Contents

Serial Page
MATHEMATICS

No.
UNIT - 1 No.

1. Rational Numbers 1
Class-VIII

2. Squares and Square roots 25

3. Cubes and Cube Roots 51

4. Exponents and Powers 67

 CHAPTER
1 RATIONAL NUMBERS

1.0 RATIONAL NUMBERS

2.0 PROPERTIES OF RATIONAL NUMBERS

2.1 Closure
2.2 Commutativity
2.3 Associativity
2.4 Special numbers 0 and 1
2.5 Additive inverse of a rational number
2.6 Multiplicative inverse of a rational number
2.7 Distributive property of multiplication over addition of rational numbers

3.0 REPRESENTATION OF RATIONAL NUMBERS ON THE NUMBER LINE

4.0 RATIONAL NUMBERS BETWEEN TWO RATIONAL NUMBERS

EXERCISE-1 (ELEMENTARY)

EXERCISE-2 (SEASONED)

EXERCISE-3 (SUBJECTIVE)
 Rational Numbers

RATIONAL NUMBERS
1.0 RATIONAL NUMBERS
A number which can be expressed in the form p/q, where p,q are integers and q ¹ 0 is called a
rational number. Clearly all fractions are rational numbers. Also natural numbers, whole numbers
and integers are rational numbers. They are represented by Q.
2 -7
E.g. , , -3, 0,1 etc.
3 9
Obviously by contrast there must be numbers which cannot be expressed as fractions. They are
irrational numbers.

2.0 PROPERTIES OF RATIONAL NUMBERS

2.1 Closure
(i) Addition: The sum of two rational numbers is again a rational number.
-2 7 (-2) + 7 5
E.g. + = = , a rational number..
9 9 9 9
Thus, rational numbers are closed under addition. That is, for any two rational numbers a
and b, a + b is also a rational number.
(ii) Subtraction: The difference of two rational numbers is again a rational number.

E.g. 2 7 16 - 21 -5 a rational number..

- = = ,
3 8 24 24
Thus, rational numbers are closed under subtraction. That is, for any two rational numbers
a and b, a – b is also a rational number.
(iii) Multiplication : The product of two rational numbers is again a rational number,

5 æ -3 ö -15 -5
E.g. ´ ç ÷ = = , a rational number..
6 è 4 ø 24 8
Thus, rational numbers are closed under multiplication. That is, for any two rational numbers
a and b, a × b is also a rational number.
(iv) Division : When a rational number is divided by a non-zero rational number, a rational
number is obtained.

3 æ -4 ö 3 æ -7 ö -21
E.g. ¸ç ÷ = ´ç ÷ = , a rational number..
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5 è 7 ø 5 è 4 ø 20
But, division of a rational number by zero is not defined.
Thus, rational numbers are not closed under division. However, if we exclude zero then the
collection of, all other rational numbers is closed under division.

2.2 Commutativity
(i) Addition : The sum of two rational numbers remains the same even if the order in which
they are added is changed.

5 æ -9 ö 50 - 63 -13
E.g. +ç ÷ = =
7 è 10 ø 70 70

æ -9 ö 5 -63 + 50 -13
ç ÷+ = =
è 10 ø 7 7 70

1
Class VIII : Mathematics

5 æ -9 ö æ -9 ö 5
So, +ç ÷=ç ÷+
7 è 10 ø è 10 ø 7
Thus, addition of rational numbers is commutative. That is, for any two rational numbers
a and b, a + b = b + a.
(ii) Subtraction : The difference of two rational numbers is not the same if the order in which
they are subtracted is changed.
3 4 21 - 20 1
E.g. - = =
5 7 35 35

4 3 20 - 21 -1 3 4 4 3
- = = So, - ¹ -
7 5 35 35 5 7 7 5
Thus, the subtraction of rational numbers is not commutative.
(iii) Multiplication : The product of two rational numbers remains the same even if the order
in which they are multiplied is changed. That is, for any two rational numbers a and b,
a × b = b × a.

2 æ -3 ö -6 -2
E.g. ´ç ÷ = =
9 è 7 ø 63 21

æ -3 ö 2 -6 -2 2 æ -3 ö æ -3 ö 2
ç ÷´ = = So, ´ ç ÷ = ç ÷ ´
è 7 ø 9 63 21 9 è 7 ø è 7 ø 9
Thus, the product of rational numbers is commutative.
(iv) Division : The quotient of two rational numbers is not the same if the order in which they
are divided is changed.

3 æ -4 ö 3 æ -7 ö -21
E.g ¸ç ÷ = ´ç ÷ =
5 è 7 ø 5 è 4 ø 20

æ -4 ö æ 3 ö -4 5 -20 3 æ -4 ö æ -4 ö æ 3 ö
ç ÷ ¸ç ÷ = ´ = So,, ¸ç ÷¹ç ÷ ¸ç ÷
è 7 ø è 5 ø 7 3 21 5 è 7 ø è 7 ø è5ø
Thus, division of rational numbers is not commutative.

2.3 Associativity
(i) Addition :The sum of three or more rational numbers remains the same even if the order
in which they are grouped is changed.
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æ -3 5 ö æ -9 ö -21 + 20 æ -9 ö
E.g. ç + ÷+ç ÷= + ç ÷ = -1 + æç -9 ö÷ = -1 + (-18) = -19
è 4 7 14
ø è ø 28 è 14 ø 28 è 14 ø 28 28

-3 é 5 æ -9 ö ù -3 é 10 + (-9) ù -3 1 -21 + 2 -19

and +ê +ç ÷ú = +ê ú = + = =
4 ë 7 è 14 ø û 4 ë 14 û 4 14 28 28

æ -3 5 ö æ -9 ö -3 é 5 æ -9 ö ù
So, ç + ÷+ç ÷= + ê +ç ÷ú
è 4 7 ø è 14 ø 4 ë 7 è 14 ø û
Thus, addition of rational numbers is associative. That is, for any three rational numbers a,
b & c, a + (b + c) = (a + b) + c.

2
 Rational Numbers

(ii) Subtraction : The difference of three or more rational numbers is not the same if the order
in which they are grouped is changed.
æ 3 2 ö 5 æ 21 - 16 ö 5 5 5 15 - 140 125
E.g. ç - ÷- = ç ÷- = - = =-
è 8 7 ø 6 è 56 ø 6 56 6 168 168
3 æ 2 5 ö 3 æ 12 - 35 ö 3 æ 23 ö 63 - (-92) 155
-ç - ÷= -ç ÷ = 8 - ç - 42 ÷ = 168
=
168
8 è 7 6 ø 8 è 42 ø è ø
æ 3 2ö 5 3 æ 2 5ö
So, ç - ÷- ¹ -ç - ÷
è8 7ø 6 8 è7 6ø
Thus, the subtraction of rational numbers is not associative.
(iii) Multiplication : The product of three or more rational numbers remains the same even if
the order in which they are grouped is changed. That is, for any three rational numbers a, b
& c, a × (b × c) = (a × b) × c.

é 2 æ -1 ö ù æ -7 ö æ -2 ö æ -7 ö 14
E.g. ê 5 ´ ç 3 ÷ ú ´ ç 11 ÷ = ç 15 ÷ ´ ç 11 ÷ = 165
ë è øû è ø è ø è ø

2 é -1 æ -7 ö ù 2 7 14 é 2 æ -1 ö ù æ -7 ö 2 é -1 æ -7 ö ù
´ê ´ç ÷ú = ´ = So, ê ´ ç ÷ ú ´ ç ÷ = ´ê ´ç ÷ú
5 ë 3 è 11 ø û 5 33 165 ë 5 è 3 ø û è 11 ø 5 ë 3 è 11 ø û
Thus, the product of rational numbers is associative.
(iv) Division : The quotient of three or more rational numbers is not the same if the order in
which they are grouped is changed.

æ 3 2ö 5 æ 3 7 ö 5 21 5 21 6 63
E.g. ç ¸ ÷ ¸ = ç ´ ÷ ¸ = 16 ¸ 6 = 16 ´ 5 = 40
è 8 7 ø 6 è 8 2ø 6

3 æ2 5ö 3 æ2 6ö 3 12 3 35 35
¸ç ¸ ÷ = ¸ç ´ ÷ = ¸ = ´ =
8 è7 6ø 8 è7 5ø 8 35 8 12 32

æ 3 2ö 5 3 æ 2 5ö
So, ç ¸ ÷ ¸ ¹ ¸ç ¸ ÷
è8 7ø 6 8 è7 6ø
Thus, the division of rational numbers is not associative.

