Chapter 1

REAL NUMBER SYSTEM
No World without Water
No Mathematics without Numbers

1.1 Introduction
In the development of science, we should know about the properties and
operations on numbers which are very important in our daily life. In the earlier classes
we have studied about the whole numbers and the fundamental operations on them.
Now, we extend our study to the integers, rationals, decimals, fractions and powers in
this chapter.

Numbers
In real life, we use Hindu Arabic numerals - a system which consists of the
symbols 0 to 9. This system of reading and writing numerals is called, “Base ten
system” or “Decimal number system”.

1.2 Revision
In VI standard, we have studied about Natural numbers, Whole numbers,
Fractions and Decimals. We also studied two fundamental operations addition and
subtraction on them. We shall revise them here.

Natural Numbers
Counting numbers are called natural numbers. These numbers start with
smallest number 1 and go without end. The set of all natural numbers is denoted by
the symbol ‘N’.
N = "1, 2, 3, 4, 5, ... , is the set of all natural numbers.

Whole numbers
Natural numbers together with zero (0) are called whole numbers. These
numbers start with smallest number 0 and go without end. The set of all whole
numbers is denoted by the symbol ‘W’.
W = "0, 1, 2, 3, 4, 5, ... , is the set of all whole numbers.
2

Real Number System
Integers
The whole numbers and negative numbers together
are called integers. The set of all integers is denoted by Z.
Z = "... - 2, - 1, 0, 1, 2, ... , is the set of all integers
(or) Z = "0, ! 1, ! 2, ... , is the set of all Integers.

Ramanujan, the greatest
Mathematician was born
at Erode in Tamil Nadu.

1.3 Four Fundamental Operations on Integers
(i) Addition of Integers
Sum of two integers is again an integer.
For example,
i)

10 + ^- 4h = 10 - 4 = 6

ii)

8 + 4 = 12

iii)

6+0 = 6

iv)

6 + 5 = 11

v)

4+0=4

(ii) Subtraction of integers
To subtract an integer from another integer, add the additive inverse of the
second number to the first number.
For example,
i)

5 – 3 = 5 + (addditive inverse of 3) = 5 + (– 3) = 2.

ii)

6 – (– 2) = 6 + (addditive inverse of (– 2)) = 6 + 2 = 8.

iii)

(– 8) – (5) = (– 8) + (– 5) = – 13.

iv)

(– 20) – (– 6) = – 20 + 6 = – 14.

(iii) Multiplication of integers
In the previous class, we have learnt that multiplication is repeated addition in
the set of whole numbers. Let us learn about it now in the set of integers.
Rules :
1. The product of two positive integers is a positive integer.
2. The product of two negative integers is a positive integer.
3. The product of a positive integer and a negative integer is a negative
integer.
3

Chapter 1
Example
i)
ii)

5 # 8 = 40
1) 0
2) 9
3) – 5
4) – 11

^- 5h # ^- 9h = 45

iii)

^- 15h # 3 =-^15 # 3h =- 45

iv)

12 # ^- 4h =-^12 # 4h =- 48

× (– 10)
× (– 7)
× (– 10)
×6

=
=
=
=

Activity
Draw a straight line on the ground. Mark the middle point of the line as ‘0’
(Zero). Stand on the zero. Now jump one step to the right on the line. Mark it as + 1.
From there jump one more step in the same direction and mark it as + 2. Continue
jumping one step at a time and mark each step (as + 3, + 4, + 5, ...). Now come back to
zero position on the line. Move one step to the left of ‘0’ and mark it as – 1. Continue
jumping one step at a time in the same direction and mark the steps as – 2, – 3, – 4,
and so on. The number line is ready. Play the game of numbers as indicated below.
i) Stand on the zero of the number line facing right side of 0. Jumping two
steps at a time. If you continue jumping like this 3 times, how far are you
from ‘0’ on number line?
ii) Stand on the zero of number line facing left side of 0. Jump 3 steps at a
time. If you continue jumping like this 3 times, how far are you from ‘0’
on the number line?
Activity
×
–6
–5
3

4
– 24

–6

–3

2

7

15

– 40
21

Example 1.1
Multiply (– 11) and (– 10).
Solution
– 11 × (– 10) = (11 × 10) = 110
Example 1.2
Multiply (– 14) and 9.
Solution
(– 14) × 9 = – (14 × 9) = – 126
4

8

Real Number System
Example 1.3
Multiplication of integers through
number patterns
Multiplying a negative integer by
another negative integer :
Eg. To explain (-2) × (-2) = 4
through number pattern.
Activity :
(+2) × (+1) = 2 ( Reduce the multiplicand each time by one)

Find the value of 15 × 18
Solution
15 × 18 = 270
Example 1.4
The cost of a television set is `5200.
Find the cost of 25 television sets.

(+1) × (+1) = 1
(0) × (+1) = 0
(–1) × (+1) = –1
(–2) × (+1) = –2

Solution
The cost one television set = `5200

Reduce the multiplier each time
by one (–2) × (0) = 0

` The cost of 25 television set = 5200 × 25

= `130000

(–2) × (–1) = 2
(–2) × (–2) = 4

Exercise 1.1
1. Choose the best answer:
		 i) The value of multiplying zero with any other integer is a
			 (A) positive integer
(B) negative integer (C) 1
(D) 0
2
		 ii) – 15 is equal to
			 (A) 225
(B) – 225
		 iii) – 15 × (– 9) × 0 is equal to

(C) 325

			 (A) – 15
(B) – 9
(C) 0
		 iv) The product of any two negative integers is a
			 (A) negative integer
(B) positive integer
			 (C) natural number
(D) whole number
2. Fill in the blanks:
		
		
		
		

i)
ii)
iii)
iv)

(D) 425
(D) 7

The product of a negative integer and zero is _________.
_________ × ^- 14h = 70
^- 72h # _________ = - 360
0 # ^- 17h = _________.

3. Evaluate:
		 i) 3 # ^- 2h
ii) ^- 1h # 25
iii) ^- 21h # ^- 31h
		iv) ^- 316h # 1
v) (– 16) × 0 × (– 18) vi) ^- 12h # ^- 11h # 10
vii) ^- 5h # ^- 5h
viii) 5 # 5
ix) ^- 3h # ^- 7h # ^- 2h # ^- 1h
		 x) ^- 1h # ^- 2h # ^- 3h # 4 			
xi) 7 # ^- 5h # ^9h # ^- 6h
		xii) 7 # 9 # 6 # ^- 5h
		
xiii) 10 × 16 × (– 9)
		xiv) 16 × (– 8) × (– 2)			
xv) (– 20) × (– 12) × 25
		xvi) 9 × 6 × (– 10) × (– 20)
5

Chapter 1
4. Multiply
i)
ii)
iii)
iv)
v)
vi)
5.

^- 9h and 15
^- 4h and ^- 4h

13 and 14
^- 25h with 32
^- 1h with ^- 1h
^- 100h with 0
The cost of one pen is `15. What is the cost of 43 pens?

6. A question paper contains 20 questions and each question carries 5 marks. If a
student answered 15 questions correctly, find his mark?
7. Revathi earns ` 150 every day. How much money will she have in 10 days?
8. The cost of one apple is `20. Find the cost of 12 apples?

(iv) Division of integers
We know that division is the inverse operation of multiplication.
We can state the rules of division as follows:
Positive integer
= Positive number
Positive integer
Negative integer
= Positive number
Negative integer
Positive integer
= Negative number
Negative integer
Negative integer
= Negative number
Positive integer

a) 0 =

10
c) - 3 =
-3

b) 9 =

-3
10
d) -2 =

Division by zero
Division of any number by zero (except 0) is meaningless because division by
zero is not defined.
Example 1.5
Divide 250 by 50.
Solution
Divide 250 by 50 is 250 = 5.
50

6

Real Number System
Example 1.6
Divide (– 144) by 12.
Solution
Divide (– 144) by 12 is - 144 = – 12.
12

Example 1.7
Find the value
Solution

15 # ^- 30h # ^- 60h
.
2 # 10

15 # ^- 30h # ^- 60h
= 27000 = 1350.
2 # 10
20

Example 1.8
A bus covers 200 km in 5 hours. What is the distance covered in 1 hour?
Solution
Distance covered in 5 hours = 200 km.
` Distance covered in 1 hour = 200 = 40 km.
5

Exercise 1.2
1. Choose the best answer:
i) Division of integers is inverse operation of
(A) addition
(B) subtraction
ii) 369 ÷ ............ = 369.

(C) multiplication (D) division

(A) 1
(B) 2
iii) – 206 ÷ ............ = 1.

(C) 369

(D) 769

(A) 1
(B) 206
iv) – 75 ÷ ............ = – 1.

(C) – 206

(D) 7

(C) – 75

(D) 10

(A) 75
2.
i)
iv)
vii)
ix)

(B) – 1

Evaluate
^- 30h ' 6
^- 49h ' 49
6^- 6h + 7 @ ' 6^- 3h + 2 @
67 + 13 @ ' 62 + 8 @

ii) 50 ' 5
iii) ^- 36h ' ^- 9h
v) 12 '6^- 3h + 1 @
vi) 6^- 36h ' 6 @ - 3
viii) 6^- 7h + ^- 19h@ ' 6^- 10h + ^- 3h@
x) [7 + 23] ÷ [2 + 3]

3. Evaluate
i)

^- 1h # ^- 5h # ^- 4h # ^- 6h

2#3

40 # ^- 20h # ^- 12h
ii) 8 # 5 # 4 # 3 # 10 iii)
4#5#6#2

4 # ^- 6h

4. The product of two numbers is 105. One of the number is (– 21). What is the
other number?
7

Chapter 1
Properties of Addition of integers
(i) Closure Property
Observe the following examples:
1.

19 + 23 = 42 			

2.

- 10 + 4 =- 6 				

3.

18 + (- 47) =- 29

In general, for any two integers a and b, a + b is an integer.		
Therefore the set of integers is closed under addition.

(ii) Commutative Property
Two integers can be added in any order. In other words, addition is commutative
for integers.
We have
So,

8 + ^- 3h = 5 and ^- 3h + 8 = 5

8 + ^- 3h = ^- 3h + 8

In general, for any two integers a and b we can say, a + b = b + a
Therefore addition of integers is commutative.
Are the following equal?
i) ^5h + ^- 12h and ^- 12h + ^5h
ii) ^- 20h + 72 and 72 + ^- 20h

(iii) Associative Property
Observe the following example:
Consider the integers 5, – 4 and 7.
Look at

5 + [(– 4) + 7] = 5 + 3 = 8 and
[5 + (– 4)] + 7 = 1 + 7 = 8

Therefore,

5 + [(– 4) + 7] = [5 + (– 4)] + 7

Are the following pairs of
expressions equal?
i) 7 + ^5 + 4h, ^7 + 5h + 4
ii) ^- 5h + 6^- 2h + ^- 4h@,
   6^- 5h + ^- 2h@ + ^- 4h

In general, for any integers a, b and c, we can say, a + ^b + ch = ^a + bh + c .
Therefore addition of integers is associative.
8

Real Number System
(iv) Additive identity
When we add zero to any integer, we get the same integer.
Observe the example: 5 + 0 = 5.
In general, for any integer a, a + 0 = a.
Therefore, zero is the additive identity for integers.

i) 17 + ___ = 17
ii) 0 + ___ = 20
iii) – 53 + ___ = – 53

Properties of subtraction of integers.
(i) Closure Property
Observe the following examples:
i) 5 - 12 =- 7 				
ii) ^- 18h - ^- 13h =- 5
From the above examples it is clear that subtraction of any two integers is again
an integer. In general, for any two integers a and b, a - b is an integer.
Therefore, the set of integers is closed under subtraction.

(ii) Commutative Property
Consider the integers 7 and 4. We see that
7-4 = 3
4 - 7 =- 3
` 7-4 ! 4-7

In general, for any two integers a and b
a-b !b-a

Therefore, we conclude that subtraction is not commutative for integers.

(iii) Associative Property
Consider the integers 7, 4 and 2
7 - (4 - 2) = 7 - 2 = 5
(7 - 4) - 2 = 3 - 2 = 1
` 7 - (4 - 2) ] (7 - 4) - 2

In general, for any three integers a , b and c
a - (b - c) ! (a - b) - c.

Therefore, subtraction of integers is not associative.
9

Chapter 1
Properties of multiplication of integers
(i) Closure property
Observe the following:
– 10 × (– 5) = 50
40 × (– 15) = – 600
In general, a × b is an integer, for all integers a and b.
Therefore, integers are closed under multiplication.

(ii) Commutative property
Observe the following:
Are the following
pairs equal?
5 × (– 6) = (– 6) × 5
i) 5 × (– 7), (– 7) × 5
Therefore, multiplication is commutative for integers. ii) 9 × (– 10), (– 10) × 9
5 × (– 6) = – 30 and (– 6) × 5 = – 30

In general, for any two integers a and b, a × b = b × a.

(iii) Multiplication by Zero
The product of any nonzero integer with zero is zero.
Observe the following:
i)
0 × 0 = _____
ii) – 100 × 0 = _____
0 × x = _____
iii)

5×0 = 0
–8×0 = 0
In general, for any nonzero integer a
a×0 = 0×a=0

(iv) Multiplicative identity
Observe the following:
5#1 = 5
1 # (- 7) = - 7

i) (– 10) × 1 = ___
This shows that ‘1’ is the multiplicative identity for integers. ii) (– 7) × ___ = – 7
iii) ___ × 9 = 9
In general, for any integer a we have
a # 1 =1 # a = a
10

Real Number System
(v) Associative property for Multiplication
Consider the integers 2, – 5, 6.
Look at
62 # ^- 5h@ # 6 = - 10 # 6

=- 60 and
2 # 6^- 5h # 6 @= 2 # ^- 30h
= - 60

Thus 62 # ^- 5h@ # 6 = 2 # 6^- 5h # 6 @
So we can say that integers are associative under multiplication.
In general, for any integers a, b, c, (a × b) × c = a × (b × c).

(vi) Distributive property
Consider the integers 12, 9, 7.
Look at

Are the following equal?

12 # ^9 + 7h = 12 # 16 = 192

1. 4 # ^5 + 6h and ^4 # 5h + ^4 # 6h

^12 # 9h + ^12 # 7h = 108 + 84 = 192

Thus 12 # ^9 + 7h = ^12 # 9h + ^12 # 7h

2. 3 # ^7 - 8h and ^3 # 7h + ^3 # ^- 8hh
3. 4 # ^- 5h and ^- 5h # 4

In general, for any integers a, b, c.
		

a # ^ b + ch = ^ a # bh + ^ a # ch .

Therefore, integers are distributive under multiplication.

Properties of division of integers
(i) Closure property
Observe the following examples:
(i)

15 ' 5 = 3 		

(ii)

^- 3h ' 9 = - 3 = - 1

(iii)

9
7
7 ' 4 = 		
4

3

From the above examples we observe that integers are not closed under
division.

11

Chapter 1
(ii) Commutative Property
Observe the following example:
8 ÷ 4 = 2 and
4÷8= 1
2
` 8÷4! 4÷8
We observe that integers are not commutative under division.

(iii) Associative Property
Observe the following example:
12 ' (6 ' 2) = 12 ' 3 = 4
(12 ' 6) ' 2 = 2 ' 2 = 1
` 12 ' (6 ' 2) ! (12 ' 6) ' 2

From the above example we observe that integers are not associative under
division.
Divide the class into groups each group has to complete the given table
using their own examples and then write true (or) false.
Properties of
Closure Property
Commutative
Associative
Integers
property
property
Addition
Subtraction
Multiplication
Division

1.4 Fractions
Introduction
In the earlier classes we have learnt about fractions which included proper,
improper and mixed fractions as well as their addition and subtraction. Now let us see
multiplication and division of fractions.
Recall :
Proper fraction: A fraction is called a proper fraction if its
Denominator > Numerator.
Example: 3 , 1 , 9 , 5

4 2 10 6

Improper fraction: A fraction is called an improper fraction if its
Numerator > Denominator.
Example : 5 , 6 , 41 , 51
4 5 30 25
Mixed fraction : A fraction consisting of a natural number and a proper fraction
is called a mixed fractions.
Example: 2 3 , 1 4 , 5 1
4

5

7

Think it : Mixed fraction = Natural number+ Proper fraction
12

Real Number System
Discuss : How many numbers are there from 0 to 1.
Recall : Addition and subtraction of fractions.
Example (i)
Simplify: 2 + 3
5

5

Solution
2 + 3 = 2+3 = 5 = 1
5 5
5
5

Example (ii)
Simplify: 2 + 5 + 7
3

12

24

Solution
2 + 5 + 7 = 2#8+5#2+7#1
24
3 12 24
16
10
7
+
+
				
=
24
		
= 33 =1 3
24
8

		

Example (iii)
Simplify: 5 1 + 4 3 + 7 5
4

4

8

Solution
5 1 + 4 3 + 7 5 = 21 + 19 + 61
4
4
8
4
4
8
		 = 42 + 38 + 61 = 141
8
8
		 = 17 5
8

Example (iv)
Simplify: 5 - 2
7

7

Solution
5 - 2 = 5-2 = 3.
7 7
7
7

Example (v)
Simplify: 2 2 - 3 1 + 6 3
3

6

4

Solution
2 2 - 3 1 + 6 3 = 8 - 19 + 27
3
6
4
3
6
4
13

All whole numbers are
fractional numbers with
1 as the denominator.

