the window, doing the young girl's
hair. On other days she was absent.
The new young neighbour's daily routine could be
seen through the window she cleaned the rice paddy;
split nuts, put the cushions in the sun to air them. In
the afternoons while the men were all at world some of
the women slept and others played cards. The girl sat
on the terrace and read. Sometimes she wrote. One day
there was hindrance. She was writing when the elderly
woman snatched the unfinished letter from her hands.
Thereafter the girl was not to be seen on the terrace.
Sometimes during the day sounds came from the house
indicating that a massive argument was going on inside.
A few days passed. One evening Ruchira noticed
the girl standing on the terrace in tears. The evening
prayer was in progress. As she did daily, the girl bowed
several times in prayer. Then she wentdownstairs. That
night Ruchira wrote a letter. She went out and posted
it that very instant. But as she lay in bed that night,
sheprayedfervently that her offer of friendship wouldn't
reach its destination. Ruchira then left for Madhupur
and returned when it was time for college to start. She
found the house opposite in darkness, locked. They had
left.
When she stepped into her room she found the desk
piled with letters one had a local stamp on it with her
name and address in unfamiliar handwriting. She
quickly read it. They continued to write to each other
for the next twenty years.
11. Why did Ruchira write a letter to her new
neighbour?
(a) She wanted to offer her, her help.
(b) She wanted to be friends with her.
(c) To apologize for her family's behaviour
towards her family.
(d) To encourage her to continue learning to read
and write.
12. Which of the following can be said about Ruchira?
A. She used to spy on her neighbours because
she didn't trust them.
B. She was at home because she was studying.
C. She did not speak to her neighbours because
they did not own property.
(a) None (b) Only B
(c) Both A & B (d) Only C

at +91 8800734161, 011- 65023618
13. How did the new young neighbour spend her
days?
(a) She was busy writing letters to Ruchira.
(b) She used to daydream about her past
experiences.
(c) She would attend to the needs of the widow.
(d) She spent her time learning to read and write.
14. Why was the young neighbour prevented from
sitting on the terrace?
(a) She used to while away her time instead of
working
(b) The old woman could no longer keep an eye
on her.
(c) She had not finished writing the letter she was
asked to.
(d) She had been writing a letter which she wasn't
supposed to.
15. What was the major argumentin the house about?
(a) There were too many people living there,
which resulted in arguments.
(b) The young girl was insisting on attending
college.
(c) The young girl had been wasting her time
instead of working.
(d) The old woman did not guard the young girl
closely.
16. Which of the following is TRUE in the context of
the passage?
(a) The young girl was very devout and prayed
everyday.
(b) Only two letters were exchanged between the
two girls.
(c) The new young neighbour was a servant.
(d) The afternoon was a time to relax for everyone.
17. Why did the young girl wish that the letter would
not reach its destination?
A. She was going away and would not be able to
see if her neighbour was glad to receive it.
B. She was afraid that it would lead to a quarrel
between the two families.
C. She was afraid that her neighbour would be
angry when she received her letter.
(a) None
(b) Only A
(c) Only C
(d) Both B & C
Directions (Q.18-20): Choose the word which is
most nearly the SAME in meaning as the world
printed in bold as used in the passage.
18. Hindrance
(a) handicapped (b) delay
(c) interruption (d) difficult
19. Offered
(a) forward (b) willing
(c) volunteered (d) provided
meaning of the word piled as used in the passage.
(a) low (b) empty
(c) blank (d) nothing
Directions (Q. 21-27): Read the following passage
it. Certain words are printed in bold to help you to
locate them while answering some of the
questions.
The yearly festival was close at hand. The store
room was packed with silk fabrics, gold ornaments, clay
bowls full of sweet curd and platefuls of sweetmeats.
The orders had been placed with shops well in advance.
The mother was sending out gifts to everyone.
The eldest son, a government servant, lived with
his wife and children in far off lands. The second son
hadleft homeatan early age. Asa merchant he travelled
all over the world. The other sons had split up over petty
squabbles, and they now lived in homes of their own.