6 -2 3 9 5
Illustration 1. Rearrange suitably and add : + + + +
11 7 7 11 7
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Solution Let us rearrange the numbers suitably to make addition easier.

6 9 6 + 9 15
+ = =
11 11 11 11
-2 3 5 -2 + 3 + 5 6 15 6 105 + 66 171
+ + = = Now,, + = =
7 7 7 7 7 11 7 77 77

-16 5 21 22
Illustration 2. Rearrange suitably and multiply : ´ ´ ´
7 11 -16 40
-16 5 21 22
Solution ´ ´ ´
7 11 -16 40
æ - 16 3 21 ö æ 1 5 ö
21
3 1 3 1 3
= çç ´ ÷÷ ´ ç ´
22 ÷ = æç ö÷ ´ æç ö÷ = ´ =
è 7 1 - 16 ø çè 11 1 40 8 4 ÷ 1
è ø è ø 4 1 4 4
ø

3
Class VIII : Mathematics

1. Complete the following table.

Closed under
Number
Addition Subtraction Multiplication Division
(i) R ational number Yes Yes … .. No
(ii) Integers … .. Yes … .. No
(iii) W hole numbers … .. … .. Yes … ..
(iv) Natural number … .. No … .. … ..

2. Complete the following table.

For Rational Number

Closure Commutative Associative
(i) Addition ….. ….. …..
(ii) Subtraction ….. ….. …..
(iii) Multiplication ….. ….. …..
(iv) Division ….. ….. …..

2.4 Special numbers 0 and 1

If we add 0 to any rational number, the sum is the number itself.
5 5 5
E.g. +0=0+ =
6 6 6
Zero is called the additive identity of rational numbers.
If we multiply a rational number by 1, then the product is the number itself.

-9 æ -9 ö -9
E.g. ´1 = 1´ ç ÷ =
11 è 11 ø 11
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One is called multiplicative identity of rational numbers.

2.5 Additive inverse of a rational number

a a a -a
If is a rational number, then there exists a rational number – such that, + æç ö÷ = 0 .
b b b è b ø
a a
and – are said to be the additive inverse or negative of each other..
b b
6 æ -6 ö 6 - 6
E.g. +ç ÷= =0
11 è 11 ø 11
6 6
Thus, and – are the additive inverse of each other..
11 11

4
 Rational Numbers

2.6 Multiplicative inverse of a rational number

a b a b a b
If is a rational number, then there exists a rational number such that, ´ = 1 . and
b a b a b a
are said to be multiplicative inverse or reciprocal of each other.
-2 5 -2 -5
E.g. for , the multiplicative inverse is - , so that çæ ÷ö ´ çæ ÷ö = 1 .
5 2 è 5 ø è 2 ø
Note that zero has no reciprocal and 1 is the multiplicative inverse of itself.

2.7 Distributive property of multiplication over addition of rational numbers

In general, for rational number a,b and c, a(b + c) = ab + ac.
This property is known as distributive property of multiplication over addition.
7 2 4
E.g. consider 3 rational numbers, - , and
8 3 5

7 æ 2 4 ö -7 é 10 + 12 ù
- ´ç + ÷ =
8 è 3 5 ø 8 ëê 15 ûú

-7 22 -77 æ -7 2 ö æ -7 4 ö
= ´ = and ç ´ ÷+ç ´ ÷
8 15 60 è 8 3ø è 8 5ø

æ -7 ö æ -7 ö -35 + (-42) -77

= ç ÷+ç ÷ = =
è 12 ø è 10 ø 60 60

-7 æ 2 4 ö æ -7 2 ö æ -7 4 ö
So, ´ç + ÷ = ç ´ ÷+ç ´ ÷.
8 è 3 5ø è 8 3ø è 8 5ø

-7
Illustration 3. If x = , verify that – (– x) = x.
8
-7
Solution x=
8
é æ -7 ö ù
So, (–(–x)) = - ê - çè ÷ø ú
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ë 8 û

é7 ù -7
= -ê ú = = x. So, – (– x) = x.
ë8û 8

Illustration 4. Write the additive inverse of

6 -7 8 -5 17
(i) (ii) (iii) (iv) (v)
8 11 -17 -9 1
Solution Additive inverse is a number with the same magnitude but opposite sign. So, the additive
inverse is.
-6 7 8 -5 -17
(i) (ii) (iii) (iv) (v)
8 11 17 9 1

5
Class VIII : Mathematics

3
1. Is 0.7 the multiplicative inverse of 1 ?
7

2. Solve using distributive property of multiplication.

æ -2 5 ö æ -2 7 ö
ç ´ ÷+ç ´ ÷
è 3 6ø è 3 6ø

2 3 5 3 1
3. - ´ + – ´ =?
3 5 2 5 6

3.0 REPRESENTATION OF RATIONAL NUMBERS ON THE NUMBER LINE

To represent a rational number on the number line, divide each unit length on the number line into
as many parts as the denominator of the rational number and move as many steps starting from
O, on the number line as the numerator (towards the right for positive rational numbers and towards
the left for negative rational numbers).
E.g. Look at the following number line

Negative rational Positive rational

numbers num bers

–1 –4 –3 –2 –1 0 1 2 3 4 1
5 5 5 5 5 5 5 5
1
  Here, each unit length is divided into 5 equal parts. Similarly, to represent , the number line may
8
be divided into eight equal parts as shown :

0 1
1
We use the number to name the first point of this division. The second point of division will
8
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2 3
be labelled , the third point , and so on as shown below :
8 8

0 1 2 3 4 5 6 7 8
8 8 8 8 8 8 8 8

Any rational number can be represented on the number line in this way.

6
 Rational Numbers

Illustration 5. Which rational number do the letters X, Y and Z represent on the following number
line ?

–1 –4 X –2 Y 0 1 2 Z 4 1
5 5 5 5 5
Solution The points X and Y lie between 0 and – 1. The distance between 0 and – 1 is divided
into 5 equal parts.
-1 -3
So, Y represent and X represent .
5 5
The point Z lies between 0 and 1. The distance between 0 and 1 is divided into 5 equals
parts.
3
So, Z represents .
5
Illustration 6. Represent the following on the number line : (i) 3/7 (ii) 8/–5
Solution (i) 3/7 will lie between 0 and 1 on the number line.

1 2 3 4 5 6 7
7 7 7 7 7 7 7

0 A B C D E F 1

For 3/7, the denominator is 7 so, divide the distance between 0 and 1 into 7 equal
parts. The points A,B,C,D,E and F do this. The point C represents 3/7.
8 -8 3
(ii) = = -1 lies between – 1 and – 2 on the number line.
-5 5 5

–10 –9 –8 –7 –6 –5
5 5 5 5 5 5

–2 A B C D –1 0
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As the denominator is 5, we will divide the distance between – 1 and – 2 into

-8
5 equal parts. The points A,B,C and D do this. The point B represents , -1
5
-5
is same as and – 2 is -10 .
5 5

1. Write the rational number for each point labelled.

0 A 2 3 B C 6 7 D E 10
5 5 5 5 5 5

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Class VIII : Mathematics

4.0 RATIONAL NUMBERS BETWEEN TWO RATIONAL NUMBERS

You already know that between two whole numbers or two integers, only a definite number of whole
number exists.
But, it isn't the same for 2 rational numbers. There are innumerable rational numbers between any
two rational numbers.

Illustration 7. How many rational numbers lie between 0 and 1 ?