Chapter 1
= 32 - 38 + 81

12
75
=
=6 1
12
4

(i) Multiplication of a fraction by a whole number

Fig. 1.1

Observe the pictures at the (fig.1.1 ) . Each shaded part is 1 part of a circle. How
8
much will the two shaded parts represent together?
They will represent 1 + 1 = 2 # 1 = 2 = 1
8

8

8

8

4

To multiply a proper or improper fraction with the whole number:
we first multiply the whole number with the numerator of the fraction, keeping
the denominator same. If the product is an improper fraction, convert it as a mixed
fraction.
To multiply a mixed fraction by a whole number, first convert the mixed fraction
to an improper fraction and then multiply.
Therefore, 4 # 3 4 = 4 # 25 = 100 = 14 2
7

7

7

7

Find :
i) 2 # 4

5
iii) 4 # 1
5

Find :
ii) 8 # 4

i) 6 # 7 2

5
iv) 13 # 6
11

3
2
ii) 3 # 7
9

(ii) Fraction as an operator ‘of’
From the figure (fig. 1.2) each shaded portion represents 1 of 1. All the three
3
shaded portions together will represent 1 of 3.
3

14

Real Number System

Fig. 1.2

Combining the 3 shaded portions we get 1.
Thus, one-third of 3 = 1 # 3 = 1.
3

We can observe that ‘of’ represents multiplication.
Prema has 15 chocolates. Sheela has 1 rd of the number of chocolates what
3

Prema has. How many chocolates Sheela has?
As, ‘of’ indicates multiplication, Sheela has 1 # 15 = 5 chocolates.
3

Example 1.9
Find : 1 of 2 1
4

5

Solution
1 of 2 1 = 1 # 2 1
4
4
5
5
1
11
		 = #
4 5
		 = 11
20

Example 1.10
In a group of 60 students 3 of the total number of students like to study
10
Science, 3 of the total number like to study Social Science.
5

(i) How many students like to study Science?
(ii) How many students like to study Social Science?
15

Chapter 1
Solution
Total number of students in the class = 60
(i) Out of 60 students, 3 of the students like to study Science.
10

Thus, the number of students who like to study Science = 3 of 60

10
= 3 # 60 = 18 .
10

		

(ii) Out of 60 students, 3 of the students like to study Social Science.
5

Thus, the number of students who like to study Social Science
= 3 of 60

		

5
= 3 # 60 = 36.
5

		
Exercise 1.3
1. Multiply :
6# 4
5
v) 2 # 7
3
ix) 4 # 14
7

i)

ii) 3 # 3

iii) 4 # 4

iv) 15 # 2

vi) 5 # 8

vii) 11 # 7

viii) 5 # 12

7

2

8

10

4

6

x) 18 # 4
3

2. Find :
1 of 28
2
v) 8 of 216
6
ix) 5 of 35
7

i)

ii) 7 of 27

3
vi) 4 of 32
8
x) 1 of 100
2

iii) 1 of 64

iv) 1 of 125
4
5
vii) 3 of 27 viii) 7 of 100
9
10

3. Multiply and express as a mixed fraction :
5#51
4
v) 7 # 7 1
2

i)

ii) 3 # 6 3

5
vi) 9 # 9 1
2

iii) 8 # 1 1

5

iv) 6 # 10 5

7

4. Vasu and Visu went for a picnic. Their mother gave them a baggage of 10 one
litre water bottles. Vasu consumed 2 of the water Visu consumed the remaining
5
water. How much water did Vasu drink?
16

Real Number System
(iii) Multiplication of a fraction by a fraction
Example 1.11
Find 1 of 3 .
5

8

Solution
1 of 3 = 1 × 3 = 3
5
8
5
8
40

Example 1.12
Find 2 × 3 .
9

2

Solution
2 × 3 = 1
9
2
3

Example 1.13
th
Leela reads 1 of a book in 1 hour. How much of the book will she read in 3 1
4
2
hours?

Solution
The part of the book read by leela in 1 hour = 1

4
1
So, the part of the book read by her in 3 hour = 3 1 # 1
2
2 4
7
						
= #1
2 4
						
= 7#1
4#2
						
=7
8
7
1
` Leela reads part of a book in 3 hours.
8
2

Exercise 1.4
1. Find :
i)

10 of 5
5
10

ii) 2 of 7
3
8

iii) 1 of 7
3
4

iv) 4 of 7
8
9

4 of 9
vi) 1 of 2
9
7
9
4
2. Multiply and reduce to lowest form :

v)

i)

2 #32
9
3

ii) 2 # 9
9 10

v)

9#3
2 3

vi) 4 # 12
5
7

iii) 3 # 6
8 9

17

iv) 7 # 9
8 14

Find
i) 1 × 7
3
5
2
8
ii) #
3 9

Chapter 1
3.		Simplify the following fractions :
i)

2 #52
5
3

ii) 6 3 # 7
4 10

iv) 5 3 # 3 1
4
2

iii) 7 1 # 1
2

v) 7 1 # 8 1
4
4

4. A car runs 20 km. using 1 litre of petrol. How much distance will it cover using
2 3 litres of petrol.
4
5. Everyday Gopal read book for 1 3 hours. He reads the entire book in 7 days.
4
How many hours in all were required by him to read the book?

The reciprocal of a fraction
If the product of two non-zero numbers is equal to one then each number is
called the reciprocal of the other. So reciprocal of 3 is 5 , the reciprocal of 5 is 3 .
5

3

3

5

Note: Reciprocal of 1 is 1 itself. 0 does not have a reciprocal.

(iv) Division of a whole number by a fraction
To divide a whole number by any fraction, multiply that whole number by the
reciprocal of that fraction.
Example 1.14
Find (i) 6 ' 2 		

(ii) 8 ' 7

5

9

Solution
(i) 6 ÷ 2 = 6 × 5 = 15

5
2
7
9
(ii) 8 ' = 8 # = 72
9
7
7

While dividing a whole number by a mixed fraction, first convert the mixed
fraction into improper fraction and then solve it.
Example 1.15
Find 6 ÷ 3 4

Find:
i) 6 ' 5 2

5

Solution

3

ii) 9 ' 3 3

7

6 ÷ 3 4 = 6 ÷ 19 = 6 × 5 = 30 = 1 11
5

5

19

19

19

(v) Division of a fraction by another fraction
To divide a fraction by another fraction, multiply the first fraction by the
reciprocal of the second fraction.
18

Real Number System
We can now find 1 ' 3
5

7

1 ' 3 = 1 # reciprocal of 3 .
5 7
5
7
		
= 1 # 7 = 7 		
5 3
15

Find:
i) 3 ' 4 , ii) 1 ' 4 , iii) 2 3 ' 7
7

5

2

5

4

2

Exercise 1.5
1. Find the reciprocal of each of the following fractions:
5
7
33
v)
2

i)

2. Find :
i) 5 ÷ 25
3

ii) 4
9
vi) 1
9

iii) 10
7
vii) 1
13

iv) 9
4
viii) 7
5

ii) 6 ÷ 36
9

iii) 7 ÷ 14
3

iv) 1 1 ÷ 15
4

3. Find :
i) 2 ' 1
ii) 5 ' 6
iii) 2 3 ' 3
iv) 3 3 ' 8
5 4
6 7
4 5
2 3
1
4. How many uniforms can be stitched from 47 metres of cloth if each scout
4
requires 2 1 metres for one uniform?
4
5. The distance between two places is 47 1 km. If it takes 1 3 hours to cover the
2
16
distance by a van, what is the speed of the van?

1.5 Introduction to Rational Numbers
p

A rational number is defined as a number that can be expressed in the form ,
q
where p and q are integers and q ^ 0. Here p is the numerator and q is the denominator.
For example 7 , - 5 , 2 , 11 , - 3 are the rational numbers
3

7

9 - 7 11

A rational number is said to be in standard form if its denominator is positive
and the numerator and denominator have no common factor other than 1.
If a rational number is not in the standard form, then it can be reduced to the
standard form.
Example 1.16
Reduce 72 to the standard form.
54

19

Chapter 1
Solution
72
72 ' 2
We have, 54 = 54 ' 2

Aliter: 72 = 72 ' 18 = 4
54

= 36 = 36 ' 3
27
27 ' 3
12
=
= 12 ' 3
9
9'3
=4
3

54 ' 18

3

In this example, note that 18 is the highest common factor (H.C.F.) of 72 and 54.
To reduce the rational number to its standard form, we divide its numerator and
denominator by their H.C.F. ignoring the negative sign if any.
If there is negative sign in the denominator divide by " - H.C.F.".
Example 1.17
Reduce to the standard form.
(i)

18 (ii)
- 12

-4
- 16

Write in standard form.

Solution

i) - 18 , ii) - 12 , iii) 7
51

(i) The H.C.F. of 18 and 12 is 6

28

35

Thus, its standard form would be obtained by dividing by – 6.
		

18 = 18 ' ^- 6h = - 3
2
- 12
- 12 ' ^- 6h

(ii) The H.C.F. of 4 and 16 is 4.
Thus, its standard form would be obtained by dividing by – 4
- 4 = - 4 ' ^- 4h = 1
4
- 16
- 16 ' ^- 4h

1.6 Representation of Rational numbers on the Number line.
You know how to represent integers on the number line. Let us draw one such
number line.
The points to the right of 0 are positive integers. The points to left of 0 are
negative integers.
Let us see how the rational numbers can be represented on a number line.

Fig. 1.3
20

Real Number System
Let us represent the number – 1 on the number line.
4

As done in the case of positive integers, the positive rational numbers would be
marked on the right of 0 and the negative rational numbers would be marked on the
left of 0.

Fig. 1.4

To which side of 0, will you mark - 1 ? Being a negative rational number, it
4
would be marked to the left of 0.
You know that while marking integers on the number line, successive integers
are marked at equal intervals. Also, from 0, the pair 1 and – 1 is equidistant .
In the same way, the rational numbers 1 and - 1 would be at equal distance
4

4

from 0. How to mark the rational number 1 ? It is marked at a point which is one

4
1
fourth of the distance from 0 to 1. So, - would be marked at a point which is one
4
fourth of the distance from 0 to - 1.

We know how to mark 3 on the number line. It is marked on the right of 0 and
2

lies halfway between 1 and 2. Let us now mark - 3 on the number line. It lies on the
left of 0 and is at the same distance as 3 from 0.

2

2

Similarly - 1 is to the left of zero and at the same distance from zero as 1 is
2

2
1
to the right. So as done above, - can be represented on the number line. All other
2

rational numbers can be represented in a similar way.

Rational numbers between two rational numbers
Raju wants to count the whole numbers between 4 and 12. He knew there
would be exactly 7 whole numbers between 4 and 12.
Are there any integers between 5 and 6 ?
There is no integer between 5 and 6.
` Number of integers between any two integers is finite.

Now let us see what will happen in the case of rational numbers ?
Raju wants to count the rational numbers between 3 and 2 .
7

21

3

Chapter 1
For that he converted them to rational numbers with same denominators.
So 3 = 9
7

21

and 2 = 14
3

21

Now he has, 9 1 10 1 11 1 12 1 13 1 14
21

21

21

21

21

21

So 10 , 11 , 12 , 13 are the rational numbers in between 9 and 14 .
21

21 21 21 21

21

Now we can try to find some more rational numbers in between 3 and 2 .
7

3

we have 3 = 18 and 2 = 28
7

42

3

42

So, 18 1 19 1 20 1 g 1 28 . Therefore 3 1 19 1 20 1 21 1 g 1 2 .
42

42

42

42

7

42

42

42
Hence we can find some more rational numbers in between 3 and 2 .
7
3

3

We can find unlimited (infinite) number of rational numbers between any
two rational numbers.
Example 1.18
List five rational numbers between 2 and 4 .
5

7

Solution
Let us first write the given rational numbers with the same denominators.
Now, 2 = 2 # 7 = 14 and 4 = 4 # 5 = 20
5

5#7
35
7
7#5
35
So, we have 14 1 15 1 16 1 17 1 18 1 19 1 20
35
35
35
35
35
35
35
15 , 16 , 17 , 18 , 19 are the five required rational numbers.
35 35 35 35 35

Example 1.19
Find seven rational numbers between - 5 and - 8 .
3
7
Solution
Let us first write the given rational numbers with the same denominators.
Now, - 5 =- 5 # 7 =- 35 and - 8 =- 8 # 3 =- 24
3

3#7
21
7
7#3
21
So, we have - 35 1 - 34 1 - 33 1 - 32 1 - 31 1 - 30
21
21
21
21
21
21
		
1 - 29 1 - 28 1 - 27 1 - 26 1 - 25 1 - 24
21
21
21
21
21
21
` The seven rational numbers are - 34 , - 33 , - 32 , - 31 , - 30 , - 29 , - 28 .
21
21
21
21
21
21
21

(We can take any seven rational numbers)
22

Real Number System
Exercise 1.6
1. Choose the best answer :
3 is called a
i)
8
		 (A) positive rational number
(B) negative rational number		
		 (C) whole number
(D) positive integer
ii)
The proper negative rational number is
		
iii)
		
iv)
		
		

(A) 4
3

(B) - 7
-5
Which is in the standard form?

(C) – 10
9

(B) – 1
(A) – 4
12
12
A fraction is a
(A) whole number
(C) odd number

(C)

1
- 12

(D) 10
9
(D) - 7
14

(B) natural number
(D) rational number

2. List four rational numbers between:
ii) 1 and 4
		 i) - 7 and - 2
5
3
2
3

iii) 7 and 8
4
7

3. Reduce to the standard form:
		 i) - 12
16
		 iv) 70
42

ii) - 18
48
v) 4
8

iii) 21
- 35

4. Draw a number line and represent the following rational numbers on it.
		 i) 3
4
		 iv) 6
5

iii) - 8
3

ii) - 5
8
v) – 7
10

5. Which of the following are in the standard form:
		 i) 2
3
		 iv) - 1
7

ii) 4
16
v) 4
7

iii) 9
6

1.7 Four Basic Operations on Rational numbers
You know how to add, subtract, multiply and divide on integers. Let us now
study these four basic operations on rational numbers.

(i) Addition of rational numbers
Let us add two rational numbers with same denominator.