The relatives were spread all across the world. They
rarely visited. The youngest son, left in the company of
a servant, was soon bored left her and stood at the door
all day long, waiting and watching. His mother, thrilled
and excited, loaded the presents on trays and plates,
covered them with colourful kerchiefs, and sent them
off with maids and servants. The neighbours looked on.
The day came to an end. All the presents had been
sent off.
The child came back into the house and dejectedly
said to his mother, "Maa, you gave present to everyone,
but you didn't give me anything !"
His mother laughed, "I have given all the gifts
away to everyone, now see what's left for you." She
kissed him on the forehead.
The child said in a tearful voice, "Don't I get a gift?"
"You'll get it when you go far away."
"But when I am close to you, don't I get something
from your own hands?"
His mother reached out her arms and drew him to
her "This is all I have in my own hands. It is the most
precious of all."

at +91 8800734161, 011- 65023618
21. Why did the woman's second son travel?
(a) He was restless by nature.
(b) He did not want to stay at home.
(c) He was rich and could afford to travel.
(d) His job prevented him from taking leave
22. Why did the woman's eldest son not attend the
festival?
(a) He was not on good terms with his youngest
brother who lived at home.
(b) He had quarrelled with his mother.
(c) His wife did not allow him to return home.
(d) His job prevented him from taking leave
23. How did the woman prepare for the fesitval?
A. She bought expensive gifts for her children
and neighbours.
B. She ordered her servents to prepare sweets
and food well in advance.
C. She made sure that her youngest child was
looked after so that he wouldn't be bored.
(a) None (b) Only A
(c) Only B (d) Both A & B
24. What did the youngest child do while his mother
was busy?
(a) He waited for a chance to steal some
sweetmeats
(b) He pestered his mother to give him a present.
(c) He stood at the door with the servants.
(a) Only A (b) Only B
(c) Both A & C (d) Only C
25. Which of the following can be said about the
woman?
(a) She was a widow who had brought up her
childern single handedly.
(b) She was not a good mother since her childern
had left home at an early age.
(c) She enjoyed sending her family gifts at festival
time.
(d) She gave expensive presents to show that she
was wealthy.
26. What did the boy receive from his mother?
(a) She taught him the value of patience
(b) She encouraged him to grow up and live
independently like his brother.
(c) She showed him the importance of giving
expensive gifts.
(d) She gave him a hug to express her love.
27. Which of the following is TRUE in the context of
the passage?
(a) The woman usually ignored her youngest son.
(b) The woman's eldest son lived abroad.
(c) Thememers of the woman's family didnot care
about her.
(d) The woman made all the preparations herself
since she did notwant to burden the servants.
Directions (Q. 28 30): Choose the word which is
most nearly the SAME in meaning as the word
printed in bold as used in the passage.
28. Left
(a) gone (b) quit
(c) remaining (d) disappeared
29. Packed
(a) filled (b) squeezed
(c) crowd (d) collected
meaning of the word dejectedly as used in the
passage.
(a) calmly (b) happily
(c) willingly (d) fortunately
Direction (31-36): Read the following two passages
them. Certain words/expressions are given in bold
in Passage I to help you locate them while
answering some of the questions.
Ghanshyam Das Birla was a great architect of
India's industrial growth. He started his career in
Kolkata at the beginning of 20th century. He set up
many industries. He entered the field of business during
the days of the First World War and established himself
after the war years. First, he established a cotton mill
in Sabzi Mandi, Delhi, followed by Keshoram Cotton
Mills and Birla Jute Mills around 1920. The Keshoram
Mills were set up with the efforts of Andrew Yule. In
1919, with an investment of Rs 50 lakhs, Birla Brothers
Limited was formed and thereafter a mill was set up in
Gwalior.
Mr. Birlarealisedthatpolitical freedom fromBritish
rule was imperative for the industrial growth of India.
In 1920, he came into contactwith Gandhiji and became
his disciple. In the decade of the 30's he set up sugar
paper mills. From 1943 to 1946, with the stock exchange
gaining ground, Birla Brothers ventured into the
areas of cars, cotton, machinery and man-made fabrics.