Solution
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

Divide the number line into 10 equal parts between the points 0 and 1. We can easily
see that, there are 9 points between 0 and 1,
1 2 3 4 5 6 7 8 9
i.e., , , , , , , , , .
10 10 10 10 10 10 10 10 10

0 —1 —2 —3 —4 —
5 —6 — 7 —8 —9 1
10 10 10 10 10 10 10 10 10
1 2
Again, divide the number line between and into 10 equal parts, we get
10 10
11 12 13 14 15 16 17 18 19
, , , , , , , , .
100 100 100 100 100 100 100 100 100
11 12
We can further divide the number line between the points and into 10 equal
100 100
parts. We get
111 112 113 114 115 116 117 118 119
, , , , , , , , .
1000 1000 1000 1000 1000 1000 1000 1000 1000
If we go on increasing the divisions between two rational numbers, we can
accommodate an infinite number of rational numbers.
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-1 1
Illustration 8. Find any 5 rational numbers between and .
2 2
-1 -5 1 5
Solution Let us write as and as .
2 10 2 10
The rational numbers between these are
-4 -3 -2 -1 0 1 2 3 4
, , , , , , , and .
10 10 10 10 10 10 10 10 10
Write any 5 rational numbers for the answer.
-4 -3 -2 -1 2
The numbers are , , , , .
10 10 10 10 10

8
 Rational Numbers

1 3
Illustration 9. Find any two rational numbers between and .
4 5
Solution First, find the equivalent rational numbers of both the given rational numbers by
converting their denominators into a common denominator.
LCM of 4 and 5 is 20.
1 1´ 5 5 3 3 ´ 4 12
= = and = =
4 4 ´ 5 20 5 5 ´ 4 20

1æ 5 ö 3 æ 12 ö 6 7
Two rational numbers between ç= ÷ and ç= ÷ aree and
4 è 20 ø 5 è 20 ø 20 20
Alternate method
We can also use the idea of mean to find rational numbers between any two given
rational numbers.
The number that is midway between the two given rational numbers, a and b
a+b
(a < b) is the mean of a and b, i.e. .
2

1 1
Illustration 10.Find a rational number that is midway between and .
5 4
Solution We find the mean of the given rational numbers.

æ1 1ö æ4 + 5ö 9
ç + ÷ ç ÷
è5 4ø = è 20 ø = 20 = 9 ´ 1 = 9
2 2 2 20 2 40

1 1
1. Find 3 rational numbers between and .
8 2
–2 1
2. Find ten rational numbers between & .
5 2
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x
• In mathematics, a Dyadic Rational is a rational number of the form , where x is an integer, and
2y
y is a natural number. So in a Dyadic Rational, the denominator is 2 or 4 or 8 and so on.
• A rational number can be expressed as a terminating or a non-terminating recurring decimal.

22
• Irrational number can never be expressed as fraction. Therefore is only an approx value of p.
7

9
Class VIII : Mathematics

SOME WORKED OUT ILLUSTRATIONS

Illustration 1
Using appropriate properties find :

2 æ 3ö 1 3 1 2
´ç- ÷ - ´ + ´
5 è 7 ø 6 2 14 5
Solution

2 æ 3ö 1 3 1 2 2 æ 3ö 1 2 1 3
´ç- ÷ - ´ + ´ = ´ç- ÷ + ´ - ´
5 è 7 ø 6 2 14 5 5 è 7 ø 14 5 6 2

2 æ 3ö 2 1 1 3 2 éæ 3 ö 1 ù 1 3
= ´ç- ÷ + ´ - ´ = ´ êç - ÷ + ú - ´
5 è 7 ø 5 14 6 2 5 ë è 7 ø 14 û 6 2

2 æ -6 + 1 ö 3 2 -5 1 -1 1
= ´ç ÷- = ´ - = -
5 è 14 ø 12 5 14 4 7 4

-4 - 7 -11
= = .
28 28

Illustration 2
Write the additive inverse of each of the following:
2 -5
(i) (ii)
8 9
Solution

2 æ -2 ö -2
(i) The additive inverse of is ç ÷ = .
8 è 8 ø 8

-5 æ -5 ö 5
(ii) The additive inverse of is – ç ÷= .
9 è 9 ø 9

Illustration 3
Find the multiplicative inverse of the following :
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-13 -5 -3
(i) – 13 (ii) (iii) ´
19 8 7
Solution
1
(i) The multiplicative inverse of – 13 is (– 13)–1 =
-13
-1
-13 æ -13 ö 19
(ii) The multiplicative inverse of is ç ÷ = .
19 è 19 ø -13
-5 -3 -5 ´ -3 15
(iii) We have, ´ = =
8 7 8´7 56
-1
15 æ 15 ö 56
The multiplicative inverse of is ç ÷ =
56 è 56 ø 15

10
 Rational Numbers

Illustration 4
8 1
Is the multiplicative inverse of – 1 ? Why or why not ?
9 8
Solution
8 1
No, is not the multiplicative inverse of – 1 .
9 8
8 1 8 -9
Because ×–1 = × = – 1 ¹ 1.
9 8 9 8

Illustration 5
Write
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution
(i) We know that there is no rational number which when multiplied with 0, gives 1. Therefore,
the rational number 0 has no reciprocal.
(ii) We know that the reciprocal of 1 is 1 and the reciprocals of – 1 is – 1. 1 and – 1 are the
only rational numbers which are their own reciprocals.
(iii) The rational number 0 is equal to its negative.

Illustration 6
Represent these numbers on the number line.
7 -5
(i) (ii)
4 6
Solution
1
(i) For 7/4, we make 7 markings of distance each on the right of zero and starting from 0.
4
7
The seventh marking is .
4
1 P 2
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0 1 2 3 4 5 6 7 8
— — — — — — — —
4 4 4 4 4 4 4 4
7
The point P represents the rational number .
4

-5 1
(ii) For , we make 5 markings of distance each on the left of zero and starting from 0. The
6 6
-5 -5
fifth marking . The point P represents the rational number .
6 6
1 P

–6 –5 –4 –3 –2 –1 0
— — — — — —
6 6 6 6 6 6

11
Class VIII : Mathematics

Illustration 7
-2 -5 -9
Represent , , on the number line.
11 11 11
Solution
-2 -5 -9
To represent , , on a number line draw a number line and mark a point O on it to represent
11 11 11
zero. Now mark a point P representing integers – 1 on the left side of O on the number line.
–9 –5 –2
— — —
–1 11 11 11
P J I H G F E C A O
Divide the segment OP into eleven equal parts. Let A, B, C, D, E, F, G, H, I, J be the points of division
-2
so that OA = AB = BC = ... = JP. By construction, OB is two eleventh of OP so B represents .
11
-5 -9
OE is five eleventh of OP so E represents and OI is nine-eleventh of OP so I represents .
11 11

Illustration 8
-2 1
Find ten rational numbers between and .
5 2
Solution
Converting the given rational numbers with the same denominators.
-2 -2 ´ 4 -8 1 1 ´ 10 10
= = and, = =
5 5´4 20 2 2 ´ 10 20
We known that – 8 < – 7 < – 6 ... < 10
-8 -7 -6 10
Þ < < < ... <
20 20 20 20
-2 1
Thus, we have the following ten rational number between and :
5 2

-7 -6 -5 -4 -3 -2 -1 1 2
, , , , , , ,0, and
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20 20 20 20 20 20 20 20 20

Illustration 9
Find five rational numbers between
2 4 -3 5
(i)and (ii) and
3 5 2 3
Solution
(i) Converting the given rational numbers with the same denominators
2 2 ´ 5 10 4 4 ´ 3 12
= = and = =
3 3 ´ 5 15 5 5 ´ 3 15
2 10 10 ´ 4 40
Also , = = =
3 15 15 ´ 4 60

12
 Rational Numbers

4 12 12 ´ 4 48
and, = = =
5 15 15 ´ 6 60
We know that
40 < 41 < 42 < 43 < 44 < 45 < 46 < 47 < 48
40 41 42 47 48
Þ < < < ... < <
60 60 60 60 60
2 4
Thus, we have the following five rational numbers between and :
3 5
41 42 43 44 45
, , , and .
60 60 60 60 60

41 47
Note : We may take any five numbers given above from to
60 60
-3 -3 ´ 3 -9
(ii) Converting the given rational numbers with the same denominators = =
2 2´ 3 6
5 5 ´ 2 10
and, = =
3 3´ 2 6
We known that
–9 < –8 < –7 < –6 < ... < 0 < 1 < 2 < 8 < 9 < 10
-9 -8 -7 -6 0 1 2 8 9 10
Þ < < < < ... < < < < ... < < < .
6 6 6 6 6 6 6 6 6 6

-3 5 -8 -7 0 1 2
Thus, we have the following five rational numbers between and : , , , and
2 3 6 6 6 6 6

Illustration 10
-3 7 3 -9 13 13
Simplify : + + + + +
10 15 -20 10 15 -20
Solution
Re-arranging and grouping the numbers in pairs such that each group contains a pair of rational
numbers with equal denominators, we have
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

-3 7 3 -9 13 13
+ + + + +
10 15 -20 10 15 -20

æ -3 -9 ö æ 7 13 ö æ 3 13 ö æ -3 -9 ö æ 7 13 ö æ -3 -13 ö
= ç + ÷+ç + ÷+ç + ÷ = ç + ÷+ç + ÷+ç + ÷
è 10 10 ø è 15 15 ø è -20 -20 ø è 10 10 ø è 15 15 ø è 20 20 ø

(-3) + (-9) 7 + 13 (-3) + (-13)