23

Chapter 1
Example 1.20
Add 9 and 7 .
5

5

Solution
9 + 7 = 9+7
5 5
5

		

= 16 .
5

Let us add two rational numbers with different denominators.
Example 1.21
Simplify: 7 + ` - 5 j
3

4

Solution
7 + -5
3 ` 4 j
= 28 - 15
12
13
=
12

(L.C.M. of 3 and 4 is 12)

Example 1.22
Simplify : - 3 + 1 - 5 .
4

2

6

Solution
- 3 + 1 - 5 = (- 3 # 3) + (1 # 6) - (5 # 2) (L.C.M. of 4,2 and 6 is 12)
12
4
2 6
		 = - 9 + 6 - 10
12
		 = - 19 + 6 = - 13
12
12

(ii) Subtraction of rational numbers
Example 1.23
Subtract : 8 from 10
7

3

Solution:
10 - 8
3
7

= 70 - 24 = 46
21
21

Example 1.24
6 - - 10
35 ` 35 j

Simplify
Solution:

6 - - 10 = 6 + 10 = 16
35 ` 35 j
35
35
24

Real Number System
Example 1.25
Simplify : `- 2 7 j - `3 6 j
35
35
Solution
77 111
7
6
`- 2 35 j - `3 35 j = -35 - 35

		 = - 77 - 111 = - 188 =- 5 13
35

35

35

Example 1.26
The sum of two rational numbers is 1. If one of the numbers is 5 , find the
20

other.
Solution
Sum of two rational numbers = 1
Given number + Required number = 1
5 + Required number = 1
20

Required number = 1 - 5

20
20
-5
		 =
20
15
		 =
= 3
20
4
3
` Required number is .
4

7 - 5 , ii) 5 - 7 ,
35 35
6 12
iii) 7 - 3 , iv) `3 3 j - `2 1 j ,
3 4
4
4
v) `4 5 j - `6 1 j
7
4

i)

Exercise1.7
1. Choose the best answer :
i) 1 + 2 is equal to
3
3
		 (A) 2
(B) 3
ii) 4 – 9 is equal to
5 5
		 (A) 1
(B) 3
iii) 5 1 + 1 10 is equal to
11
11
		 (A) 4
(B) 3
iv)

(C) 1

(D) 4

(C) – 1

(D) 7

(C) – 5

(D) 7

The sum of two rational numbers is 1. If one of the numbers is 1 , the other
2
number is
(A) 4
3

(B) 3
4

(C) - 3
4
25

(D) 1
2

Chapter 1
2. Add :
12 and 6
ii) 7 and 17
5
5
13
13
7
5
7
8
iv) - and v) and
13
13
3
4
9
10
3
vii)
viii)
and and - 7
7
3
6
2
x) 4 , - 7 and - 8
5
10
15
3. Find the sum of the following :

i)

-3 + 7
4 4
iv) - 7 + 9
8 16
vii) 11 + `- 7 j
13
2
ix) 7 + `- 10 j + `- 7 j
9
18
27

i)

iii) 8 and 6
7

7
5
vi) - and 7
7
6
8
9
ix)
, and 1
4 7
28

ii) 9 + 15

iii) - 3 + 6
6
6
4 11
v) 4 + 7
vi)`- 6 j + `- 14 j
5 20
13
26
viii) `- 2 j + 5 + `- 7 j
5
12
10
x) 6 + `- 7 j + `- 9 j
3
6
12

4. Simplify :

i)
iv)

7 - 5
35 35
3
1
`3 4 j - `2 4 j

ii) 5 - 7
6

iii) 7 - 3

12

3

4

v) `4 5 j - `6 1 j
7

4

5. Simplify :

i)

2
5
`1 11 j + `3 11 j

ii) `3 4 j - `7 3 j
5
10

iii) `- 1 2 j + `- 3 5 j + `6 3 j
11
11
11
v)

iv) `- 3 9 j + `3 2 j + `6 5 j
10
5
20

4
3
`- 3 5 j + `2 8 j

vi) `- 1 5 j + `- 2 7 j
12
11

vii) `9 6 j + `- 11 2 j + `- 5 7 j
viii) `7 3 j + `- 10 7 j
7
10
3
21
42
6. The sum of two rational numbers is 17 . If one of the numbers is 5 , find the
4
2
other number.
7. What number should be added to 5 so as to get 49 .
6
30
8. A shopkeeper sold 7 3 kg, 2 1 kg and 3 3 kg of sugar to three consumers in a
4
2
5
day. Find the total weight of sugar sold on that day.
9. Raja bought 25 kg of Rice and he used 1 3 kg on the first day, 4 1 kg on the
4
2
second day. Find the remaining quantity of rice left.
10. Ram bought 10 kg apples and he gave 3 4 kg to his sister and 2 3 kg to his
5
10
friend. How many kilograms of apples are left?
26

Real Number System
(iii) Multiplication of Rational numbers
To find the product of two rational numbers, multiply the numerators and
multiply the denominators separately and put them as new rational number. Simplify
the new rational number into its lowest form.
Example 1.27
Find the product of ` 4 j and ` - 22 j .
8
- 11
Solution
4
22
` 11 j # ` -8 j
=` - 4 j # ` - 22 j = 88
11
8
88
=1

Example 1.28
Find the product of `- 2 4 j and `- 3 2 j .
15
49
Solution
4
2
`- 2 15 j # `- 3 49 j

= ` - 34 j # ` - 149 j
15
49

		 = 5066 = 6 656
735

735

Example 1.29
The product of two rational numbers is 2 . If one of the numbers is 1 , find the
2
9
other rational number.
Solution
The product of two rational numbers =
One rational number =
` Given number # required number

=

1 # required number
2

=

required number =

2
9
1
2
2
9
2
9
2#2 =4
9 1
9

` Required rational number is 4 .
9

Multiplicative inverse (or reciprocal) of a rational number
If the product of two rational numbers is equal to 1, then one number is called
the multiplicative inverse of other.
27

Chapter 1
7 # 23 = 1
23
7
` The multiplicative inverse of 7 is 23 .
23
7
23
Similarly the multiplicative inverse of
is 7 .
7
23
8
12
ii) `
#
=1
12 j ` - 8 j
` The multiplicative inverse of ` - 8 j is ` 12 j .
12
-8

i)

Find
1) 7 # 9 , 2) 11 # 24
8

12
12 33
3) `- 1 1 j # `- 7 2 j
4
3

(iv) Division of rational numbers
To divide one rational number by another rational number, multiply the first
rational number with the multiplicative inverse of the second rational number.
Example 1.30
Find ` 2 j ' ` - 5 j .
3

10

Solution
2
5 = 2 ' -1
` 3 j ' `10 j
3 ` 2 j
		 = 2 # (- 2) = - 4
3
3
Example 1.31

Find 4 3 ' 2 3 .
7

8

Solution
4 3 ' 2 3 = 31 ' 19
7
8
7
8
		 = 31 # 8 = 248
7 19
133
		 = 1 115
133

Exercise 1.8
1. Choose the best answer :
		 i) 7 × 13 is equal to
13
7
			

(A) 7

(B) 13

(C) 1

		 ii) The multiplicative inverse of 7 is
8
7
8
(B)
(C) - 7
			 (A)
8
7
8
		 iii) 4 × ` - 22 j is equal to
8
- 11
			

(A) 1

(B) 2

(C) 3
28

(D) – 1
(D) - 8
7

(D) 4

Real Number System
		 iv) – 4 ÷ 9 is equal to
9
36
			 (A) - 16
(B) 4
9

(C) 5

(D) 7

2. Multiply :
ii) - 7 and 5
		 i) - 12 and 6
5
5
13
13
3
7
6
iii)
iv)
and
and 44
9
8
11
22
50
28
5
v)
vi)
and
and 4
7
10
6
15
3. Find the value of the following :
9 # - 10 # 15
5
4
18
		 iii) 1 1 # 2 2 # 9 3
5
5
10

		 i)

ii) - 8 # - 5 # - 30
4
6
10
iv) - 3 4 # - 2 1 # 9 1
15
5
5

v) 3 # 9 # 10
6 7
4

4. Find the value of the following :
		 i) - 4 ' 9
ii) 3 ' ` - 4 j
9
4
5
10
8
7
3
		 iii) `
iv) - 9 ' 1 3
'
35 j 35
4
40
5. The product of two rational numbers is 6. If one of the number is 14 , find the
3
other number.
6. What number should be multiply 7 to get 21 ?
2
4

1.8 Decimal numbers
(i) Represent Rational Numbers as Decimal numbers
You have learnt about decimal numbers in the earlier classes. Let us briefly
recall them here.
All rational numbers can be converted into decimal numbers.
For Example
(i)

(ii)

(iii)
(iv)

1 = 1'8
8
` 1 = 0.125
8
3 = 3'4
4
` 3 = 0.75
4
1
3 = 16 = 3.2
5
5
2 = 0.6666g Here 6 is recurring without end.
3
29

Chapter 1
(ii) Addition and Subtraction of decimals
Example 1.32
Add 120.4, 2.563, 18.964
Solution
120.4
2.563
18.964
141.927
Example 1.33
Subtract 43.508 from 63.7
Solution
63.700
( – ) 43.508
20.192
Example 1.34
Find the value of 27.69 – 14.04 + 35.072 – 10.12.
Solution
27.690

– 14.04

35.072

– 10.12

62.762

– 24.16

The value is

62.762
– 24.16
38.602

38.602.

Examples 1.35
Deepa bought a pen for `177.50. a pencil for `4.75 and a notebook for
`20.60. What is her total expenditure?
Solution
Cost of one pen = ` 177.50
Cost of one pencil = ` 4.75
Cost of one notebook = ` 20.60
` Deepa’s total expenditure = ` 202.85

30

Real Number System
(iii) Multiplication of Decimal Numbers
Rani purchased 2.5 kg fruits at the rate of `23.50 per kg. How much money
should she pay? Certainly it would be `(2.5 × 23.50). Both 2.5 and 23.5 are decimal
numbers. Now, we have come across a situation where we need to know how to
multiply two decimals. So we now learn the multiplication of two decimal numbers.
Let us now find 1.5 × 4.3
Multiplying 15 and 43. We get 645. Both, in 1.5 and 4.3, there is 1 digit to the
right of the decimal point. So, count 2 digits from the right and
put a decimal point. (since 1 + 1 = 2)
While multiplying 1.43 and 2.1, you will first multiply
143 and 21. For placing the decimal in the product obtained,
you will count 2 + 1 = 3 digits starting from the right most
digit. Thus 1.43 × 2.1 = 3.003.

i) 2.9 × 5
ii) 1.9 × 1.3
iii) 2.2 × 4.05

Example 1.36
The side of a square is 3.2 cm. Find its perimeter.
Solution
All the sides of a square are equal.
Length of each side = 3.2 cm.
Perimeter of a square = 4 × side
Thus, perimeter = 4 × 3.2 = 12.8 cm.

Perimeter of a
square = 4 × side

Example 1.37
The length of a rectangle is 6.3 cm and its breadth is 3.2 cm. What is the area
of the rectangle?
Solution
Length of the rectangle
Breadth of the rectangle
Area of the rectangle
		

=
=
=
=

6.3 cm
3.2 cm.
( length) × (breath)
6.3 × 3.2 = 20.16 cm2

Multiplication of Decimal number by 10, 100 and 1000
Rani observed that 3.7 = 37 , 3.72 = 372 and 3.723 = 3723 Thus, she
10
100
1000
found that depending on the position of the decimal point the decimal number can be
converted to a fraction with denominator 10 , 100 or 1000. Now let us see what would
happen if a decimal number is multiplied by 10 or 100 or 1000.
31

Chapter 1
For example,
3.23 × 10 = 323 × 10 = 32.3
100

Decimal point shifted to the right by one place since
10 has one zero over one.
3.23 × 100 = 323 × 100 = 323
100

Decimal point shifted to the right by two places since
100 has two zeros over two.

i)

0.7 × 10

ii) 1.3 × 100
iii) 76.3 × 1000

3.23 × 1000 = 323 × 1000 = 3230
100

Exercise 1.9
1. Choose the best answer :
i) 0.1 × 0.1 is equal to
		 (A) 0.1

(B) 0.11

(C) 0.01

(D) 0.0001

(C) 0.05

(D) 0.0005

(C) 0.0001

(D) 0.1

(C) 2

(D) 3

ii) 5 ÷ 100 is equal to
		 (A) 0.5

(B) 0.005

iii) 1 × 1 is equal to
10 10
		 (A) 0.01
(B) 0.001
iv) 0.4 × 5 is equal to
		 (A) 1
2. Find :
(i) 0.3 × 7

(B) 0.4

(ii) 9 × 4.5

(iii) 2.85 × 6

(iv) 20.7 × 4

(v) 0.05 × 9

(vi) 212.03 × 5

(vii) 3 × 0.86

(viii) 3.5 × 0.3

(ix) 0.2 × 51.7

(x) 0.3 × 3.47

(xi) 1.4 × 3.2

(xii) 0.5 × 0.0025

(xiii) 12.4 × 0.17 (xiv) 1.04 × 0.03
3. Find :
(i) 1.4 × 10
(v) 32.3 × 100

(ii) 4.68 × 10

(iii) 456.7 × 10

(vi) 171.4 × 100

(vii) 4.78 × 100

(iv) 269.08 × 10

4. Find the area of rectangle whose length is 10.3 cm and breadth is 5 cm.
5. A two-wheeler covers a distance of 75.6 km in one litre of petrol. How much
distance will it cover in 10 litres of petrol?

32

Real Number System
(iv) Division of Decimal Numbers
Jasmine was preparing a design to decorate her classroom. She needed a few
colourd strips of paper of length 1.8 cm each. She had a strip of coloured paper of
length 7.2 cm. How many pieces of the required length will she get out of this strip?
She thought it would be 7.2 cm. Is she correct?
1.8

Both 7.2 and 1.8 are decimal numbers. So we need to know the division of
decimal numbers .
For example,
141.5 ' 10 = 14.15
141.5 ' 100 = 1.415
141.5 ' 1000 = 0.1415
To get the quotient we shift the point in the decimal
number to the left by as many places as there are zeros
over 1.

Find:
i) 432.5 ÷ 10
ii) 432.5 ÷ 100
iii) 432.5 ÷ 1000

Example 1.38
Find 4.2 ÷ 3.
Solution
4.2 ÷ 3 = 42 ' 3 = 42 # 1

10
10 3
		 = 42 # 1 = 1 # 42
10 # 3
10 # 3
		 = 1 # 42 = 1 # 14
10
3
10
		 = 14 = 1.4
10

Find:
i) 85.8 ÷ 3
ii) 25.5 ÷ 5

Example 1.39
Find 18.5 ÷ 5.
Solution
First find 185 ÷ 5. We get 37.
There is one digit to the right of the decimal point in 18.5.
Place the decimal point in 37 such that there would be one digit
to its right. We will get 3.7.

33

Find:
i) 73.12 ÷ 4
ii) 34.55 ÷ 7

Chapter 1
Division of a Decimal Number by another Decimal number
Example 1.40
Find 17.6 .
0.4

Find :

Solution

i) 9.25

We have
17.6 ÷ 0.4 = 176 ' 4
10 10
		 = 176 # 10 = 44.
10
4

0.5
ii) 36
0.04
iii) 6.5
1.3

Example 1.41

A car covers a distance of 129.92 km in 3.2 hours. What is the distance covered
by it in 1 hour?
Solution
Distance covered by the car = 129.92 km.
Time required to cover this distance = 3.2 hours.
So, distance covered by it in 1 hour = 129.92 = 1299.2 = 40.6km.
3.2

32

Exercise 1.10
1. Choose the best answer :
		 i)
			
		 ii)
			

0.1 ÷ 0.1 is equal to
(A) 1

(B) 0.1

1 is equal to
1000
(A) 0.01
(B) 0.001

(C) 0.01

(D) 2

(C) 1.001

(D) 1.01

		 iii) How many apples can be bought for `50 if the cost of one apple is `12.50?
			

(A) 2

(B) 3

(C) 4

(D) 7

		 iv) 12.5 is equal to
2.5
			 (A) 4
(B) 5

(C) 7

(D) 10

2. Find :
(i) 0.6 ÷ 2

(ii) 0.45 ÷ 5

(iii) 3.48 ÷ 3

(iv) 64.8 ÷ 6

(v) 785.2 ÷ 4

(vi) 21.28 ÷ 7

3. Find :
(i) 6.8 ÷ 10

(ii) 43.5 ÷ 10		(iii) 0.9 ÷ 10

(iv) 44.3 ÷ 10

(v) 373.48 ÷ 10
34

(vi) 0.79 ÷ 10

Real Number System
4. Find :
(i) 5.6 ÷ 100

(ii) 0.7 ÷ 100		(iii) 0.69 ÷ 100

(iv) 743.6 ÷ 100

(v) 43.7 ÷ 100

(vi) 78.73 ÷ 100

(i) 8.9 ÷ 1000

(ii) 73.3 ÷ 1000

(iii) 48.73 ÷ 1000

(iv) 178.9 ÷ 1000
6. Find :
(i) 9 ÷ 4.5

(v) 0.9 ÷ 1000

(vi) 0.09 ÷ 1000

5. Find :

(iv) 40.95 ÷ 5

(ii) 48 ÷ 0.3		(iii) 6.25 ÷ 0.5
(v) 0.7 ÷ 0.35		(vi) 8.75 ÷ 0.25

7. A vehicle covers a distance of 55.2 km in 2.4 litres of petrol. How much distance
will it cover in one litre of petrol?
8. If the total weight of 11 similar bags is 115.5 kg, what is the weight of 1 bag?
9. How many books can be bought for `362.25, if the cost of one book is `40.25?
10. A motorist covers a distance of 135.04 km in 3.2 hours. Find his speed?
11. The product of two numbers is 45.36. One of them is 3.15. Find the other
number?