United Commercial Bank was set up during this period.
Prior to this, he had established Ruby, Asiatic Insurance
Co. and Inland Air Service.
After independence, the Birlas expanded their
business and started production in many fields. Near

at +91 8800734161, 011- 65023618
Mirzapur, he, in collaboration with Caesar, an American
friend, set up an aluminium plant Hindalco in record
time. He bought the Century Mill from Sir Chunnilal
V.Mehta, the cousin of Sir Purshottam Das Thakur. He
also boughttea estates and started cement and fertiliser
factories. He established a new style of management.
In his birth place Pilani, and at many other places he
started many educational institutions. To his credit go
many temples, planetariums and hospitals. In 1983, he
died while in London, but not before seeing his business
flourish as one of the topmostestablishments in India.
During the decades of 70's and 80's, Birla Brothers was
among the topmost Industrial Houses in India.
Anna Saheb Karve's life was an simple and clean
as that of an ancient ascetic (Rishi Muni). He was liberal
in his views and sympathetic in his attitude. He was
deeply moved by the troubles and tribulations of women
flok.
What Anna Saheb did for the upliftment of women
in the last decade of the nineteenth century was beyond
the imagination of the people. He created an awakening
among women through his writing in Kesari. He was a
staunch supporter of widow remarriage. On March 11,
1893, he set an example for society by taking the bold
step of marrying a widow. The couple dedicated their
lives to social work and reform. He established an
orphanage in Pune.
Anna Saheb realised the importance of education
for women. He spared no effort in starting the Bharatiya
Mahila Vidyapeeth, the first university for women. In
1921, he travelled to Europe and America and met
famous people like Albert Einstein and formulated his
opinions regarding work. In 1958, he was awarded the
'Bharat Ratna' in recognition of his services in
educational and social reforms.
31. Ghanshyam Das Birla was better known to India
as
(a) a freedom fighter against British rule
(b) an architect of temples in various cities
(c) a pioneer of India's industrial growth
(d) an architect of the textile industry in India
32. Ghanshyam Das Birla's first industrial venture
was
(a) a textile related industry
(b) a cotton mill, Gwalior
(c) Keshoram Cotton Mills, Delhi
(d) car manufacturing company
33. What was Andrew Yule's role in Ghanshyam Das
Birla's career?
(a) He helped Ghanshyam Das Birla in setting
up cotton mills in Sabzi Mandi, Delhi.
(b) He invested Rs. 50 lakhs in Keshoram Cotton
Mills
(c) He extended a lot of help to Ghanshyam Das
Birla in his second venture.
(d) He helped Ghanshyam Das Birla in the field
of business before the First World War.
34. What was Ghanshyam Das Birla's view about the
British rule in India?
(a) The British rule was very atrocious.
(b) The British rule was counter productive to
India's industrial growth.
(c) The British rule was not a hindrance to India's
industrial growth.
(d) The industrial growth of India had nothing to
do with gaining freedom from the British.
35. Which of the following can be inferred from the
passage ?
A. Ghanshyam Das Birla could set up a large
number of diverse industriesdue toBritish rule
in India.
B. Ghanshyam Das Birla, with his family
members, started various industries due to
flourishing of the Stock Exchange.
C. Ghanshyam Das Birla was impressed by
Gandhiji's philosopy.
(a) Both A and B (b) Both A and C
(c) All the three (d) Only B and C
36. Which of the following statements is FALSE in
the context of the passage?
A. Ghanshyam Das Birla unfortunately died
before realising the progress his business had
made.
B. Hindalco was set up in a very short time span.
C. Ghanshyam Das Birla's management style
was different from the traditional one.
(a) Only A (b) Only B
(c) Only C (d) Both A and B
Directions (Q. 37-38): Choose the word/group of
words which is/are most nearly the SAME in
meaning to the word/group of words given in bold
as used in the passage.
37. Imperative
(a) trial (b) dispensable
(c) inadequate (d) unavoidable
38. Gaining Ground
(a) obtaining land (b) making advances
(c) losing heavily (d) grounding due to losses

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