= + +
10 15 20

-12 20 -16 -6 4 -4
= + + = + + [Expressing each rational in lowest terms]
10 15 20 5 3 5

æ -6 -4 ö 4 (-6) + (-4) 4 -10 4 4 -6 + 4 -2

= ç + ÷+ = + = + = -2 + = =
è 5 5 ø 3 5 3 5 3 3 3 3

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Class VIII : Mathematics

Illustration 11

æ -3 4 ö æ 9 -10 ö æ 1 3 ö
Simplify : ç ´ ÷+ç ´ ÷-ç ´ ÷
è 2 5ø è5 3 ø è2 4ø
Solution

æ -3 4 ö æ 9 -10 ö æ 1 3 ö -3 ´ 4 9 ´ (-10) 1 ´ 3
ç ´ ÷+ç ´ ÷-ç ´ ÷ = + -
è 2 5 ø è 5 3 ø è 2 4 ø 2´ 5 5´3 2´ 4

-3 ´ 2 3 ´ (-2) 3 -6 -6 3
= + - = + -
1´ 5 1´ 1 8 5 1 8
-6 -6 -3 (-6) ´ 8 + (-6) ´ 40 + (-3) ´ 5
= + + =
5 1 8 40
-48 + (-240) + (-15) -303
= =
40 40

Illustration 12
Represent the following on the number line.
-5 -17
(i) (ii)
9 -10
Solution
-5
(i) will lie between 0 and –1 on the number line.
9
–9 –8 –7 –6 –5 –4 –3 –2 –1
— — — — — — — — —
9 9 9 9 9 9 9 9 9

-1 P Q R S T U V W 0
Since the denominator is 9, we will divide the distance between 0 and –1 into 9 equal parts.
-5
The points P, Q, R, S, T, U, V and W do this. The point S represents .
9

-17 17 7
(ii) = =1 lies between 1 and 2 on the number line.
-10 10 10
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

10 11 12 13 14 15 16 17 18 19 20
10 10 10 10 10 10 10 10 10 10 10

1 E F G H I J K L M 2

As the denominator is 10, we divide the distance between 1 and 2 into 10 equal parts. The
17
point E, F, G, H, I, J, K, L and M do this. The point K represent . The representation by
10
the other points is also shown.

14
 Rational Numbers

1
1. The additive inverse of is
9

1 1
(A) 9 (B) –9 (C) (D) -
9 9
2. How many rational numbers exist between any two distinct rational numbers ?
(A) 2 (B) 3 (C) 11 (D) infinite

3 6
3. If of a number is 22, what is of that
11 11
(A) 6 (B) 11 (C) 12 (D) 44
7
4. What number should be subtracted from –4 to get .
8

25 25 39 39
(A) - (B) (C) - (D)
8 8 8 8

4 4
5. What number should be added to - to get .
3 3

4 8 3
(A) 0 (B) (C) (D)
3 3 4

7 -16
6. ´ 0=0
11 21
(A) + (B) – (C) × (D) ¸

7. If a is reciprocal of b, then the reciprocal of b is _____

1
(A) a (B) b (C) (D) None of these
a

9
8. Which of the following is (are) greater than x, when x = ?
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

11

1 x +1 x +1
I. II. III.
x x x -1
(A) I only (B) I and II only (C) I and III only (D) II and III only

9. Arrange the following fractions in ascending order

3 4 7 1
, , ,
7 5 9 2

4 7 3 1 3 1 7 4 4 7 1 3 1 3 7 4
(A) , , , (B) , , , (C) , , , (D) , , ,
5 9 7 2 7 2 9 5 5 9 2 7 2 7 9 5

15
Class VIII : Mathematics

10.
æ 9 5ö æ 9 5ö
ç ´ ÷ - ç- ´ ÷ =
è 10 3 ø è 10 3 ø
(A) 0 (B) 1 (C) 2 (D) 3

æ8 4ö æ 8 3ö æ 4 3ö
11. ç ´ ÷ - ç- ´ ÷ = - - - - -´ç + ÷
è9 7ø è 9 7ø è7 7ø

8 8 4 4
(A) (B) - (C) (D) -
9 9 9 9

12. (17 × 12)–1 = 17–1 × _____

1 1
(A) (B) 12 (C) 17 (D)
12 17

-4 0 -4
13. =
7 9 7
(A) + (B) – (C) + / – (D) ×

æ -9 5 ö æ -9 4 ö æ5 4ö
14. ç ´ ÷-ç ´ ÷ = _______´ ç - ÷
è 13 6 ø è 13 6 ø è6 6ø

9 18 18 9
(A) (B) (C) - (D) -
13 13 13 13

7 æ 17 14 ö æ 7 14 ö æ 7 ö
15. ´ç + ÷ = ç ´ ÷ + ç ´ _______ ÷
8 è 13 13 ø è 8 13 ø è 8 ø

17 17 34 34
(A) - (B) (C) (D) -
13 13 13 13

16. The product of two positive rational numbers is always ______

(A) Positive (B) Can be Positive or Negative
(C) Negative (D) None of these
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

17. The product of a positive rational number and a negative rational number is always _____
(A) Positive (B) Can be Positive or Negative
(C) Negative (D) None of these

18. The product of two negative rational numbers is always ______

(A) Positive (B) Can be Positive or Negative
(C) Negative (D) None of these

19. The reciprocal of a negative rational number is _____

(A) Positive (B) Can be Positive or Negative
(C) Negative (D) None of these

16
 Rational Numbers

20. The reciprocal of a positive rational number is _____

(A) Positive (B) Can be Positive or Negative
(C) Negative (D) None of these

21. The product of a rational number and its reciprocal is _____

1
(A) 1 (B) –1 (C) 0 (D)
2

4 4
22. + = _____
-18 18
4
(A) 0 (B) 1 (C) –1 (D)
9

6
23. ´ ______ = 1
-12

12 12
(A) - (B) (C) 1 (D) 0
6 6

17
24. + ______ = 0
23

17 17
(A) (B) 0 (C) 1 (D) -
23 23

1
25. Reciprocal of , a ¹ 0 is ____
a

1
(A) 1 (B) (C) a (D) None of these
a2
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

17
Class VIII : Mathematics

2 3
1. The cost of 7 metres of rope is 12 . Then the cost per metre of the rope is
3 4

51 61
(A) Rs. 1 (B) Rs. 1 (C) Rs. 2 (D) None of tehse
92 92

2. Which of the following is correct ?

2 41 4 2 4 41 2 41 4 41 2 4
(A) > > (B) < < (C) < < (D) < <
3 60 5 3 5 60 3 60 5 60 3 5

15 5
3. If - ´ k = - , then k is equal to
28 7

4 3 5 3
(A) (B) (C) (D)
3 4 3 5

1 æ 3 -5 ö 1 -5
4. If ´ç + ÷ = ´x + y´ , then
2 è 4 12 ø 2 12

3 1 1 3 3 5
(A) x = , y= (B) x = ,y= (C) x = ,y= (C) None of these
4 2 2 4 2 12

a+b 2 1
5. If k = , where a = , b = , then k is equal to
a-b 5 2
(A) 9 (B) 10 (C) –9 (D) –10

37
6. If x < - , then x can be
9
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

(A) –1 (B) –2 (C) –3 (D) None of these

-8 -5
7. (Fill in the blank)
9 9
(A) > (B) < (C) = (D) None of these

4 æ 5 -8 ö æ 4 ö -8
8. If - ´ç ´ ÷ = ç- ´ k÷´ , then k is equal to
5 è7 9 ø è 5 ø 9

5 5 4
(A) - (B) (C) (D) None of these
7 7 7

18
 Rational Numbers

1 1
9. - = .............
9 3

2 2 2 1
(A) - (B) (C) - (D)
3 3 9 3

3 æ 3ö æ 2ö 5 19
10. If + ç- ÷ + ç- ÷ + + k = - , then k is equal to
4 è 5ø è 3ø 8 120

2 2 4 4
(A) (B) - (C) (D) -
15 15 15 15

æ 7 15 ö æ 1ö æ1 1ö
11. The value of ç - ´ ÷ - ç 1 ´ ÷ + ç ´ ÷ is equal to
è 18 -7 ø è 4ø è2 4ø

17 17 17 48
(A) (B) (C) (D)
26 24 48 17

1 1
12. The sum of - and - is ............
8 8

1 1
(A) (B) - (C) 0 (D) 1
4 4

13. The multiplicative inverse of –1 is

(A) 1 (B) –1 (C) 10 (D) –10

æ1 1ö
14. - ç + ÷ = ...........
è 2 2ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

-1
(A) 1 (B) 0 (C) –1 (D)
4

1
15. The multiplicative inverse of is
8

1
(A) –8 (B) 8 (C) - (D) 1
8

19
Class VIII : Mathematics

Very short answer type questions

1. Check whether the following statements are true or false. If true, then state the property illustrated
by the statement.