1.9 Powers
Introduction
Teacher asked Ramu, “Can you read this number 2560000000000000?”
He replies, “It is very difficult to read sir”.
“The distance between sun and saturn is 1,433,500,000,000 m. Raja can you
able to read this number?” asked teacher.
He replies, “Sir, it is also very difficult to read”.
Now, we are going to see how to read the difficult numbers in the examples
given above.

Exponents
We can write the large numbers in a shortest form by using the following
methods.
10 = 101
100 = 101 × 101 = = 102
1000 = 101 ×101 × 101 = 103
35

Chapter 1
Similarly,
21 # 21 = 22
21 # 21 # 21 = 23
21 # 21 # 21 # 21 = 24

a # a = a2 [read as ‘a’ squared or ‘a’ raised to the power 2]
a # a # a = a3 [read as ‘a’ cubed or ‘a’ raised to the power 3]
a # a # a # a = a4 [read as ‘a’ raised to the power 4 or the 4th power of ‘a’]
gggggggg
gggggggg

a # a # ... m times = am [read as ‘a’ raised to the power m or mth power of ‘a’]
Here ‘a’ is called the base, ‘m’ is called the exponent (or) power.
Note: Only a2 and a3 have the special names “a squared’ and “a cubed”.
` we can write large numbers in a shorter form using exponents.

Example 1.42
Express 512 as a power .
Solution
We have

512 = 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 × 2

So we can say that

512 = 29

Example: 1.43
Which one is greater 25 , 52 ?
Solution
We have

25 = 2 # 2 # 2 × 2 × 2 = 32

and

52 = 5 # 5 = 25

Since 32 > 25.
Therefore 25 is greater than 52.

36

Real Number System
Example: 1.44
Express the number 144 as a product of powers of prime factors.
Solution
144 = 2 # 2 # 2 # 2 × 3 # 3
		 = 24 # 32
Thus, 144 = 24 # 32
Example 1.45
(i) 45 (ii) (-4)5

Find the value of
Solution
(i)

45

=

4#4#4#4#4

		 = 1024.
(ii) (–4)5

=

(– 4) # (– 4) # (– 4) # (– 4) # (– 4)

		 = – 1024.
Excercise 1.11
1. Choose the best answer :
i)		 – 102 is equal to
		 (A) – 100

(B) 100

(C) – 10

(D) 10

(C) 10

(D) – 10

(C) an

(D) am + n

(C) 0

(D) 3

ii)		 (– 10)2 is equal to
		 (A) 100

(B) – 100

iii) a × a × a × ..... n times is equal to
		 (A) am

(B) a–n

iv) 1033 × 0 is equal to
		 (A) 103

(B) 9

2. Find the value of the following :
(ii) 33		
(i) 28
(iv) 123
(v) 134		

(iii) 113
(vi) 010

3. Express the following in exponential form :
(i) 7 # 7 # 7 # 7 # 7 × 7		

(ii) 1 # 1 # 1 # 1 # 1

(iii) 10 # 10 # 10 # 10 # 10 # 10		

(iv) b # b # b # b # b

		(v) 2 # 2 # a # a # a # a		
37

(vi) 1003 × 1003 × 1003

Chapter 1
4. Express each of the follwing numbers using exponential notation. (with smallest
base)
(i) 216
(ii) 243
(iii) 625
(iv) 1024
(v) 3125
(vi) 100000
5. Identify the greater number in each of the following :
(ii) 26 , 62
(iii) 32 , 23
(i) 45 , 54
(iv) 56 , 65
(v) 72 , 27
(vi) 47 , 74
6. Express each of the following as product of powers of their prime factors :
(i) 100
(ii) 384
(iii) 798
(iv) 678
(v) 948
(vi) 640
7. Simplify :
(ii) 0 # 104
(iii) 52 # 34
(i) 2 # 105
(iv) 24 # 34
(v) 32 # 109
(vi) 103 # 0
8. Simplify :
(ii) (– 1)10
(iii) (– 3)2 # (– 2)3
(i) (– 5)3
(iv) (– 4)2 # (– 5)3 (v) (6)3 # (7)2
(vi) (– 2)7 # (– 2)10

Laws of exponents
Multiplying powers with same base
1)

32 # 34

= (3 # 3) # (3 # 3 # 3 × 3)

		 = 31 # 31 # 31 # 31 # 31 # 31
		 = 36
2)

(– 5)2 # (– 5)3

= [(– 5) # (– 5) ] # [(– 5) # (– 5) # (– 5)]

		 = (– 5)1 # (– 5)1 # (– 5)1 # (– 5)1 # (– 5)1
		 = (– 5)5
3)

a2 # a5 = (a # a) # (a # a # a # a # a)

		 = a1 # a1 # a1 # a1 # a1 # a1 # a1 = a7
From this we can generalise that for
any non-zero integer a, where m and n are
whole numbers a m # a n = a m + n
i) 25 # 27

ii) 43 # 44

iii) p3 # p5 iv) ^- 4h100 # ^- 4h10

38

Real Number System
Dividing powers with the same base
We observe the following examples:
7
27 ÷ 25 = 25

i)

2
		 = 2 # 2 # 2 # 2 # 2 # 2 # 2 = 22
2#2#2#2#2
(- 5) 4
ii)
(- 5) 4 ' (- 5) 3 =
(- 5) 3

		 = (- 5) # (- 5) # (- 5) # (- 5) = - 5
(- 5) # (- 5) # (- 5)

From these examples, we observe: In general, for any non-zero integer ‘a’,
a m ' a n = a m - n where m and n are whole numbers and m > n.
a m ' a m = a m - m = a0 = 1 .

Power of a power
Consider the following:
(i)

(33)2

=

33 × 33 		

=

33+3 = 36

(ii)

(22)3

=

22 × 22 × 22

=

22+2+2 = 26

From this we can generalise for any non-zero integer ‘a’
^a mhn = a mn , where m and n are whole numbers.

Example: 1.46
Write the exponential form for 9 × 9 × 9 × 9 by taking base as 3.
Solution
We have

9 × 9 × 9 × 9 = 94

We know that

9 = 3×3

Therefore

94 = (32)4

= 38

Exercise 1.12
1. Choose the best answer :
i) am × ax is equal to
(A) am x

(B) am + x

x

(C) am – x

(D) a m

(C) 0

(D) 1010

ii) 1012 ÷ 1010 is equal to
(A) 102

(B) 1
39

If n = m

Chapter 1
iii) 1010 × 102 is equal to
(A) 105

(B) 108

(C) 1012

(D) 1020

(B) 212

(C) 220

(D) 210

iv) (22)10 is equal to
(A) 25

Using laws of exponents, simplify in the exponential form.
2. i) 35 # 33 # 34

		 ii) a3 # a2 # a7
		 iii) 7 x # 72 # 73
		 iv) 100 # 102 # 105
		 v) 56 # 52 # 51
3. i) 510 ' 56

4. i) ^34h3

		 ii) a6 ' a2

		 ii) ^25h4

		 iii) 1010 ' 100

		 iii) ^45h2

		 iv) 46 ' 44

		 iv) ^40h10

		 v) 33 ' 33

		 v) ^52h10
Multiplication of fractions pictorially

Step 1 :
		Take a transparent sheet of paper.
Step 2 :
		Draw a rectangle 16 cm by 10 cm and divide it vertically in to 8 equal parts.
Shade the first 3 parts. The shaded portion represents 3/8 of the rectangle.
Step 3 :
		Draw another rectangle of the same size and divide it horizontally into 5 equal
parts. Shade the first 2 parts. The shaded portion represents 2/5 of the rectangle.
Step 4 :
		Place the first transparent sheet on the top of the second sheet so that the two
rectangles coincide.
		We find that,
Total number of squares				
= 40
Number of squares shaded vertically and horizontally = 6
6
			
` 3#2 =
8 5
40
40

Real Number System

1. Natural numbrs N = {1, 2, 3, ...}
2. Whole numbers W = {0, 1, 2, ...}
3. Integers Z = {..., – 3, – 2, – 1, 0, 1, 2, 3, ...}
4. The product of two positive integers is a positive integer.
5. The product of two negative integers is a positive integer.
6. The product of a positive integer and a negative integer is a negative integer.
7. The division of two integers need not be an integer.
8. Fraction is a part of whole.
9. If the product of two non-zero numbers is 1 then the numbers are called the
reciprocal of each other.
10. a × a × a × ... m times = am
		 (read as ‘a’ raised to the power m (or) the mth power of ‘a’)
11. For any two non-zero integers a and b and whole numbers m and n,
		 i)

am an = am+n

a m = a m - n , where m > n
an
mn
		 iii) ^a mhn = a
		 ii)

		 iv) (– 1)n = 1, when n is an even number
				

(– 1)n = – 1, when n is an odd number

41

Chapter 2

ALGEBRA
2.1 Algebraic Expressions
(i) Introduction
In class VI, we have already come across simple algebraic expressions like
x + 10, y – 9, 3m + 4, 2y – 8 and so on.
Expression is a main concept in algebra. In this chapter you are going to learn
about algebraic expressions, how they are formed, how they can be combined, how to
find their values, and how to frame and solve simple equations.

(ii) Variables, Constants and Coefficients
Variable
A quantity which can take various numerical values is known as a variable (or
a literal).
Variables can be denoted by using the letters a, b, c, x, y, z, etc.

Constant
A quantity which has a fixed numerical value is called a constant.
For example, 3, - 25, 12 and 8.9 are constants.
13

Numerical expression
A number or a combination of numbers formed by using the arithmetic
operations is called a numerical expression or an arithmetic expression.
For example, 3 + (4 × 5), 5 – (4 × 2), (7 × 9) ÷ 5 and (3 × 4) – (4 × 5 – 7) are
numerical expressions.

Algebraic Expression
An algebraic expression is a combination of variables and constants connected
by arithmetic operations.

42

Algebra
Example 2.1
		

Statement

Expressions

(i)

5 added to y

y+5

(ii)

8 subtracted from n

n–8

(iii)

12 multiplied by x

12 x

(iv)

p divided by 3

p
3

Term
A term is a constant or a variable or a product of a constant and one or more
variables.
3x2, 6x and – 5 are called the terms of the expression 3x2 + 6x - 5 .
A term could be
(i) a constant
(ii) a variable
(iii) a product of constant and a variable (or variables)
(iv) a product of two or more variables
In the expression 4a2 + 7a + 3, the terms are 4a2, 7a and 3. The number of
terms is 3.
In the expression - 6p2 + 18pq + 9q2 - 7, the terms are - 6p2, 18pq, 9q2 and
– 7. The number of terms is 4.
Find the number of terms :
(i) 8b			

(iv) 7x2 y - 4y + 8x - 9

(ii) 3p – 2q		

(v) 4m2 n + 3mn2

(iii) a2 + 4a - 5

Coefficient
The coefficient of a given variable or factor in
a term is another factor whose product with the given
variable or factor is the term itself.

In the term 6xy, the
factors are 6, x, y, 6x,
6y, xy and 6xy.

If the coefficient is a constant, it is called a constant coefficient or a numerical
coefficient.

43

Chapter 2
Example 2.2
In the term 5xy,
coefficient of xy is 5 (numerical coefficient),
coefficient of 5x is y,
coefficient of 5y is x.
Find the numerical coefficient in
(i) 3z

(ii) 8ax

(iii) ab

(v) 1 mn

(vi) - 4 yz

(iv) – pq
Example 2.3
In the term – mn2 ,
coefficient of mn2 is – 1,
coefficient of – n2 is m ,
coefficient of m is – n .
2

2

7

An algebraic box contains cards that
have algebraic expressions written on it. Ask
each student to pick out a card from the box
and answer the following :
l Number of terms in the expression
l Coefficients of each term in the expression
l Constants in the expression

S.No.

Expression

1.

10 – 2y

2.

11 + yz

3.

yn2 + 10

4.

- 3m2 y + n

Term which
contains y

Coefficient
of y

yz

z

44

Algebra
Exercise 2.1
1. Choose the correct answer:
(i) The numerical coefficient in - 7xy

is

(B) x
(C) y
		 (A) - 7
(ii) The numerical coefficient in - q is

(D) xy

		 (A) q
(B) - q
(iii) 12 subtracted from z is

(C) 1

(D) - 1

		 (A) 12 + z
(B) 12z
(iv) n multiplied by - 7 is

(C) 12 - z

(D) z - 12

(C) 7
		 (A) 7n
(B) - 7n
n
(v) Three times p increased by 7 is
(A) 21p

(B) 3p - 7

(C) 3p + 7

(D) - 7
n
(D) 7 - 3p

2. Identify the constants and variables from the following:
		 a, 5, - xy, p, - 9.5
3. Rewrite each of the following as an algebraic expression
		
(i) 6 more than x
		
(ii) 7 subtracted from - m
		
(iii) 11 added to 3 q
		
(iv) 10 more than 3 times x
		
(v) 8 less than 5 times y
4. Write the numerical coefficient of each term of the expression 3y2 - 4yx + 9x2 .
5. Identify the term which contains x and find the coefficient of x
(i) y2 x + y
(iii) 5 + z + zx

(ii) 3 + x + 3x2 y
(iv) 2x2 y - 5xy2 + 7y2

6. Identify the term which contains y2 and find the coefficient of y2
(i) 3 - my2

(ii) 6y2 + 8x

(iii) 2x2 y - 9xy2 + 5x2

(iii) Power
If a variable a is multiplied five times by itself then it is written as
a # a # a # a # a = a5 (read as a to the power 5). Similarly, b # b # b = b3 (b to the
power 3) and c # c # c # c = c4 (c to the power 4). Here a, b, c are called the base and
5, 3, 4 are called the exponent or power.
Example 2.4
(i) In the term - 8a2 , the power of the variable a is 2
(ii) In the term m, the power of the variable m is 1.
45

Chapter 2
(iv) Like terms and Unlike terms
Terms having the same variable or product of variables with same powers
are called Like terms. Terms having different variable or product of variables with
different powers are called Unlike terms.
Example 2.5
(i) x, ­ 5x, 9x are like terms as they have the same variable x
(ii) 4x2 y, - 7yx2 are like terms as they have the same variable x2 y
Example 2.6
(i) 6x, 6y are unlike terms
(ii) 3xy2, 5xy, 8x, - 10y are unlike terms.
Identify the like terms and unlike terms:
(i) 13x and 5x
(iv) 36mn and - 5nm
(v) - 8p2 q and 3pq2
(ii) - 7m and - 3n
(iii) 4x2 z and - 10zx2
To identify the variables, constants, like terms and unlike terms
Make a few alphabetical cards x, y, z, ... numerical cards 0, 1, 2, 3, ... and cards
containing operations + , –, × , ' out of a chart paper and put it in a box. Call each
student and ask him to do the following activity.
l

Pick out the variables

l

Pick out the constants

l

Pick out the like terms

l

Pick out the unlike terms

(v) Degree of an Algebraic expression
Consider the expression 8x2 - 6x + 7. It has 3 terms 8x2, - 6x and 7 .
In the term 8x2, the power of the variable x is 2.
In the term - 6x , the power of the variable x is 1.
The term 7 is called a constant term or an independent term.
The term 7 is 7 # 1 = 7x0 in which the power of the variable x is 0.
In the above expression the term 8x2 has the highest power 2. So, the degree
of the expression 8x2 – 6x + 7 is 2.
2
2
Consider the expression 6x y + 2xy + 3y .
2
In the term 6x y , the power of variable is 3.
(Adding the powers of x and y we get 3 (i.e.) 2 + 1 = 3).
In term 2xy , the power of the variable is 2.
2
In term 3y , the power of the variable is 2.
46

Algebra
2

2

2

So, in the expression 6x y + 2xy + 3y , the term 6x y has the highest power 3.
So the degree of this expression is 3.
Hence, the degree of an expression of one variable is the highest value of the
exponent of the variable. The degree of an expression of more than one variable is the
highest value of the sum of the exponents of the variables in different terms.
Note: The degree of a constant is 0.
Example 2.7
The degree of the expression: (i) 5a2 - 6a + 10 is 2
(ii) 3x2 + 7 + 6xy2 is 3
(iii) m2 n2 + 3mn + 8 is 4