-2 æ -2 ö -2 -4 7 7 æ -4 ö
(i) +0 =0+ç ÷ = (ii) - = -ç ÷
3 è 3 ø 3 5 8 8 è 5 ø

2. If each of the given statements illustrates the property of addition of rational numbers, then find
the value of x.
-8 4
(i) +0= x (ii) +x=0
17 5

3. Find the additive inverse of the given rational numbers.

24 -16
(i) (ii)
25 -29

4. Find the multiplicative inverse of the given rational numbers.

5 -17 -15 -29
(i) + (ii) (iii)
3 11 27 -13

5. State the property illustrated by the statement.

4 4 -2 5 5 æ -2 ö
(i) ´1 = 1´ (ii) ´ = ´ç ÷
5 5 7 9 9 è 7 ø

-5 æ -1 ö -1 æ -5 ö é 2 æ -9 ö ù 3 2 æ -9 3 ö
(iii) ´ç ÷= ´ç ÷ (iv) ê ´ ç ÷ ú ´ = ´ ç ´ ÷
6 è -19 ø -19 è 6 ø ë 3 è 16 ø û 4 3 è 16 4 ø

6. Each of the following statement is true for rational numbers. Find the value of x.

13 12 13 é 6 æ -8 ö ù æ -5 6 ö æ -8 ö
(i) ´x = ´ (ii) x ´ ê ´ ç ÷ú = ç ´ ÷´ç ÷
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

10 13 10 ë 7 è 11 ø û è 6 7 ø è 11 ø

3 æ 8 ö 3 8 3 4 4 9 4
(iii) ´ç - x÷ = ´ - ´ (iv) ¸ = ´x
11 è 23 ø 11 23 11 17 7 -11 7

Short answer type questions

7. Using appropriate properties, find

6 -4 3 11 9 6 3 3
(i) + + + (ii) + + +
17 5 5 17 -11 7 7 11
4 2 -4 1 -5 1 3 -5 1 3
(iii) + + + + (iv) + + + +
7 5 5 7 7 8 5 8 5 8

20
 Rational Numbers

8. Solve by suitable rearrangement.

9 8 -24 33
(i) ´ ´ ´
12 11 27 40
4 -5 26 28
(ii) ´ ´ ´
7 13 30 40

-5 5 26 21
(iii) ´ ´ ´
13 7 30 50

16
9. Is 0.21 the multiplicative inverse of 4 ?
21

10. Use the distributive property of multiplication and solve.

æ -7 4 ö æ -7 1 ö
(i) ç ´ ÷+ç ´ ÷
è 12 5 ø è 12 5 ø

æ 16 14 ö æ 16 9 ö
(ii) ç ´ ÷+ç ´ ÷
è 21 23 ø è 21 23 ø

æ -11 6 ö æ -11 7 ö
(iii) ç ´ ÷+ç ´ ÷
è 14 13 ø è 14 13 ø

4 5 1 21
11. Subtract the sum of – and – from the sum of and - .
7 3 2 22

Long answer type questions

12. Draw a number line and represent the following on it.

-7 14 -11
(i) (ii) (iii)
8 10 6
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

13. Find the rational number midway between the given numbers.
-1 -1 1 3
(i) and (ii) and
2 3 2 4

14. Find any 3 rational numbers between the given numbers.

1 1
(i) 4 and 5 (ii) – 6 and – 7 (iii) 5 and 6
2 2
-5 6 4
(iv) and (v) - and 1
6 9 7

15. Use the method of mean to find 3 rational numbers between – 1 and – 2.

21
Class VIII : Mathematics

High order thinking skills (HOTS)

9
16. A tiger can cross a 13 feet wide canal. A lion can cross a canal whose width is of the canal
13
crossed by the tiger. A hare can cross the canal, which is 1/3 of the canal crossed by the lion. Find
the width of the canal crossed by the hare.

17. A dishonest shopkeeper uses 990 g weight instead of 1 kg. Neha went to the shopkeeper and asked
5
him to give kg sugar. Find what fraction of sugar was she cheated of..
2

3 4 3
18. of a wall was completed on day 1, of the wall was completed on day 2, of the wall was
10 14 15
1
completed on day 3. If the total length of the wall is 513 m, find.
3
(i) What part of the wall was constructed on day 4?
(ii) How many metres of the wall was constructed on day 4?

10
19. Mohan, Sohan and Mahesh are standing in a queue. Mohan's height is of the height of Sohan,
9
Sohan's height is 3/4 of the height of Mahesh. If Mahesh's height is 180 cm, find the height of Sohan
and Mohan.

20. A gardener has to lay grass in a circular ground, leaving a space of 5 m all around it. If the area
59400 2 22
of the ground with grass is m , find the area of the ground without grass. (Use p = )
21 7

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

22
 Rational Numbers

ANSWERS
CHECK POST-1

1. (i) Yes, (ii) Yes, Yes (iii) Yes, No , No (iv) Yes, Yes, No

For R ational N um bers

C lo su re C o m m u tative A ss ociative
(i) A d d ition Ye s Ye s Ye s
2. (ii) S ubtraction Ye s No No
(iii) M ultiplication Ye s Ye s Ye s
(iv ) D iv ision No No No

CHECK POST-2
4
1. YES 2. - 3. 2
3
CHECK POST-3

1 4 8 9
1. A= , B = , C = 1, D = , E =
5 5 5 5

CHECK POST-4
5 23 13 -7 -6 -5 -4 -3 -2 -1 1 2
1. , & 2. , , , , , , , 0, ,
16 64 32 20 20 20 20 20 20 20 20 20

EXERCISE-1

Que. 1 2 3 4 5 6 7 8 9 10
Ans. D D D C C C A B B D
Que. 11 12 13 14 15 16 17 18 19 20
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

Ans. A A C D B A C A C A
Que. 21 22 23 24 25
Ans. A A A D C

EXERCISE-2

Que. 1 2 3 4 5 6 7 8 9 10
Ans. B C A A C D B B C D
Que. 11 12 13 14 15
Ans. B B B C B

23
Class VIII : Mathematics

EXERCISE-3

Very short answer type questions

1. (i) True, Additive identity (ii) False

8 4
2. (i) - (ii) -
17 5

24 -16
3. (i) - (ii)
25 29

33 27 -13
4. (i) (ii) - (iii)
4 15 -29
5. (i) Multiplicative identity
(ii) Commutative property of multiplication
(iii) Commutative property of multiplication
(iv) Associative property of multiplication

12 5 4 11
6. (i) (ii) - (iii) (iv) -
13 6 17 9

Short answer type questions

4 57 2 27
7. (i) (ii) (iii) - (iv)
5 77 5 40

2 2 1
8. (i) - (ii) - (iii) - 9. Yes
5 15 10

7 16 11 412
10. (i) - (ii) (iii) - 11.
12 21 14 231

Long answer type questions

5 5
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\1.Rational Number\Theory

13. (i) - (ii)

12 8

High order thinking skills (HOTS)

1 3
16. 3 feet 17. 18. (i) (ii) 110 m
100 14

3
19. Sohan = 135 cm, Mohan = 150cm 20. 1021 sqm
7

24
 SQUARE AND
CHAPTER 2
SQUARE ROOTS

1.0 SQUARE NUMBER / PERFECT SQUARE

2.0 PROPERTIES OF SQUARE NUMBERS

3.0 ONE'S DIGIT IN SQUARE OF A NUMBER

4.0 SOME INTERESTING PATTERNS

4.1 Adding triangle numbers
4.2 Number between square numbers
4.3 Adding odd numbers
4.4 A sum of consecutive natural numbers
4.5 Product of two consecutive even or odd natural numbers
4.6 Some more pattern in square numbers

5.0 FINDING THE SQUARE OF A NUMBER

5.1 Other patterns in Squares
5.2 Pythagorean triplets

6.0 SQUARE ROOTS

6.1 Some properties of square roots
6.2 Finding square root through repeated subtraction
6.3 Finding square root through prime factorisation
6.4 Finding the square root by division method

7.0 SQUARE ROOT OF DECIMALS

8.0 ESTIMATING SQUARE ROOT

EXERCISE-1 (ELEMENTARY)
EXERCISE-2 (SEASONED)
EXERCISE-3 (SUBJECTIVE)
 Squares and Square Roots

SQUARES AND SQUARE ROOTS

1.0 SQUARE NUMBER / PERFECT SQUARE

2
If a natural number m can be expressed as n , where n is also a natural number, then m is a square
2
number or a perfect square. i.e. If m = n where m, n are natural numbers
Then m is a square number or a perfect square.
2 2 2 2
E.g.(i) 36 = 2 × 2 × 3 × 3 = 2 × 3 = (2 × 3) = 6
Hence, 36 is a perfect square of 6.
(ii) 50 = 2 × 5 × 5
Hence, 2 is unpaired, so, 50 is not a perfect square.

2.0 PROPERTIES OF SQUARE NUMBERS

Following table shows the square of numbers from 1 to 20.
Number Square Number Square
1 1 11 121
2 4 12 144
3 9 13 169
4 16 14 196
5 25 15 225
6 36 16 256
7 49 17 289
8 64 18 324
9 81 19 361
10 100 20 400
Study the square numbers in the above table. All these numbers end with 0, 1, 4, 5, 6 or 9 at unit's
place. None of these end with 2, 3, 7 or 8 at unit's place. So, we can say that a number that ends in
2, 3, 7 or 8 is never a perfect square.
Thus, by just looking at the numbers 152, 453, 1657 and 2348, we can say that these are not
perfect squares.

Illustration 1. Which of the following numbers are perfect squares?

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(i) 1057 (ii) 7928 (iii) 784

Solution (i) 1057
The number ends with 7
Þ it is not a perfect square.
(ii) 7928
The number ends with 8
so it cant be perfect square.
(iii) 784
The number ends with 4
Þ It can be square number so, we check it
784 = 2 × 2 × 2 × 2 × 7 × 7
2 2
= (2 × 2 × 7) = (28)
Hence, 784 is a perfect square.

25
Class VIII : Mathematics

Illustration 2. Without doing any calculation, find the numbers which are surely not perfect squares:
(i) 153 (ii) 257 (iii) 408 (iv) 441
Solution We know that a number ending in 2,3,7 or 8 is not a perfect square. Therefore,
(i) 153 is not a perfect square.
(ii) 257 is not a perfect square.
(iii) 408 is not a perfect square
(iv) 441 may be a perfect square.
So, 153, 257 and 408 are surely not perfect squares.

3.0 ONE'S DIGIT IN SQUARE OF A NUMBER

(i) The ones digit in the square of number can be determined if the ones digit of the number is
known.
If the number Square of the number Examples
ends in would end in

2 2
1 or 9 1 11 = 121 ; 19 = 361
2 2
2 or 8 4 12 = 144 ; 18 = 324
2 2
4 or 6 6 14 = 196 ; 16 = 256
2 2
3 or 7 9 13 = 169 ; 17 = 289
2
5 5 15 = 225
0 0 202 = 400

(ii) The number of zeros at the end of a perfect square is always even and is double the number
of zeros at the end of the given number.
E.g.
2
10 = 100 Two zero s
2
50 = 2500
O ne zero
2
{ 600 = 360000}
Two zero s Fo ur zeros
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(iii) The square of an even number is always an even number and square of an odd number is
always an odd number.

Even number Odd numbers

and and
their squares their squared
2 2
2 =4 5 = 25
2 2
4 = 16 11 = 121
2 2
12 = 144 13 = 169

26
 Squares and Square Roots

4.0 SOME INTERESTING PATTERNS

4.1 Adding triangle numbers
The sum of two consecutive triangular numbers is a square number

• • • •
• • • • • •
• • • • • •
• • • •
1 3 6 10

• • • • • • • • •
• • • • • • • • •
1+3=4=2 2 • • • • • • •
3+6 =9=3
2
• • • 2
•
6+1 0=16=4

4.2 Number between Square Numbers

There are '2n' non perfect square numbers between the squares of two consecutive natural numbers
n and (n + 1).
2 2
Between 1 (=1) and 2 (= 4) ® 2, 3
2 × 1 = 2 non square numbers exist.
2 2
Between 2 (= 4) and 3 (= 9) ® 5, 6, 7, 8
2 × 2 = 4 non square numbers exist.
2 2
Between 3 (= 9) and 4 (= 16) ® 10, 11, 12, 13, 14, 15
2 × 3 = 6 non square numbers exist.
4.3 Adding Odd Numbers
Consider the following
2
1 One odd number =1 =1
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

2
1+3 Sum of first 2 odd number =4 =2
2
1+3+5 Sum of first 3 odd number =9 =3
2
1+3+5+7 Sum of first 4 odd number = 16 = 4
2
1+3+5+7+9 Sum of first 5 odd number = 25 =5
2
So, we can say that the sum of first n odd natural numbers is n .
or, we can say if the number is a square number, it has to be the sum of successive odd numbers
starting from 1.
4.4 A sum of consecutive natural numbers
The square of an odd natural number 'n' can be expressed as the sum of two consecutive natural
n2 - 1 n2 + 1
numbers and
2 2

27
Class VIII : Mathematics

n2 - 1 n2 + 1
2
Then, n = +
2 2
2
3 =9=4 + 5
¯ ¯

32 - 1 32 + 1
,
2 2
2
5 = 25 = 12 + 13
¯ ¯

52 - 1 52 + 1
,
2 2
2
11 = 121 = 60 + 61
¯ ¯

112 - 1 112 + 1
,
2 2

4.5 Product of two consecutive even or odd natural numbers

If (n + 1) and (n – 1) are two consecutive even or odd numbers then their product i.e. (n + 1)
2
(n – 1) is n – 1.
2
11 × 13 = 143 = 12 – 1

Also, 11 × 13 = (12 – 1) (12 + 1)

2
Þ 11 × 13 = (12 – 1) (12 + 1) = 12 – 1

Similarly,
2
13 × 15 = (14 – 1) (14 + 1) = 14 – 1

4.6 Some more pattern in square numbers

Observe the squares of numbers 1, 11, 111, .... etc.

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

2
1 = 1
2
11 = 1 2 1
2
111 = 1 2 3 2 1
2
1111 = 1 2 3 4 3 2 1
2
11111 = 1 2 3 4 5 4 3 2 1
2
111111 = 1 2 3 4 5 6 5 4 3 2 1

28
 Squares and Square Roots

Another interesting pattern :

2
7 = 49
2
67 = 4489
2
667 = 444889
2
6667 = 44448889
2
66667 = 4444488889
2
666667 = 444444888889

Illustration 3. Using the given pattern, find the missing numbers:

2 2 2 2
1 +2 +2 =3
2 2 2 2
2 +3 +6 =7
2 2 2 2
3 + 4 + 12 = 13
2 2 2
4 + 5 + __ = 21
2 2 2
5 + __ + 30 = 31
2 2
6 + 7 + __ = __
Solution The missing number are as below :
2 2 2 2
4 + 5 + 20 = 21
2 2 2 2
5 + 6 + 30 = 31
2 2 2 2
6 + 7 + 42 = 43

1. What will be the 'One's digit' in the square of the following numbers.
(i) 109 (ii) 77 (iii) 34 (iv) 26 (v) 9018
(vi) 10000 (vii) 225 (viii) 1243 (ix) 962 (x) 2122

2. How many zeros are there in the square of the given numbers ?
(i) 670 (ii) 800 (iii) 625
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

3. Find whether the square of the following numbers are even or odd.
(i) 79 (ii) 824 (iii) 158 (iv) 727

4. What will be the unit digit of the squares of the following numbers ?
(i) 3853 (ii) 1234 (iii) 26387

5. The following numbers are obviously not perfect squares. Give reason.
(i) 65000 (ii) 89722 (iii) 222000

6. Without adding find the sum.

(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

29
Class VIII : Mathematics

5.0 FINDING THE SQUARE OF A NUMBER

It is easy to find the square of small numbers like 2, 3, 6, 7, 9 etc. but we cannot find the square of
large numbers so quickly.
E.g. There is a way to find this without having to multiply 47 × 47
47 = 40 + 7
2
47 = (40 + 7) (40 + 7)
= 40 (40 + 7) + 7 (40 + 7)
= 1600 + 280 + 280 + 49
= 2209
5.1 Other patterns in squares
Consider the following pattern
2
25 = 625 = (2 × 3) hundreds + 25
2
35 = 1225 = (3 × 4) hundreds + 25
2
75 = 5625 = (7 × 8) hundreds + 25
Generalising it, consider a number with unit digit 5 i.e. a5
2 2
(a 5) = (10 a + 5)
= 10 a (10a + 5) + 5 (10 a + 5)
2
= 100 a + 50 a + 50 a + 25
= 100 a (a + 1) + 25
= a (a + 1) hundred + 25
5.2 Pythagorean triplets
Consider the following,
2 2 2
3 + 4 = 9 + 16 = 25 = 5
The collection of numbers 3, 4 and 5 is known as pythagorean triplet.
6, 8, 10 is also a pythagorean triplet since
2 2 2
6 + 8 = 36 + 64 = 100 = 10
Generalising it, for any natural number m > 1,
we have
2 2 2 2 2
(2m) + (m – 1) = (m + 1)
2 2
So, 2m, m – 1 and m + 1 form a pythagorean triplet.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