(vi) Value of an Algebraic expression
We know that an algebraic expression has variables and a variable can take any
value. Thus, when each variable takes a value, the expression gives some value.
For example, if the cost of a book is ` x and if you are buying 5 books, you
should pay ` 5x. The value of this algebraic expression 5x depends upon the value of
x which can take any value.
If x = 4, then 5x = 5 # 4 = 20 .
If x = 30, then 5x = 5 # 30 = 150.
So to find the value of an expression, we substitute the given value of x in the
expression.
Example 2.8
Find the value of the following expressions when x = 2.
(i) x + 5

(ii) 7x - 3

(iii) 20 - 5x2

Solution : Substituting x = 2 in
(i)
(ii)

x+5 = 2+5=7
7x – 3 = 7 (2) – 3

			 = 14 – 3 = 11
(iii)

20 – 5x2 = 20 – 5 (2)2

			 = 20 – 5 (4)
			 = 20 – 20 = 0
47

Chapter 2
Example 2.9
Find the value of the following expression when a =- 3 and b = 2 .
(i) a + b

(iii) a2 + 2ab + b2

(ii) 9a - 5b

Solution Substituting a =- 3 and b = 2 in
(i)

a+b = –3+2=–1

(ii)

9a – 5b = 9 (– 3) – 5 (2)

			 = – 27 – 10 = – 37
(iii) a2 + 2ab + b2 = (- 3) 2 + 2 (– 3) (2) + 22
			 = 9 – 12 + 4 = 1

1. Find the value of the following expressions when p =- 3
(i) 6p - 3 		

2p2 - 3p + 2

(ii)

2. Evaluate the expression for the given values
3

x

5

6

10

x- 3

3. Find the values for the variable
x
2x

6

14

28

42

Exercise 2.2
1. Choose the correct answer
(i) The degree of the expression 5m2 + 25mn + 4n2 is
		 (A) 1
(B) 2
(C) 3
(D) 4
(ii) If p = 40 and q = 20 , then the value of the expression ^ p - qh + 8 is
		 (A) 60
(B) 20
(C) 68
2
2 2
(iii) The degree of the expression x y + x y + y

(D) 28
is

		 (A) 1
(B) 2
(C) 3
(D) 4
(iv) If m = - 4 , then the value of the expression 3m + 4 is
		 (A) 16

(C) - 12

(B) 8

48

(D) - 8

Algebra
(v) If

p=2

and q = 3, then the value of the expression (p + q) - ^ p - qh is

(A) 6
(B) 5
(C) 4
2. Identify the like terms in each of the following:

(D) 3

(i) 4x, 6y, 7x
(ii) 2a, 7b, - 3b
(iii) xy, 3x2 y, - 3y2, - 8yx2
(iv) ab, a2 b, a2 b2, 7a2 b
(v) 5pq, - 4p, 3q, p2 q2, 10p, - 4p2, 25pq, 70q, 14p2 q2
3. State the degree in each of the following expression:
(ii)
(iii)
(i) x2 + yz 		
15y2 - 3
(iv) a2 b2 - 7ab 		
(v)
1 - 3t + 7t2

6x2 y + xy

4. If x =- 1, evaluate the following:
(ii)
(i) 3x - 7 		
-x + 9

(iii)

3x2 - x + 7

5. If a = 5 and b =- 3, evaluate the following:
(ii)
(i) 3a - 2b 		
a2 + b2 		

(iii)

4a2 + 5b - 3

2.2 Addition and subtraction of expressions
Adding and subtracting like terms
Already we have learnt about like terms and unlike terms.
The basic principle of addition is that we can add only like terms.
To find the sum of two or more like terms, we add the numerical coefficient of
the like terms. Similarly, to find the difference between two like terms, we find the
difference between the numerical coefficients of the like terms.
There are two methods in finding the sum or difference between the like terms
namely,
(i) Horizontal method
(ii) Vertical method
(i) Horizontal method: In this method, we arrange all the terms in a horizontal
line and then add or subtract by combining the like terms.
Example 2.10
Add 2x and 5x.
Solution:
2x + 5x = ^2 + 5h # x

= 7 # x = 7x

Divide the entire class into 5 groups. Ask
the students of the each group to take out the
things from their pencil boxes and seggregate
them. Now ask them to list out the number
of pens, pencils, erasers... from each box and
also the total of each .
49

Chapter 2
(ii) Vertical method: In this method, we should write the like terms vertically
and then add or subtract.
Example 2.11
Add 4a and 7a.
Solution:

4a
+

7a
11 a

Example 2.12
Add 7pq, - 4pq and 2pq .
Solution:

Horizontal method

Vertical method

7pq - 4pq + 2pq

7 pq

= ^7 - 4 + 2h # pq

– 4 pq

=5 pq

+ 2 pq

		

5 pq

Example 2.13
Find the sum of 5x2 y, 7x2 y, - 3x2 y, 4x2 y .
Solution:

Horizontal method
5x2 y + 7x2 y - 3x2 y + 4x2 y

Vertical method
5x2 y

=^5 + 7 - 3 + 4h x2 y

+ 7x2 y

= 13x2 y

- 3x2 y

		

+ 4x2 y

		

13x2 y

Example 2.14
Subtract 3a from 7a.
Solution:

Horizontal method
7a - 3a = ^7 - 3h a

Vertical method
7a

=4a

+3a
(- )   (Change of sign)

		
		

4a
50

Algebra

When we subtract a number from another number, we add the additive inverse
to the earlier number. i.e., while subtracting 4 from 6 we change the sign of 4
to negative (additive inverse) and write as 6 - 4 = 2.
Note: Subtracting a term is the same as adding its inverse. For example
subtracting + 3a is the same as adding – 3a.
Example 2.15
(i)

Subtract - 2xy from 9 xy .

Solution:

9 xy
– 2 xy
(+)

(change of sign)

11 xy
(ii) Subtract
8p2 q2 from - 6p2 q2

Solution:

- 6p2 q2

+ 8p2 q2
(–)
- 14p2 q2

Unlike terms cannot be added or subtracted the way like terms are added or
subtracted.
For example when 7 is added to x we write it as x + 7 in which both the terms
7 and x are retained.
Similarly, if we add the unlike terms 4xy and 5, the sum is 4xy + 5. If we
subtract 6 from 5pq the result is 5pq- 6.
Example 2.16
Add 6a + 3 and 4a - 2 .
Solution:

51

Chapter 2
		

= 6a + 4a + 3 – 2

		

= 10a + 1

(grouping like terms)

Example 2.17
Simplify : 6t + 5 + t + 1
Solution

= 6t + t + 5 + 1 (grouping like terms)
		
		
= 7t + 6
Example 2.18
Add 5y + 8 + 3z and 4y - 5
Solution
5y + 8 + 3z + 4y - 5
= 5y + 4y + 8 - 5 + 3z
= 9y + 3 + 3z 		

(grouping like terms)
(The term 3z will remain as it is.)

Example 2.19
Simplify the expression 15n2 - 10n + 6n - 6n2 - 3n + 5
Solution
Grouping like terms we have
15n2 - 6n2 - 10n + 6n - 3n + 5
= ^15 - 6h n2 + ^- 10 + 6 - 3h n + 5

= 9n2 + ^- 7h n + 5
= 9n2 - 7n + 5
Example 2.20
Add 10x2 - 5xy + 2y2, - 4x2 + 4xy + 5y2
Solution

and

3x2 - 2xy - 6y2 .

10x2 - 5xy + 2y2
- 4x2 + 4xy + 5y2
+ 3x2 - 2xy - 6y2
2

9x - 3xy + y

2

52

Add:
(i) 8m - 7n, 3n - 4m + 5
(ii) a + b, - a + b
(iii) 4a2, - 5a2, - 3a2, 7a2

Algebra
Example 2.21
Subtract 6a - 3b from - 8a + 9b .
- 8a + 9b

Solution

+ 6a - 3b

(–)

(+)

- 14a + 12b 				

Example 2.22
Subtract 2^ p - qh from

3^5p - q + 3h

3^5p - q + 3h - 2^ p - qh

Solution

Just as

= 15p - 3q + 9 - 2p + 2q
= 15p - 2p - 3q + 2q + 9

-^8 - 5h = - 8 + 5,
- 2^m - nh =- 2m + 2n
the signs of algebraic terms are
handled in the same way as signs
of numbers.

= 13p - q + 9
Example 2.23
Subtract a2 + b2 - 3ab from a2 - b2 - 3ab .
Solution
Horizontal method

Vertical method

^a2 - b2 - 3abh - ^a2 + b2 - 3abh

a2 – b2 – 3ab

		 = a2 - b2 - 3ab - a2 - b2 + 3ab
		 = - b2 - b2

a2 + b2 – 3ab
(–)

		 = - 2b2

(–)
– 2 b2

Example 2.24
If A = 5x2 + 7x + 8, B = 4x2 - 7x + 3, find 2A - B .
Solution 2A = 2^5x2 + 7x + 8h
		 = 10x2 + 14x + 16
Now 2 A – B = ^10x2 + 14x + 16h - ^4x2 - 7x + 3h
		 = 10x2 + 14x + 16 - 4x2 + 7x - 3
		 = 6x2 + 21x + 13

53

(+)

Chapter 2
Subtract:
(i) ^a - bh from ^a + bh
(ii) (5x – 3y) from (– 2x + 8y)
Example 2.25
What should be subtracted from 14b2 to obtain 6b2 ?
14b2

Solution

6b2

(–)
8b2

Example 2.26
What should be subtracted from 3a2 - 4b2 + 5ab to obtain - a2 - b2 + 6ab .
Solution
3a2 - 4b2 + 5ab
- a2 - b2 + 6ab

(+)

(+)

(–)

4a2 - 3b2 - ab

Take 30 cards written with x2, x, 1 (10 in each variety). Write on the backside of
each card any one of –x2, –x, –1.
1. Ask two students to frame 2 different expressions as told by the teacher.
2. Ask the third student to add the expressions and read out the answer.
3. Ask another student to subtract the expressions and read out the answer.
Exercise 2.3
1. Choose the correct answer :
(i) Sum of 4x, - 8x and 7x is
(A) 5x
(ii) Sum of
(A) 14 ab

(B) 4x
2ab,

4ab, - 8ab

(C) 3x

(D) 19x

(C) 2ab

(D) - 14ab

is

(B) - 2ab
54

Algebra
(iii) 5ab + bc - 3ab

is

(A) 2ab + bc

(B) 8ab + bc

2

(iv) 5y - 3y - 4y + y
(A) 9y + 4y

2

2

(D) 3ab

is
(B) 9y - 4y

(v) If A = 3x + 2

(C) 9ab

2

(C) y + 2y

2

(D) y - 2y

2

and B = 6x - 5 , then A - B is

(A) - 3x + 7
2. Simplify :

(B) 3x - 7

(C) 7x - 3

(D) 9 x + 7

(i)
(ii)
(iii)
(iv)
(v)
(vi)

6a - 3b + 7a + 5b
8l - 5l2 - 3l + l2
- z2 + 10z2 - 2z + 7z2 - 14z
p - ^ p - qh - q - ^q - ph
3mn - 3m2 + 4nm - 5n2 - 3m2 + 2n2
^4x2 - 5xy + 3y2h - ^3x2 - 2xy - 4y2h

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)

7ab, 8ab, - 10ab, - 3ab
s + t, 2s - t, - s + t
3a - 2b, 2p + 3q
2a + 5b + 7, 8a - 3b + 3, - 5a - 7b - 6
6x + 7y + 3, - 8x - y - 7, 4x - 4y + 2
6c - c2 + 3, - 3c - 9, c2 + 4c + 10
6m2n + 4mn – 2n2 + 5, n2 – nm2 + 3, mn – 3n2 – 2m2n – 4

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)

6a from 14a
- a2 b from 6a2 b
7x2 y2 from - 4x2 y2
3xy - 4 from xy + 12
m^n - 3h from n^5 - mh
9p2 - 5p from - 10p - 6p2
- 3m2 + 6m + 3 from 5m2 - 9
- s2 + 12s - 6 from 6s - 10
5m2 + 6mn - 3n2 from 6n2 - 4mn - 4m2

3. Add :

		
4. Subtract :

5. (i) What should be added to 3x2 + xy + 3y2 to obtain 4x2 + 6xy?
		 (ii) What should be subtracted from 4p + 6q + 14 to get - 5p + 8q + 20?
		 (iii) If A = 8x - 3y + 9,

B =- y - 9 and C = 4x - y - 9 find A + B - C.

6. Three sides of a triangle are 3a + 4b - 2,
perimeter?

55

a - 7 and 2a - 4b + 3. What is its

Chapter 2
7. The sides of a rectangle are 3x + 2 and 5x + 4 . Find its perimeter.
8. Ram spends `4a+3 for a shirt and ` 8a - 5 for a book. How much does he spend
in all?
9. A wire is 10x - 3 metres long. A length of 3x + 5 metres is cut out of it for use.
How much wire is left out?
10. If A = p2 + 3p + 5 and B = 2p2 - 5p - 7 , then find
		 (i) 2A + 3B		

(ii) A- B

11. Find the value of P - Q + 8 if P = m2 + 8m and Q =- m2 + 3m - 2 .

1. Algebra is a branch of Mathematics that involves alphabet, numbers and
mathematical operations.
2. A variable or a literal is a quantity which can take various numerical values.
3. A quantity which has a fixed numerical value is a constant.
4. An algebraic expression is a combination of variables and constants connected
by the arithmetic operations.
5. Expressions are made up of terms.
6. Terms having the same variable or product of variables with same powers are
called Like terms. Terms having different variable or product of variables with
different powers are called Unlike terms.
7. The degree of an expression of one variable is the highest value of the exponent
of the variable. The degree of an expression of more than one variable is the
highest value of the sum of the exponents of the variables in different terms

56

Geometry

GEOMETRY
Geometry is a branch of Mathematics that deals with the properties of various
geometrical shapes and figures. In Greek the word “Geometry” means “Earth
Measurement”. Geometry deals with the shape, size, position and other geometrical
properties of various objects. Geometry is useful in studying space, architecture,
design and engineering.

3.1. Revision
Basic Geometrical concepts:
In earlier classes you have studied about some geometrical concepts. Let us
recall them.

Point
A fine dot made with a sharp pencil may be
taken as roughly representing a point. A point has a
position but it has no length, breadth or thickness. It
is denoted by the capital letters. In the figure A, B, C,
D are points.

Fig. 3.1

Line
A line is traced out by a moving point.
If the point of a pencil is moved over a sheet
of paper, the trace left represents a line. A line
has length, but it has no breadth. A line has no
Fig. 3.2
end points. A line AB is written as AB . A line
may be named with small letters l, m, n, etc. we read them as line l, line m, line n etc.
A line has no end points as it goes on endlessly in both directions.

Ray
A ray has a starting point but has no end point. The starting
point is called the initial point.
Here OA is called the ray and it is written as OA . That is
the ray starts from O and passes through A.
57

Fig. 3.3

Chapter 3
Line Segment
Let AB be a straight line.
Two points C and D are taken on it. CD is a part
of AB. CD is called a line segment, and is written as CD .
A line segment has two end points.

Fig. 3.4

Plane
A plane is a flat surface which extends indefinitely in all directions. The upper
surface of a table, the blackboard and the walls are some examples of planes.

3.2. Symmetry
Symmetry is an important geometrical concept commonly seen in nature and is
used in every field of our life. Artists, manufacturers, designers, architects and others
make use of the idea of symmetry. The beehives, flowers, tree leaves, hand kerchief,
utensils have symmetrical design.

Fig. 3.5

Symmetry refers to the exact match in shape and
size between two halves of an object. If we fold a picture
in half and both the halves-left half and right half - match
exactly then we say that the picture is symmetrical.
For example, if we cut an apple into two equal
halves, we observe that two parts are in symmetry.

Tajmahal in Agra is a
symmetrical monument.
Fig. 3.6
58

Geometry
A butterfly is also an example of a symmetrical form. If a line is drawn down
the centre of the butterfly’s body, each half of the butterfly looks the same.

Fig. 3.7

Symmetry is of different types. Here we discuss about
1. Line of symmetry or axis of symmetry
2. Mirror symmetry
3. Rotational symmetry

1. Line of symmetry
In the Fig 3.8 the dotted lines divide the figure into two identical parts. If figure
is folded along the line, one half of the figure will coincide exactly with the other half.
This dotted line is known as line of symmetry.
When a line divides a given figure into two
equal halves such that the left and right halves matches
exactly then we say that the figure is symmetrical
about the line. This line is called the line of symmetry
or axis of symmetry.

Fig. 3.8

Activity 1:
Take a rectangular sheet of
paper. Fold it once lengthwise, so
that one half fits exactly over the
other half and crease the edges.
Now open it, and again fold it once
along its width.

Fig. 3.9

59

Chapter 3
In this paper folding,
You observe that a rectangle has two lines of symmetry.
Discuss: Does a parallelogram have a line of symmetry?
Activity 2:
One of the two set squares in your geometry box has angle of
0
0
0
measure 30 , 60 , 90 . Take two such identical set squares. Place them
side by side to form a ‘kite’ as shown in the Fig. 3.10.
How many lines of symmetry does the shape have?
You observe that this kite shape figure has one line of symmetry
about its vertical diagonal.
Fig. 3.10

Activity 3:
For the given regular polygons find the lines of symmetry by using paper folding
method and also draw the lines of symmetry by dotted lines.

Fig. 3.11

In the above paper foldings, you observe that
(i) An equilateral triangle has three lines of
symmetry.
(ii) A square has four lines of symmetry
(iii) A regular pentagon has five lines of
symmetry.

A polygon is said to be
regular if all its sides are
of equal length and all
its angles are of equal
measure.

(iv) A regular hexagon has six lines of symmetry.
Each regular polygon has as many lines of symmetry as it has sides.

60

Geometry

Identify the regular polygon
A circle has many lines of
symmetry.
Some objects and figures have no line of symmetry.
Make a list of English
alphabets which have
no line of symmetry
Fig. 3.12

The above figures have no line of symmetry; because these figures are not
symmetrical. We can say that these figures are asymmetrical.
To reflect an object means to produce its mirror image.

2. Mirror line symmetry
When we look into a mirror we see our image is behind the
mirror. This image is due to reflection in the mirror. We know that
the image is formed as far behind the mirror as the object is in front
Fig. 3.13
of it.
In the above figure if a mirror is placed along the line at the
middle, the half part of the figure reflects through the mirror creating the remaining
identical half. In other words, the line where the mirror is placed divides the figure into
two identical parts in Fig. 3.13. They are of the same size and one side of the line will
have its reflection exactly at the same distance on the other side. Thus it is also known
as mirror line symmetry.
While dealing with mirror reflection, we notice that the left-right changes as
seen in the figure.
Example 3.1
The figure shows the reflection of the
mirror lines.
61

Chapter 3
Exercise 3.1
1. Choose the correct answer :
		 i)
			
			

An isosceles triangle has
(A) no lines of symmetry
(C) three lines of symmetry

(B) one line of symmetry
(D) many lines of symmetry

		 ii) A parallelogram has
			 (A) two lines of symmetry
			 (C) no lines of symmetry

(B) four lines of symmetry
(D) many lines of symmetry

		 iii) A rectangle has
			 (A) two lines of symmetry
			 (C) four lines of symmetry

(B) no lines of symmetry
(D) many lines of symmetry

		 iv) A rhombus has
			 (A) no lines os symmetry
			 (C) two lines of symmetry

(B) four lines of symmetry
(D) six lines of symmetry

			v) A scalene triangle has
			 (A) no lines of symmetry
			 (C) one line of symmetry

(B) three lines of symmetry
(D) many lines of symmetry

2. Which of the following have lines of symmetry?

		 How many lines of symmetry does each have?
3. In the following figures, the mirror line (i.e. the line of symmetry) is given in
dotted line. Complete each figure performing reflection in the dotted (mirror) line.

62

Geometry
4. Complete the following table:
Shape

Rough figure

Number of lines of symmetry

Equilateral triangle
Square
Rectangle
Isosceles triangle
Rhombus

5. Name a triangle which has
(i) exactly one line of symmetry.
(ii) exactly three lines of symmetry.
(iii) no lines of symmetry.
6. Make a list of the capital letters of English alphabets which
(i) have only one line of symmetry about a vertical line.
(ii) have only one line of symmetry about a horizontal line.
(iii) have two lines of symmetry about both horizontal and vertical line
of symmetry.

3.3 Rotational Symmetry
Look at the following figures showing the shapes that we get, when we rotate
0
0
about its centre ‘O’ by an angle of 90 or 180

Fig. 3.14

Fig. 3.15
63

Chapter 3

Fig. 3.16
0

In the case of a square, we get exactly the same shape after it is rotated by 90
while in the case of a rectangle, we get exactly the same shape after it is rotated by
180° such figures which can be rotated through an angle less than 360° to get the same
shape are said to have rotational symmetry.

Angle of Rotation
The minimum angle through which the figure has to be rotated to get the original
figure is called the angle of rotation and the point about which the figure is rotated is
known as centre of rotation.
Activity 4:
Take two card board sheets and cut off
one equilateral triangle in each sheet such that
both the triangles are identical. Prepare a circle
on a card board and mark the degrees from 0 to
360 degree in the anticlockwise direction. Now
place one triangle exactly over the other and put
a pin through the centres of the figures. Rotate
the top figure until it matches with the lower
figure.
You observe that the triangle has been
rotated through an angle 120°.
Again rotate the top figure until it matches with the lower figure for the second
time. Now you observe that the top of figure has been rotated through an angle 240°
from the original position.
Rotate the top figure for the third time to match with the lower figure. Now the
top triangle has reached its original position after a complete rotation of 360° From
the above activity you observe that an equilateral triangle has angle of rotation 120°.
64

Geometry

Fig. 3.17

Angle of rotation of a hexagon

Fig. 3.18

In the above Fig. 3.15 to 3.18.
We get exactly the same shape of square, rectangle, equilateral triangle and
0
0
0
0
hexagon after it is rotated by 90 , 180 , 120 , 60 respectively.
Thus the angle of rotation of
0

(i)

a square is 90

(ii)

a rectangle is 180

(iii)

an equilateral triangle is 120

(iv)

a hexagon is 60

0

0

0

Order of rotational symmetry
The order of rotational symmetry is the number that tell us how many times a
figure looks exactly the same while it takes one complete rotation about the centre.
Thus if the angle of rotation of an object is x
Its order of rotational symmetry = 360
0
x

In Fig. 3.15 to 3.18.
65

0

Chapter 3
The order of rotational symmetry of
0

(i) a square is
(ii) a rectangle is
(iii) an equilateral triangle is

360 = 4
0
90
0
360 = 2
0
180
0
360 = 3
120
0

360 = 6.
0
60

(iv) a hexagon is
Example 3.2

The objects having no line of symmetry can have
rotational symmetry.
Have you ever made a paper wind mill? The paper wind
mill in the picture looks symmetrical. But you do not find any
line of symmetry. No folding can help you to have coincident
halves. However if you rotate it by 90° about the the centre,
the windmill will look exactly the same. We say the wind mill
has a rotational symmetry.

In a full turn, there are four positions (on rotation through the angles
0
0
0
90 , 180 270 and 360 ) in which the wind mill looks exactly the same. Because of
this, we say it has a rotational symmetry of order 4.
0

AcActivity 5:tivity: 5
As shown in figure cut out a card board or paper triangle. Place
it on a board and fix it with a drawing pin at one of its vertices. Now
0
rotate the triangle about this vertex, by 90 at a time till it comes to
its original position.

66

Geometry
You observe that, for every 90 you have the following figures (ii to v).
0

(i)

(ii)

(iii)

(iv)

(v)

The triangle comes back to its original position at position (v) after rotating
0
0
through 360 . Thus the angle of rotation 0 of this triangle is 360 and the order of
rotational symmetry of this triangle is 3600 = 1 .
360

Exercise 3.2
1. Choose the correct answer:
		 i)

		

The angle of rotation of an equilateral triangle is

(A)

60

0

(B) 90

0

(C) 120

0

(D) 180

0

		 ii) The order of rotational symmetry of a square is

		

(A)

2

(B) 4		

(C) 6		

(D) 1.

0

		 iii) The angle of rotation of an object is 72 then its order of rotational symmetry is

		

(A)

1

(B) 3		

(C) 4		

(D) 5

		 iv) The angle of rotation of the letter ‘S’ is

		

(A)

90

0

(B) 180

0

(C) 270

0

(D) 360

0

		 v) The order of rotational symmetry of the letter ‘V’ is one then its angle of
rotation is

		

(A)

60

0

(B) 90

0

(C) 180

67

0

(D) 360

0

Chapter 3

2. The following figures make a rotation to come to the new position about a given
centre of rotation. Examine the angle through which the figure is rotated.

(i)

(ii)

(iii)

(iv)

3. Find the angle of rotation and the order of rotational symmetry for the
following figures given that the centre of rotation is ‘O’.

(i)

(ii)

(iii)

(iv)

4. A circular wheel has eight spokes.

		 What is the angle of rotation and the order of rotation?

3.3 Angle
Two rays starting from a common point form an angle.
In + AOB, O is the vertex, OA and OB are the two rays.
Fig. 3.19

Types of angles
(i) Acute angle:
An angle whose measure is greater than 0° but less than
90 is called an acute angle.
0

0

0

0

0

Example: 15 , 30 , 60 , 75 , In Fig. 3.20 + AOB = 30
is an acute angle.
68

0

Fig. 3.20

Geometry
(ii) Right angle
0

An angle whose measure is 90 is called a right angle.
0

In Fig. 3.21 + AOB = 90 is a right angle.
Fig. 3.21

(iii) Obtuse angle
0

An angle whose measure is greater than 90 and less
0
than 180 is called an obtuse angle.
0

0

0

Example: 100 , 110 , 120 , 140

0

0

In Fig. 3.22 + AOB = 110 is an obtuse angle.
Fig. 3.22

(iv) Straight angle
When the rays of an angle, are opposite rays forming
a straight line. The angle thus formed is a straight angle and
0
0
whose measure is 180 In Fig. (3.23) + AOB = 180 is a straight
angle.

Fig. 3.23

(v) Reflex angle
0

An angle whose measure is more than 180 but less
0
than 360 is called a reflex angle. In Fig. 3.24 + AOB = 220°
is a reflex angle.

Fig. 3.24

(vi) Complete angle
In Fig. 3.25
The angle formed by OP and OQ is one complete
0
circle, that is 360 .Such an angle is called a complete angle

Fig. 3.25

Related Angles
(i) Complementary angles
0

If the sum of the measures of two angle is 90 , then the
two angles are called complementary angles. Here each angle
is the complement of the other.
0

0

The complement of 30 is 60 and the complement of
0

60 is 30

0

69

Fig. 3.26

Chapter 3

(ii) Supplementary angles
If the sum of the measures of two angle is
180 , then the two angles are called supplementary
angles. Here each angle is the supplementary of
the other.
0

0

Fig. 3.27

0

The supplementary angle of 120 is 60 and the supplementary angle of 60° is 120

Identify the following pairs
of angles are complementary
or supplementary

Fill in the blanks.

(a) 80 and 10 _____

(b) Complement of 30° is ___

(b) 70 and110 _____

(c) Supplement of 60 is ____

0

0

0

(a) Complement of 85 is____

0

0

0

0

0

(d) Supplement of 90 is_____

(c) 40 and50 ______
0

0

(d) 95 and85 _______
0

0

(e) 65 and 115 ______
0

0

Intersecting lines

Fig. 3.28

Look at the Fig. 3.28. Two lines l1 and l2 are shown. Both the lines pass through
a point P. We say l1 and l2 intersect at P. If two lines have one common point, they are
called intersecting lines. The common point ‘P’ is their point of intersection.
70

Geometry
Angles in intersecting lines
C

When two lines intersect at a point
angles are formed.

O
In Fig. 3.29 the two lines AB
and CD intersect at a point ‘O’, + COA,
+ AOD, + DOB, + BOC are formed.
Among the four angles two angles are
Fig. 3.30
Fig. 3.29
acute and the other two angles are obtuse.
But in figure 3.30 if the two intersecting lines are perpendicular to each other then the
four angles are at right angles.

Adjacent angles
If two angles have the same vertex and a common ray,
then the angles are called adjacent angles.
In Fig. 3.31 + BAC and + CAD are adjacent angles
(i.e . + x and + y) as they have a common ray AC, a common
vertex A and both the angle + BAC and + CAD are on either
side of the common ray AC .

+ ROP and

Fig. 3.31

Look at the following figure

+ QOP

are not adjacent angle.
Why?

Open a book looks like the above
figure. Is the pair of angles are
adjacent angles?

(i) Adjacent angles on a line.
When a ray stands on a straight line two angles are
formed. They are called linear adjacent angles on the line.
Fig. 3.32
71

Chapter 3
In Fig. 3.32 the ray OC stands on the line AB. + BOC and + COA are the two
adjacent angles formed on the line AB. Here ‘O’ is called the common vertex, OC is called
the common arm. The rays OA and OB lie on the opposite sides of the common ray OC.
Two angles are said to be linear adjacent angles on a line if they have a
common vertex, a common ray and the other two rays are on the opposite sides
of the common ray.

(ii) The sum of the adjacent angles on a line is 180°

Fig. 3.33

Fig.3.34
0

In Fig. 3.33 + AOB = 180 is a straight angle.
In Fig. 3.34 The ray OC stands on the line AB. + AOC and + COB are adjacent
0
angles. Since + AOB is a straight angle whose measure is 180
+ AOC + + COB = 180

0

From this we conclude that the sum of the adjacent angles on a line is 180

0

Note 1: A pair of adjacent angles whose non common rays are opposite rays.
Note 2: Two adjacent supplementary angles form a straight angle.

Are the angles marked 1 and 2 adjacent?
If they are not adjacent, Justify your answer.
72

Geometry

A vegetable chopping board

A pen stand

The chopping blade makes a linear pair of angles with the board.
The pen makes a linear pair of angles with the stand.
Discuss :
(i) Can two adjacent acute angles form a linear pair?
(ii) Can two adjacent obtuse angles form a linear pair?
(iii) Can two adjacent right angles form a linear pair?
(iv) Can an acute and obtuse adjacent angles form a linear pair?

(iii) Angle at a point
In Fig. 3.35, four angles are formed at the point ‘O’. The
0
sum of the four angles formed is 360 .
(i.e) + 1 + + 2 + + 3 + + 4 = 360

O

0

Fig. 3.35

(iv) Vertically opposite angles
If two straight lines AB and CD intersect at a point
‘O’. Then + AOC and + BOD form one pair of vertically
opposite angles and + DOA and + COB form another pair
of vertically opposite angles.

Fig. 3.36

The following are some real life
example for vertically
Opposite angles

73

Chapter 3
Activity 6: Draw two lines ‘l’ and ‘m’, intersecting at a point ‘P’ mark + 1,
+ 2, + 3 and + 4 as in the Fig. 3.37.
Take a trace copy of the figure on a transparent sheet. Place the copy on the
original such that + 1 matches with its copy, + 2, matches with its copy.. etc...
Fix a pin at the point of intersection of two lines ‘l’ and ‘m’ at P. Rotate the copy
0
by 180 . Do the lines coincide again?

Fig. 3.37

You find that + 1 and + 3 have interchanged their positions and so have + 2
and + 4. (This has been done without disturbing the position of the lines).
Thus + 1 = + 3 and + 2 = + 4.
From this we conclude that when two lines intersect, the vertically opposite
angles are equal.
Now let us try to prove this using Geometrical idea.

1

Let the lines AB and CD intersect at ‘O’ making angles + 1,
+ 2, + 3 and + 4.
0

Now + 1 = 180 - + 2 " (i)
0

( Since sum of the adjacent angle on a line 180 )
0

+ 3 = 180 - + 2 " (ii)
0

( Since sum of the adjacent angle on a line 180 ).
From (i) and (ii)
+ 1 = + 3 and similarly we prove that + 2 = + 4.

Example 3.3
In the given figure identify
(a) Two pairs of adjacent angles.
(b) Two pairs of vertically opposite angles.

74

Fig. 3.38

Geometry
Solution
(a) Two pairs of adjacent angles are
(i) + EOA, + COE since OE is common to + EOA and + COE
(ii) + COA, + BOC since OC is common to + COA and + BOC
(b) Two pairs of vertically opposite angles are i) + BOC, + AOD
							

ii) + COA, + DOB.

Example 3.4
Find the value of x in the given figure.
Solution
+ BCD + + DCA = 180

			

0

0

(Since + BCA = 180 is a straight angle)
45° + x = 180°
x = 180° – 45°

		 = 135°
0

` The value of x is 135 .

Example 3.5
Find the value of x in the given figure.
Solution
+ AOD + + DOB = 180

			

0

(Since + AOB = 180 is a straight angle)
0

100 + x = 180

			
			

0

0

0

x = 180 - 100

= 80

0

0

0

` The value of x is 80 .