Illustration 4. Write a pythagorean triplet whose one member is

(i) 6 (ii) 14
2 2
Solution (i) Put m = 3 in 2m, m – 1, m + 1
2 2 2
\ 2m = 6, m – 1 = 3 – 1 = 9 – 1 = 8 and m + 1 = 9 + 1 = 10
Thus, 6, 8 and 10 are Pythagorean triplets.
2 2
(ii) Put m = 7 in 2m, m – 1, m + 1
2 2 2
\ 2m = 14, m – 1 = 7 – 1 = 49 – 1 = 48 and m + 1
2
= 7 + 1 = 49 + 1 = 50

Thus, 14, 48 and 50 are Pythagorean triplets.

30
 Squares and Square Roots

Illustration 5. Find a pythagorean triplet where one of the numbers is 12.

2
Solution If we take m – 1 = 12
2
Þ m = 12 + 1 = 13
then the value of m will not be an integer.
2
So, we try to take, m + 1 = 12
2
Þ m = 12 – 1 = 11
Again, the value of m will not be an integer.
So, let us take 2m = 12
Þ m=6
Thus, the other members of pythagorean triplet are as follows :
2 2
m – 1 = 6 – 1 = 36 – 1 = 35
2 2
m + 1 = 6 + 1 = 36 + 1 = 37
Thus, the required triplet is 12, 35 and 37

6.0 SQUARE ROOTS

Finding the square root is the inverse operation of squaring.
2
If 4 × 4 = 16 i.e. 4 = 16.
Then the square root of 16 is 4.
2
In other words, if q = p , then |p| is called the square root of q.

The square root of a number is denoted by the symbol or Ö.

for e.g. 16 = 4
2 2
But, we know, (4) = (–4) = 16
2 2
Similarly 3 = (–3) = 9

No but square root is always positive. Positive square root of a number is denoted by the symbol

For Example : 4 = 2 (not - 2) 9 = 3 (not - 3) etc.

6.1 Some Properties of Square Roots

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(i) The square root of an even perfect square is even and that of an odd perfect square is odd.
for e.g. 4 = 2, 16 = 4,
9 = 3, 25 = 5,
(ii) Since there is no number whose square is negative, the square root of a negative number is
not defined.

(iii) If a number ends in an odd number of zeroes, then it cannot have a square root which is a
natural number.

(iv) If the unit's digit of a number is 2, 3, 7 or 8, then square root of that number (in natural
numbers) is not possible.

(v) If m is not a perfect square, then there is no integer n such that square root of m is n.

31
Class VIII : Mathematics
6.2 Finding Square Root Through Repeated Subtraction

Recall the pattern formed while adding consecutive odd numbers.

2
1 + 3 + 5 + 7 + 9 = 5 = 25
2
1 + 3 + 5 + 7 + 9 + 11 = 6 = 36
2
Sum of first n odd numbers = n

The above pattern can be used to find the square root of the given number.

(i) Obtain the given perfect square whose square root is to be calculated. Let the number be a.

(ii) Subtract from it successively 1, 3, 5, 7, 9,..... till you get zero.

(iii) Count the number of times the subtraction is performed to arrive at zero. let the number be n.

(iv) Write a = n.

Illustration 6. Find the square root of 36 by successive subtractions.

Solution We have,
36 – 1 = 35
35 – 3 = 32
32 – 5 = 27
27 – 7 = 20
20 – 9 = 11
11 – 11 = 0
Clearly, we have performed subtraction six times.
\ 36 = 6
This method is the simplest method for finding square root but this method is convenient
for calculating square root of small numbers only, because for large numbers it will be
lengthy and time taking.
So, we have more methods for finding square roots : prime factorisation method and
long division method.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

6.3 Finding Square Root Through Prime Factorisation

In order to find the square root of a perfect square by prime factorization, we follow the following
steps.
(i) Obtain the given number.
(ii) Resolve the given number into prime factors by successive division.
(iii) Make pairs of prime factors such that both the factors in each pair are equal.
(iv) Take one factor from each pair and find their product.
(v) The product obtained is the required square root.

32
 Squares and Square Roots

2 7744
Illustration 7. Find the square root of 7744 by prime factorization. 2 3872
Solution Resolving 7744 into prime factors, Þ 7744 2 1936

2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 2 968
2 484
Taking one factor from each pair
2 242
7744 = 2 × 2 × 2 × 11 11 121

= 8 × 11 = 88 11 11
1

6.4 Finding the Square Root by Division Method

When the numbers are large, even the method of finding square root by prime factorisation becomes
lengthy and difficult. To overcome this problem we use long division method.

(i) Consider the following steps to find the square root of 529.

Step-1 : Place a bar over every pair of digits starting from the digit at one's place. If the number of
digits in it is odd, then the left most single digit too will have a bar. Thus we have, 5 29
Step-2 : Find the largest number whose square is less than
or equal to the number under the extreme left 2
–
2 2 2 52 9
bar (2 < 5 < 3 ). Take this number as the divisor –4
and the quotient with the number under the 1
extreme left bar as the dividend (here 5). Divide
and get the remainder (1 in this case).
2
Step-3 : Bring down the number under the next bar –
2 52 9
(i.e., 29 in this case) to the right of the –4
remainder. So the new dividend is 129. 12 9

Step-4 : Double the divisor and enter it with a blank

2
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

on its right. –
2 529
Step-5 : Guess a largest possible digit to fill the –4
4_ 12 9
blank which will also become the new digit in
the quotient, such that when the new divisor
23
is multiplied to the new quotient the product –
2 52 9
is less than or equal to the dividend. –4
In this case 42 × 2 = 84. 43 12 9
12 9
As 43 × 3 = 129 so we choose the new ×
digit as 3. Get the remainder.

Step-6 : Since the remainder is 0 and no digits are left in the given number, therefore, 529 = 23.

33
Class VIII : Mathematics
(ii) Consider the following steps to find the square root of 4096.
Step-1 : Place a bar over even pair of digits starting from the one's digit (4096) .
Step-2 : Find the largest number whose square 6
is less than or equal to the number under 6 40 96
2 2
the left-most bar (6 < 40 < 7 ). Take this – 36
number as the divisor and the number 4
under the left-most bar as the dividend.
Divide and get the remainder i.e., 4 in
this case. 6
Step-3 : Bring down the number under the next 6 40 96
– 36
bar (i.e., 96) to the right of the remainder.
496
The new dividend is 496.

Step-4 : Double the divisor and enter it with a blank on its right.
Step-5 : Guess a largest possible digit to fill
the blank which also becomes the new 6
digit in the quotient such that when the 6 40 9 6
new digit is multiplied to the new quotient – 36
the product is less than or equal to the
12 – 49 6
di vi d e nd . I n t hi s c a s e w e s e e t h a t
124 × 4 = 496.
So the new digit in the quotient is 4. Get the remainder.

Step-6 : Since the remainder is 0 and no bar left, therefore, 4096 = 64.
64
6 40 9 6
– 36
12 4 49 6
– 496
0
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

Illustration 8. Find the square root of 361.

19
1 36 1
1
Solution 29 2 61
2 61
0
Thus the square root of 361 is 19.

34
 Squares and Square Roots
Illustration 9. Find the least number that should be subtracted from 2750 to get a perfect square.
5 2
5 27 50
– 25
10 2 2 50
Solution
– 2 04
46

The remainder is 46. Therefore, 46 should be subtracted from 2750 to get a perfect
square. The square root of this number will be 52.