Example 3.6
Find the value of x in the given figure.
Solution
0

+ POR + + ROQ = 180
0
			
( Since + POQ = 180 is a straight angle)
75

Chapter 3
x + 2x = 180

		

3x = 180

		

x = 180
3
0
= 60

		

0

0

0

` The value of x is 60

0

Example 3.7
Find the value of x in the given figure.
Solution
+ BCD + + DCA = 180

0

(Since + BCA = 180 is a straight angle)

			
3x + x = 180

0

4x = 180

0

x = 180
4
0
= 45

0

0

` The value of x is 45

0

Example 3.8
Find the value of x in the given figure.
Solution
+ BCD + + DCE + + ECA = 180

0

(Since + BCA = 180 is a straight angle)

					
		

0

0

0

0

0

0

40 + x + 30 = 180

			

x + 70 = 180

0

x = 180 - 70

				

= 110

				
` The value of x is 110

0

0

0

Example 3.9
Find the value of x in the given figure.
Solution
0

0

+ BCD + + DCE + + ECA = 180 (Since + BCA = 180 straight angle).
76

Geometry
0

0

0

0

0

x + 20 + x + x + 40 = 180
3x + 60 = 180

		

0

			

3x = 180 - 60

			

3x = 120

0

0
x = 120 = 40
3

		
		

0

` The value of x is 40

0

Example 3.10
Find the value of x in the given figure.

Solution
+ BOC + + COA + + AOD + + DOE + + EOB = 360

0

(Since angle at a point is 360 )

					
2x + 4x + 3x + x + 2x = 360

0

			

12x = 360

			

x = 360
12
0
= 30

			

` The value of x is 30

0

0

0

Example 3.11
Find the value of x in the given figure.
Solution
+ BOD + + DOE + + EOA = 180
0

0

(Since + AOB = 180 is straight angle)
		

2x + x + x = 180

0

0

				

4x = 180

				

x = 180
4
0
= 45

				
		

0

` The value of x is 45

0

0

77

Chapter 3
Exercise: 3.3
1. Choose the correct answer:
		 i)

The number of points common to two intersecting line is

(A) one
		 ii)

(B) Two

0

(B) 180

0

(C) 270

0

In the figure + COA will be
0

(A) 80 		
0

(C) 100 		
		 iv)

(D) four

The sum of the adjacent angles on a line is

(A) 90 		
		 iii)

(C) three

(B) 90

0

(D) 95

0

In the figure + BOC will be
0

(A) 80 		
0

(C) 100 		

(B) 90

0

(D) 120

0

		 v)

In the figure CD is perpendicular to AB.

				

Then the value of + BCE will be
0

(B) 35

0

(D) 50

(A) 45 		
(C) 40 		

0

0

2. Name the adjacent angles in the following figures

3. Identify the vertically opposite angles in the figure:

4. Find + B if + A measures?
		 (i) 30

0

		 (ii) 80

0

		 (iii) 70°
		 (iv) 60°
		 (v) 45

0

78

(D) 360

0

Geometry
5. In figure AB and CD be the intersecting lines if
		 + DOB = 35 find the measure of the other angles.
0

6. Find the value of x in the following figures :

(i)

(ii)

(iv)

(v)

(iii)

(vi)

7. In the following figure two lines AB and CD intersect at
		 the point O. Find the value of x and y.

8. Two linear adjacent angles on a line are 4x and ^3x + 5h. Find the value of x.

79

Chapter 3

1. Symmetry refers to the exact match in shape and size between two halves of
an object.
2. When a line divides a given figure into two equal halves such that the left and
right halves matches exactly then we say that the figure is symmetrical about
the line. This line is called the line of symmetry or axis of symmetry.
3. Each regular polygon has as many lines of symmetry as it has sides.
4. Some objects and figures have no lines of symmetry.
5. Figures which can be rotated through an angle less than 360° to get the same
shape are said to have rotational symmetry.
6. The order of rotational symmetry is the number that tell us how many times
a figure looks exactly the same while it takes one complete rotation about the
centre.
7. The objects having no line of symmetry can have rotational symmetry.
8. If two angles have the same vertex and a common ray, then the angles are
called adjacent angles.
9. The sum of the adjacent angles on a line is 180°.
10. When two lines intersect, the vertically opposite angles are equal.
11. Angle at a point is 360°.

80

Practical Geometry

Practical Geometry
4.1 Introduction
This chapter helps the students to understand and confirm the concepts they
have learnt already in theoretical geometry. This also helps them to acquire some
basic knowledge in geometry which they are going to prove in their later classes. No
doubt, all the students will do the constructions actively and learn the concepts easily.
In the previous class we have learnt to draw a line segment, the parallel lines,
the perpendicular lines and also how to construct an angle.
Here we are going to learn about the construction of perpendicular bisector of a
line segment, angle bisector, some angles using scale and compass and the construction
of triangles.

Review
To recall the concept of angles, parallel lines and perpendicular lines
from the given figure.
We shall identify the points, the line segments, the angles, the parallel
lines and the perpendicular lines from the figures given below in the table.
S.
No.

1

Figures

Points
identified

Line
segment
identified

A, B,
C and
D

AB,
BC,
CD,
AD, and
BD

Angles identified

Parallel
lines

1 - +BAD (+A)
2 - +DCB (+C) AB || DC
BC || AD
3 - +DBA
4 - +CBD

81

Perpendicular lines

AB = AD
AB = BC
BC = CD
CD = AD

Chapter 4

Sl.
No.

Figures

Points
Line
iden- segment
tified identified

Angles
identified

Parallel
lines

1.

2.

4.2 Perpendicular bisector of the given line segment
(i) Activity : Paper folding
•

Draw a line segment AB on a sheet of paper.

X

•

Fold the paper so that the end point B lies on A.
Make a crease XY on the paper.
Y
X

•

Unfold the paper. Mark the point O where the line
of crease XY intersects the line AB.

O

Y

82

Perpendicular
lines

Practical Geometry
•

By actual measurement we can see that OA = OB and the line of crease XY
is perpendicular to the line AB.

The line of crease XY is the perpendicular bisector of the line AB.
The perpendicular bisector of a line segment is a
perpendicular line drawn at its midpoint.

(ii) To construct a perpendicular bisector to a given line segment.
Step 1 : Draw a line segment
AB of the given
measurement.

Step 2 : With ‘A’ as centre draw
arcs of radius more than
half of AB, above and
below the line AB.

Step 3 : With ‘B’ as centre and
with the same radius
draw two arcs. These
arcs cut the previous
arcs at P and Q.

83

Chapter 4

Step 4 : Join PQ. Let PQ intersect
AB at ‘O’.

		 PQ is a perpendicular
bisector of AB.

Mark any point on the perpendicular bisector PQ. Verify that it
is equidistant from both A and B.

The perpendicular bisector of a line
segment is the axis of symmetry for the
line segment.

Example 4.1
Draw a perpendicular bisector to the
line segment AB = 8 cm.
Solution
Step 1 : Draw the line segment
AB = 8cm.
Step 2 : With ‘A’ as centre draw
arcs of radius more than
half of AB above and
below the line AB.
Step 3 : With ‘B’ as centre draw
the arcs of same radius to
cut the previous arcs at X
and Y.
84

Can there be more than one
perpendicular bisector for the
given line segment?

Practical Geometry
Step 4 : Join XY to intersect the line AB at O.
		 XY is the perpendicular bisector of AB.
1. With PQ = 6.5 cm as diameter draw a circle.
2. Draw a line segment of length 12 cm. Using compass divide
it into four equal parts. Verify it by actual measurement.
3. Draw a perpendicular bisector to a given line segment
AC. Let the bisector intersect the line at ‘O’. Mark the
points B and D on the bisector at equal distances from O.
Join the points A, B, C and D in order. Verify whether all
lines joined are of equal length.
Think!
In the above construction mark the points B and D on the
bisector, such that OA = OB = OC = OD. Join the points A, B,
C and D in order. Then
1. Do the lines joined are of equal length?
2. Do the angles at the vertices are right angles?
3. Can you identify the figure?

4.3 Angle Bisector
(i) Activity : Paper folding
•

Take a sheet of paper and mark a point O on it.
With O as initial point draw two rays OA and
OB to make + AOB.

C

•

Fold the sheet through ‘O’ such that the rays
OA and OB coincide with each other and make
a crease on the paper.

85

Chapter 4
•

Let OC be the line of crease on the paper after
unfold. By actual measurement, + AOC and
+ BOC are equal.

•

So the line of crease OC divides the given angle
into two equal parts.

•

This line of crease is the line of symmetry for + AOB.

•

This line of symmetry for + AOB is called the angle bisector.
The angle bisector of a given angle is the line of symmetry
which divides the angle into two equal parts.

(ii) To construct an angle bisector of the given angle using scale and compass

Step 1 : Construct an angle of given
		 measure at O.

Step 2 : With ‘O ’ as centre draw an
arc of any radius to cut the
rays of the angle at A and B.

Step 3 : With ‘A’ as centre draw an
arc of radius more than half
of AB, in the interior of the
given angle.

86

Practical Geometry

Step 4 : With ‘B’ as centre draw an
arc of same radius to cut the
previous arc at ‘C’.

Step 5 : Join OC.
		 OC is the angle bisector of the
given angle.

Mark any point on the angle bisector OC. Verify that
it is equidistant from the rays OA and OB.

Example 4.2
Construct + AOB = 80° and draw its angle bisector.
Solution
Step 1 : Construct + AOB = 80° angle at
the point ‘O’ using protractor.
Step 2 : With ‘O’ as centre draw an arc
of any radius to cut the rays OA
and OB at the points X and Y
respectively.
Step 3 : With ‘X’ as centre draw an arc
of radius more than half of XY
in the interior of the angle.

87

C

Chapter 4
Step 4 : With ‘Y’ as centre draw an arc of the same radius to cut the previous
arc at C. Join OC.
		 OC is the angle bisector of the given angle 80°.

Draw an angle of measure 120° and divide
into four equal parts.
Exercise 4.1
1. Draw the line segment AB = 7cm and construct its perpendicular bisector.
2. Draw a line segment XY = 8.5 cm and find its axis of symmetry.
3. Draw a perpendicular bisector of the line segment AB = 10 cm.
4. Draw an angle measuring 70° and construct its bisector.
5. Draw an angle measuring 110° and construct its bisector.

6. Construct a right angle and bisect it using scale and compass.

1. Draw a circle with centre ‘C’ and radius 4 cm. Draw any chord AB.
Construct perpendicular bisector to AB and examine whether it passes
through the centre of the circle.
2. Draw perpendicular bisectors to any two chords of equal length in a circle.
(i) Where do they meet? (ii) Verify whether the chords are at a same
distance from the centre.
3. Plot three points not on a straight line. Find a point equidistant from them.
Hint: Join all the points in order. You get a triangle. Draw perpendicular
bisectors to each side. They meet at a point which is equidistant from the
points you have plotted. This point is called circumcentre.

88

DATA HANDLING
5.1 Introduction
Data Handling is a part of statistics. The word statistics is derived from the
Latin word “ Status”. Like Mathematics, Statistics is also a science of numbers. The
numbers referred to here are data expressed in numerical form like,
(i) Marks of students in a class
(ii) Weight of children of particular age in a village
(iii) The amount of rainfall in a region over a period of years.
Statistics deals with the methods of collection, classification, analysis and
interpretation of such data.
Any collection of information in the form of numerical figures giving the
required information is called data.

Raw data
The marks obtained in Mathematics test by the students of a class is a collection
of observations gathered initially. The information which is collected initially and
presented randomly is called a raw data.
The raw data is an unprocessed and unclassified data.

Grouped data
Some times the collected raw data may be huge in number and it gives us no
information as such. Whenever the data is large, we have to group them meaningfully
and then analyse.
The data which is arranged in groups or classes is called a grouped data.

Collection of data
The initial step of investigation is the collection of data. The collected data
must be relevant to the need.

89

Chapter 5
Primary data
For example, Mr. Vinoth, the class teacher of standard
VII plans to take his students for an excursion. He asks the Collect some possible
information from the
students to give their choice for
people in your locality.
(i) particular location they would like to go
(ii) the game they would like to play
(iii) the food they would like to have on their trip
For all these, he is getting the information directly from the students. This type
of collection of data is known as primary data.

5.2 Collecting and Organizing of Continuous Data
Secondary data
Mr. Vinoth, the class teacher of standard VII is collecting the information about
weather for their trip. He may collect the information from the internet, news papers,
magazines, television and other sources. These external sources are called secondary
data.

Variable
As far as statistics is concerned the word variable means a measurable quantity
which takes any numerical value within certain limits.
Few etxamples are (i) age, (ii) income, (iii) height and (iv) weight.

Frequency
Suppose we measure the height of students in a school. It is possible that a
particular value of height say 140 cm gets repeated. We then count the number of times
the value occurs. This number is called the frequency of 140 cm.
The number of times a particular value repeats itself is called its frequency.

Range
The difference between the highest value and the lowest value of a particular
data is called the range.
Example 5.1
Let the heights (in cm) of 20 students in a class be as follows.
120, 122, 127, 112, 129, 118, 130, 132, 120, 115
124, 128, 120, 134, 126, 110, 132, 121, 127, 118.
Here the least value is 110 cm and the highest value is 134 cm.
Range = Highest value – Lowest value
				
= 134 – 110 = 24
90

Data Handling
Class and Class Interval
In the above example if we take 5 classes say 110 - 115, 115 - 120, 120 - 125,
125 - 130, 130 - 135 then each class is known as class interval. The class interval
must be of equal size. The number of classes is neither too big nor too small. i.e The
optimum number of classes is between 5 and 10.

Class limits
In class 110 - 115, 110 is called the lower limit of the class and 115 is the upper
limit of the class.

Width (or size) of the class interval:
The difference between the upper and lower limit is called the width of the
class interval. In the above example, the width of the class interval is 115 - 110 = 5. By
increasing the class interval, we can reduce the number of classes.
There are two types of class intervals. They are (i) Inclusive form and
(ii) Exclusive form.

(i) Inclusive form
In this form, the lower limit as well as upper limit will be included in that class
interval. For example, in the first class interval 110 - 114, the heights 110 as well as 114
are included. In the second class interval 115 - 119, both the heights 115 and 119 are
included and so on.

(ii) Exclusive form
In the above example 5.1, in the first class interval 110 - 115, 110 cm is included
and 115 cm is excluded. In the second class interval 115 is included and 120 is excluded
and so on. Since the two class intervals contain 115 cm, it is customary to include
115 cm in the class interval 115 - 120, which is the lower limit of the class interval.

Tally marks
In the above example 5.1, the height 110 cm, 112 cm belongs to the class interval
110 - 115. We enter | | tally marks. Count the tally marks and enter 2 as the frequency
in the frequency column.
If five tally marks are to be made we mark four tally marks first and the fifth one
is marked across, so that | | | | represents a cluster of five tally marks.
To represent seven, we use a cluster of five tally marks and then add two more
tally marks as shown |||| ||.
91

Chapter 5
Frequency Table
A table which represents the data in the form of three columns, first column
showing the variable (Number) and the second column showing the values of the
variable (Tally mark) and the third column showing their frequencies is called a
frequency table (Refer table 5.3).
If the values of the variable are given using different classes and the frequencies
are marked against the respective classes, we get a frequency distribution. All the
frequencies are added and the number is written as the total frequency for the entire
intervals. This must match the total number of data given. The above process of
forming a frequency table is called tabulation of data.
Now we have the following table for the above data. (Example 5.1)
Inclusive form
Class Interval

Tally Marks

Frequency

110 - 114

||

2

115 - 119

|||

3

120 - 124

|||| |

6

125 - 129

||||

5

130 - 134

||||

4

Total

20

Table 5.1

Exclusive form
Class Interval

Tally Marks

Frequency

110 - 115

||

2

115 - 120

|||

3

120 - 125

|||| |

6

125 - 130

||||

5

130 - 135

||||

4

Total

20

Table 5.2
92

Data Handling
Frequency table for an ungrouped data
Example 5.2
Construct a frequency table for the following data.
5, 1, 3, 4, 2, 1, 3, 5, 4, 2
1, 5, 1, 3, 2, 1, 5, 3, 3, 2.
Solution
From the data, we observe the numbers 1, 2, 3, 4 and 5 are repeated. Hence
under the number column, write the five numbers 1, 2, 3, 4, and 5 one below the other.
Now read the number and put the tally mark in the tally mark column against
the number. In the same way put the tally mark till the last number. Add the tally
marks against the numbers 1, 2, 3, 4 and 5 and write the total in the corresponding
frequency column. Now, add all the numbers under the frequency column and write
it against the total.
Number
1
2
3
4
5

Tally Marks
||||
||||
||||
||
||||
Total

Frequency
5
4
5
2
4
20

Table 5.3

In the formation of Frequency distribution for the given data values, we should
(i)

select a suitable number of classes, not very small and also not very large.

(ii) take a suitable class - interval or class width and
(iii) present the classes with increasing values without any gaps between classes.

Frequency table for a grouped data
Example 5.3
The following data relate to mathematics marks obtained by 30 students in
standard VII. Prepare a frequency table for the data.
25, 67, 78, 43, 21, 17, 49, 54, 76, 92, 20, 45, 86, 37, 35
60, 71, 49, 75, 49, 32, 67, 15, 82, 95, 76, 41, 36, 71, 62
Solution:
The minimum marks obtained is 15.
The maximum marks obtained is 95.
93

Chapter 5
Range = Maximum value – Minimum value
		 = 95 – 15
		 = 80
Choose 9 classes with a class interval of 10. as 10 - 20, 20 - 30,g ,90 - 100. The
following is the frequency table.
Class Interval (Marks)
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
70 - 80
80 - 90
90 - 100

Tally Marks
||
|||
||||
||||
||
||||
|||| |
||
||
Total

Frequency
2
3
4
5
2
4
6
2
2
30

Table 5.4

5.2 Continuous grouped Frequency distribution Table
To find the class limits in continuous grouped frequency distribution.

Steps to do
(i) Find the difference between the upper limit of the first class and lower
limit of the second class.
(ii) Divide the difference by 2. Let the answer be x.
(iii) Subtract ‘x’ from lower limits of all the class intervals.
(iv) Add ‘x’ to all the upper limits of all the class intervals. Now the new
limits will be true class limits.
Example 5.4
Form the frequency distribution table for the following data which gives the
ages of persons who watched a particular channel on T.V.
Class Interval
(Age)
Number of
persons

10 -19

20 -29

30 - 39

40 - 49

50 - 59

60 - 69

45

60

87

52

25

12

94

Data Handling
Solution:
In this table, the classes given here have gaps. Hence we rewrite the classes
using the exclusive method.
Difference between upper limits of first class and lower limits of second class
		 = 20 – 19 = 1
Divide the difference by 2 then,
x = 1 = 0.5
2

Now subtract 0.5 from lower limits and add 0.5 to the upper limits. Now we
get continuous frequency distribution table with true class limits.
Class Interval
(Age)

Frequency
(Number of persons)

9.5 - 19.5

45

19.5 - 29.5

60

29.5 - 39.5

87

39.5 - 49.5

52

49.5 - 59.5

25

59.5 - 69.5

12
Table 5.5

Exercise 5.1
1. Choose the correct answer :
i) The difference between the highest and lowest value of the variable in the given
data. is called.
(A) Frequency
(B) Class limit
(C) Class interval (D) Range
ii) The marks scored by a set of students in a test are 65, 97, 78, 49, 23, 48, 59, 98.
The range for this data is
(A) 90
(B) 74
(C) 73
iii) The range of the first 20 natural numbers is

(D) 75

(A) 18
(B) 19
(C) 20
iv) The lower limit of the class interval 20 - 30 is

(D) 21

(A) 30
(B) 20
(C) 25
v) The upper of the class interval 50 - 60 is

(D) 10

(A) 50

(B) 60

(C) 10
95

(D) 55

Chapter 5
2. Construct a frequency table for each of the following data:
			

10, 15, 13, 12, 14, 11, 11, 12, 13, 15

			

11, 13, 12, 15, 13, 12, 14, 14, 15, 11

3. In the town there were 26 patients in a hospital.
		 The number of tablets given to them is given below. Draw a frequency table for
the data.
			

2, 4, 3, 1, 2, 2, 2, 4, 3, 5, 2, 1, 1, 2

			

4, 5, 1, 2, 5, 4, 3, 3, 2, 1, 5, 4.

4. The number of savings book accounts opened in a bank during 25 weeks are
given as below. Form a frequency table for the data:
			

15, 25, 22, 20, 18, 15, 23, 17, 19, 12, 21, 26, 30

			

19, 17, 14, 20, 21, 24, 21, 16, 22, 20, 17, 14

5. The weight (in kg) 20 persons are given below.
			

42, 45, 51, 55, 49, 62, 41, 52, 48, 64

			

52, 42, 49, 50, 47, 53, 59, 60, 46, 54

		 Form a frequency table by taking class intervals 40 - 45, 45 - 50, 50 - 55,
55 - 60 and 60 - 65.
6. The marks obtained by 30 students of a class in a mathematics test are given
below.
			

45, 35, 60, 41, 8, 28, 31, 39, 55, 72, 22, 75, 57, 33, 51

			

76, 30, 49, 19, 13, 40, 88, 95, 62, 17, 67, 50, 66, 73, 70

		 Form a grouped frequency table:
7. Form a continuous frequency distribution table from the given data.
Class Interval
(weight in kg.)

21 - 23

24 - 26

27 - 29

30 - 32

33 - 35

36 - 38

Frequency
(Number of children)

2

6

10

14

7

3

8. The following data gives the heights of trees in a grove. Form a continuous
frequency distribution table.
Class Interval
(Height in metres)

2 -4

5-7

8 - 10

11 - 13

14 - 16

Frequency
(Number of trees)

29

41

36

27

12

96

Data Handling

1. Any collection of information in the form of numerical figures giving the
required information is called data.
2. The raw data is an unprocessed and unclassified data.
3. The data which is arranged in groups (or classes) is called a grouped data.
4. The number of times a particular value repeats itself is called its frequency.
5. Range = Highest value – Lowest value.
6. The difference between the upper and the lower limit is called the width of the
class interval.

97

Answers

Answers
Chapter - 1
Exercise 1.1
1. i) D
2. i) 0
3. i) – 6
		vii) 25
		xiii) –1440
4. i) – 135
5. ` 645

ii)
ii)
ii)
viii)
xiv)
ii)
6.

B
iii) C
iv) B
–5
iii) 5
iv) 0
– 25
iii) 651
iv) – 316 v) 0
25
ix) 42
x) – 24
xi) 1890
256
xv) 6000 xvi) 10800
16
iii) 182
iv) – 800 v) 1
75 marks			
7. `1500			

ii)
ii)
viii)
ii)

A
10
2
20

vi) 1320
xii) – 1890
vi) 0
8. `240

Exercise 1.2
1. i) D
2. i) – 5
		vii) – 1
3. i) 20
4.		 – 5

iii)
iii)
ix)
iii)

C
4
2
– 400

iv) A
iv) – 1
x) 6

v) – 6

vi) – 9

v) 14
3

vi) 20

v) 288

vi) 16

Exercise 1.3
1. i) 24
5
77
		vii)
4
2. i) 14

ii) 9
7

iii) 2

iv) 3

viii) 10

ix) 8

x) 24

ii) 63

iii) 16

iv) 25

		vii) 9

viii) 70

ix) 25

x) 50

iii) 9 3
5

iv) 64 2
7

v) 52 1
2

vi) 85 1
2

iv) 7
18
iv) 9
16

v) 1

2
63
vi) 48
35

iv) 20 1
8

v) 59 13
16

ii) 19 4
3. i) 26 1
4
5
4.	  Vasu drank 4 litres.
Exercise 1.4
1. i) 1
2. i) 22
27
3. i) 2 4
15
4.	  55 km

7
iii) 7
12
12
ii) 1
iii) 1
5
4
ii) 4 29
iii) 7 1
40
2
5. 12 1 hrs
4
ii)

98

v) 9
2

vi)

Answers
Exercise 1.5
1. i) 7
5

ii) 9
4
5
		vii) 13
viii)
7
1
2. i)
ii) 1
15
54
8
35
3. i)
ii)
5
36
4.	  21 uniforms			

iii)

7
10

iv) 4
9

v)

2
33

vi) 9

iv) 1
12
7
iii) 4
iv) 1 11
12
16
5. 40 km/hour
iii)

1
6

Exercise 1.6
1. i) A

ii) C

iiii) B

iv) D

2. i) - 20 , - 19 , - 18 , - 17 			
15
15
15
15

ii) 7 , 6 , 5 , 4
6 6 6 6

		iii) 48 , 47 , 46 , 45
28 28 28 28
3. i) - 3
4

ii) - 3
8

iii) - 3
5

iv) - 5
3

1. i) C

ii) C

iii) D

iv) D

2. i) 18
5

ii) 24
13

iii) 2

iv) - 12
13

viii) – 3

ix) 24
7

x) - 13
30

ii) 4

iii) - 9
44

iv) - 5
16

v) - 1
2

5.		 i, iv, v
Exercise 1.7

		vii) - 43
21
3. i) 1

v) 13
3

vi) 19
42

v) 23
20

vi) – 1

		vii) - 69
26

viii) - 41
60

ix) - 1
27

x) 1
12

4. i) 2
35

ii) 1
4

iii) 19
12

iv) 3
2

v) - 43
28

5. i) 4 7
11

ii) – 3 1
2

iii) 1 7
11

iv) 5 3
4

v) – 1 17 vi) – 4 7
40
132

		vii) – 6 41 viii) – 3 7
42
210
6. 7 		
4
9. 18 3 kg.
4

7. 4
5

8. 13 17 kg.
20

10. 3 9 kg.
10
99

Answers
Exercise 1.8
1. i) C

ii) B

iii) A

iv) A

2. i) - 72
25

ii) - 35
169

iii) - 7
24

iv) - 12
11

3. i) - 15
4

ii) – 5

iii) 26 98 iv) 66 44 			
125
375

4. i) 16
81

ii) - 3
2

iii) - 8
7

iv) – 9 3
43

5. 9 		
7

6. 3
2

v) – 20

vi) 2
9
v) 45
28

Exercise 1.9
1. i) C

ii) C

iii) A

iv) C

2. i) 2.1

ii) 40.5

iii) 17.1

iv) 82.8

v) 0.45

vi) 1060.15

ix) 10.34

x) 1.041

xi) 4.48

xii) 0.00125

iii) 4567

iv) 2690.8 v) 3230

		vii) 2.58

viii) 1.05

		xiii) 2.108

xiv) 0.0312

3. i) 14

ii) 468

vi) 17140

		vi) 478
4.	  51.5 cm2

5. 756 km.

Exercise 1.10
1. i) A

ii) B

iii) C

iv) B

2. i) 0.3

ii) 0.09

iii) 1.16

iv) 10.8

v) 196.3

3. i) 0.68

ii) 4.35

iii) 0.09

iv) 4.43

v) 37.348 vi) 0.079

4. i) 0.056

ii) 0.007

iii) 0.0069 iv) 7.436

v) 0.437

5. i) 0.0089			

ii) 0.0733			

iii) 0.04873

		iv) 0.1789			

v) 0.0009			

vi) 0.00009

6. i) 2

ii) 160

7.	  23 km

8. 10.5 kg

iii) 12.5

iv) 8.19

v) 2

9. 9Books 10. 42.2 km/hour

vi) 3.04
vi) 0.7873

vi) 35
11. 14.4

Exercise 1.11
1. i) A

ii) A

iii) C

iv) C

2. i) 256

ii) 27

iii) 1331

iv) 1728

v) 28561

vi) 0

3. i) 76

ii) 15

iii) 06

iv) b5

v) 22a4

vi) (1003)3

4. i) 23 × 33

ii) 35

iii) 54

iv) 210

v) 55

vi) 105

5. i) 45

ii) 26

iii) 32

iv) 56

v) 27

vi) 47

100

Answers
6. i) 52 × 22

ii) 27 × 31

		 v) 22 × 3 × 79			
7. i) 200000

ii) 0

		 v) 9000000000			
8. i) – 125

iii) 21 × 31 × 1331

iv) 2 1× 3 1× 1131

vi) 27 × 51
iii) 2025

iv) 1296

vi) 0

ii) 1

iii) 72

iv) – 2000 v) 10584

1. i) A

ii) A

iii) C

iv) C

2. i) 312

ii) a12

iii) 75 + x

iv) 107

v) 59

3. i) 54

ii) a4

iii) 1010

iv) 42

v) 30 = 1

4. i) 312

ii) 220

iii) 220

iv) 1

v) 520

vi) – 131072

Exercise 1.12

Chapter - 2
Exercise 2.1
1. (i) A

(ii) D

(iii) D

(iv) B

(v) C

2. Constants: 5, – 9.5; Variables: a, – xy, p.
3. (i) x + 6

(ii) – m – 7 (iii) 3q + 11		 (iv) 3x + 10		

(v) 5y – 8

4. 3, – 4, 9
5. (i) y2 x, coefficient = y2.			
		(iii) zx, coefficient = z.			

(ii) x, coefficient = 1.
(iv) – 5xy2, coefficient = – 5y2.

6. (i) – my2, coefficient = – m.			

(ii) 6y2, coefficient = 6.

		(iii) – 9xy2, coefficient = – 9x.
Exercise 2.2
1. (i) B

(ii) D

(iii) D

(iv) D

2. (i) 4x, 7x

(ii) 7b, – 3b (iii) 3x2y, – 8yx2		

(v) A
(iv) a2b, 7a2b

		(v) 5pq, 25pq ; – 4p, 10p; 3q, 70q ; p2q2, 14 p2 q2
3. (i) 2

(ii) 2

(iii) 3

4. (i) – 10

(ii) 10

(iii) 11

5. (i) 21

(ii) 34

(iii) 82

(ii) B

(iii) A

(iv) 4

(v) 2

(iv) D

(v) A

Exercise 2.3
1. (i) C

2. (i) 13a + 2b (ii) 5l – 4l2 (iii) 16z2 – 16z
		(iv) p – q

(v) 7m2n – 4m2 – 6n2 + 4mn2			
101

(vi) x2 – 3xy + 7y2

Answers
3. (i) 2ab

(ii) 2s + t

(iii) 3a – 2b + 2p + 3q

		(iv) 5a – 5b + 4(v) 2x + 2y – 2
		(vi) 7c + 4

(vii) 3m2n + 5mn – 4n2 + 4
(ii) 7a2b

4. (i) 8a

		(v) 5n – 2mn + 3m
		(viii)

s2 – 6s – 4		

5. (i) x2 + 5xy – 3y2		

(iii) – 11x2y2			

(iv) – 2xy + 16

(vi) – 5p – 15p2		

(vii) 8m2 – 6m – 12

(ix) 9n2 – 10mn – 9m2
(ii) 9p – 2q – 6		

6. 6a – 6			

7. 16x + 12

8. `12a – 2			

9. 7x – 8 metres

10. (i) 8p2 – 9p – 11			

(ii) – p2 + 8p + 12

(iii) 4x – 3y + 9

11. 2m2 + 5m + 10
Chapter - 3
Exercise 3.1
1. (i) B

(ii) C

(iii) A

(iv) C

(v) A

2. (i) Equilateral triangle - 3 lines of symmetry (iv) Rhombus - 2 lines of symmetry
5. (i) isosceles triangle

(ii) equilateral triangle (iii) scalene triangle

Exercise 3.2
1. (i) C

(ii) B

(iii) D

(iv) B

(v) D

2. (i) 90°

(ii) 90°

(iii) 180°			

3. (i) 90°, 4

(ii) 360°, 1 (iii) 180°, 2			

(iv) 360°, 1

(ii) B

(v) D

(iv) 180°

4. 45°, 8
Exercise 3.3
1. (i) A

(iii) C

(iv) D

2. (i) + DOC, + COB; + COB, + BOA
		(ii) + QOX, + XOP; + POY, + YOQ; + YOQ, + QOX; + XOP, + POY
3. + POR, + QOS; + SOP, + ROQ
4. (i) 150°

(ii) 100°

		(iii) 110°

(iv) 120°

(v) 135°

102

Answers
5. + BOC = 145°; + AOD = 145°; + COA = 35°.
6. (i) 80°

(ii) 110°

		(iii) 20°

(iv) 80°

		(v) 36°

(vi) 45°

7. y = 120°; x = 60°			 8. x = 25°
Chapter - 5
Exercise 5.1
1. (i) D

(ii) D

(iii) B

(iv) B

103

(v) B

Answers

104