7.0 SQUARE ROOT OF DECIMALS

Consider a number 176.341. Put bars on both integral part and decimal part. In what way is putting
bars on decimal part different from integral part ? Notice for 176 we start from the unit's place
close to the decimal and move towards left. The first bar is over 76 and the second bar over 1.
For 341, we start from the decimal and move towards right. First bar is move 34 and for the second
bar we put 0 and 1 and make 3410 .
Consider the following steps to find the square root of 17.64.
Step-1 : To find the square root of a decimal number we put bars on the integral part (i.e. 17) of
the number in the usual manner. And place bars on the decimal part. (i.e.64) on every
pair of digits beginning with the first decimal place. Proceed as usual. We get 17.64 .
4
Step-2 : Now proceed in a similar manner.
2 2
4 17 6. 4
The left most bar is on 17 and 4 < 17 < 5 . Take this
– 16
number as the divisor and the number under the left- 1
most bar as the dividend, (i.e. 17). Divide and get the
remainder. 4
4 17 64
.
Step-3 : The remainder is 1. Write the number
under the next bar (i.e.64) to the right of – 16
this remainder, to get 164. 8– 1 64
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

Step-4 : Double the divisor and enter it with 4.

a blank on its right. 4 17 .6 4
Since 64 is the decimal part so put a – 16
82 16 4
decimal point in the quotient.
Step-5 : We know 82 × 2 = 164, therefore,
4 .2
the new digit is 2. 4 17 .6 4
Divide and get the remainder. – 16
82 16 4
Step 6 : Since the remainder is 0 and no bar left,
– 1 64
therefore 17.64 = 4.2 0

35
Class VIII : Mathematics
8.0 ESTIMATING SQUARE ROOT
We can also estimate square roots of the numbers by finding the number whose square is closest
to the given number. It can be illustrated by the following example.
Estimate the square root of 300.
We know that, 100 < 300 < 400.
Since. 100 = 10 and 400 = 20
so, 10 < 300 < 20
2 2
Now between 10 and 20, we know that, 17 = 289 and 18 = 324.
Thus, 17 < 300 < 18.
But, 300 is closer to 289 as compared to 324.
Therefore, 300 is approximately equal to 17.

Illustration 10.Find the squares roots of 100 and 169 by the method of repeated subtraction.
Solution From 100, we subtract successive add numbers starting from 1 as under.
100 – 1 = 99 99 – 3 = 96
96 – 5 = 91 91 – 7 = 84
84 – 9 = 75 75 – 11 = 64
64 – 13 = 51 51 – 15 = 36
36 – 17 = 19 19 – 19 = 0
and obtain 0 at 10th step.
\ 100 = 10

Illustration 11.Find the square roots of the following numbers by the Prime Factorisation Method :
3 729
(i) 729 (ii) 400 (iii) 4096
3 243
Solution (i) 3 × 3 × 3 × 3 × 3 × 3 3 81
3 27
\ 729 = 3 × 3 × 3 = 27 3 9
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

3 3
1

2 400
(ii) By prime factorisation, we get
400 = 2 × 2 × 2 × 2 × 5 × 5 2 200
2 100
\ 400 = 2 × 2 × 5
2 50
= 20
5 25
5 5
1

36
 Squares and Square Roots

2 4096
2 2048
2 1024
(iii) By prime factorisation, we get 2 512
4096 =2 × 2 × 2 × 2 × 2 × 2 2 256
×2×2×2×2×2×2 2 128
2 64
\ 4096 = 2 × 2 × 2 × 2 × 2 × 2 2 32
= 64 2 16
2 8
2 4
2 2
1

Illustration 12. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution The smallest number divisible by each one of the numbers 4, 9 and 10 is their LCM.,
which is (2 × 2 × 9 × 5), i.e., 180.
Now,180 = 2 × 2 × 3 × 3 × 5
it must be multiplied by 5.
\ Required number = 180 × 5 = 900

Illustration 13. Find the square root of each of the following numbers by division method :

(i) 2304 (ii) 4489 (iii) 3481

Solution (i) By long division, we have

48
——
4 2304
\ 2304 = 48 –16
88 704
–704
0
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(ii) By long division, we have

67
\ 4489 = 67 ——
6 4489
–36
127 889
–889
0

59
(iii) By long division, we have — —
5 3481
\ 3481 = 59 – 25
109 981
– 981
0

37
Class VIII : Mathematics

Illustration 14. Find the least number which must be subtracted from each of the following numbers
so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402 (ii) 1989
Solution (i) Let us try to find the square
root of 402.
2
This shows the (20) is less 20
than 402 by 2.So, in order 2 402
to get a perfect square, 2 –4
must be subtracted from the 40 002
given number. – 0 00
\ Required perfect square number 2
= 402 – 2 = 400
Also, 400 = 20

(ii) Let us try to find the square

44
root of 1989. — —
2 4 1 9 89
This shows that (44) is less –16
than 1989 by 53. So, in order 84 389
to get a perfect square, 53 –336
must be subtracted from the 53
given number.
\ Required perfect square number
= 1989 – 53 = 1936
Also, 1936 = 44

Illustration 15. Find the square root of 0.056 correct to 3 places of decimal.
Solution We must find the square root to 4 places of decimal and then round off to 3 places.
For that the number must have 8 places of decimal i.e. 0.056 = 0.
05 60 00 00 .

0.23 66
2 0 . 05 6 0 0 0 0 0
–4
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

43 1 60
– 1 29
4 66 31 0 0
– 2 7 96
47 2 6 30 400
– 2 83 56
473 2 1 04 4
The square root we get is 0.2366 as 6 is greater than 5, we round off to 0.237.
Ans. The square root of 0.056 correct to 3 places of decimal is 0.237.

38
 Squares and Square Roots
Illustration 16. Estimate the square root of 140.
2 2
Solution We know that 11 = 121 and 12 = 144
so, 140 will lie between 121 and 144 i.e. lie between 11 and 12.
Now 140 – 121 = 19
and 144 – 140 = 4.
So the square root will be close to 144 .
The approximate value of 140 is 12.

1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are
perfect squares or not? If the number is a perfect square then find its square root.
(i) 44 (ii) 169 (iii) 36 (iv) 74

2. Find the square root of 9604 by prime factorization method.

3. Find the square root of the following numbers by long division method.
(i) 3136 (ii) 7921 (iii) 900

4. Find the square root of the following numbers.

(i) 2.56 (ii) 51.84 (iii) 31.36

5. Find the smallest number by which 675 should be divided to make it a perfect square.

6. Check whether the following numbers are perfect squares or not

(i) 2061 (ii) 7928 (iii) 1069

7. Find the smallest number by which 162 should be multiplied to make it a perfect square.

• A number ends with 0, 1, 4, 5, 6 or 9 It may or may not a square number but a square number must
end with 0, 1, 4, 5, 6 or 9.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

• A number ending in an odd number of zeros is never a perfect square. Thus, number 1000, 25000,
64000 are not perfect squares.

• The dot pattern of a square number can be arranged as a square.

• Square roots of integers that are not perfect squares are always irrational numbers.

• The square root of decimal numbers or fractions above 1 is less than the number itself ; the square
root of 1 is 1 and the square root of decimal numbers or fractions less than 1 (except 0) is greater
than the number.

39
Class VIII : Mathematics

SOME WORKED OUT ILLUSTRATIONS

Illustration 1.
What will be the unit digit of the squares of the following numbers ?
(i) 81 (ii) 272 (iii) 799
Solution
The unit digit of the squares of the given numbers is shown against the number in the following
table:
Unit digit in
S.No. Number the square of Reason
the number
(i) 81 1 1×1=1
(ii) 272 4 2×2=4
(iii) 799 1 9 × 9 = 81

Illustration 2.
The following numbers are not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928
Solution
A number that ends either with 2, 3, 7or 8 cannot be a perfect square. Also, a number that ends
with odd number of zero(s) cannot be a perfect square.
(i) Since the given number 1057 ends with 7, so it cannot be a perfect square.
(ii) Since the given number 23453 ends with 3, so it cannot be a perfect square.
(iii) Since the given number 7928 ends with 8, so it cannot be a perfect square.

Illustration 3.
The squares of which of the following would be odd numbers ?
(i) 431 (ii) 2826 (iii) 7779
Solution
(i) The given number 431 being odd, so its square must be odd.
(ii) The given number 2826 being even, so its square must be even.
(iii) The given number 7779 being odd, so its square must be odd.

Illustration 4.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

Find the square of the following numbers.

(i) 32 (ii) 35
Solution
(i) 322 = (30 + 2)2
= 30 (30 + 2) + 2 (30 + 2)
= 302 + 30 × 2 + 2 × 30 + 22
= 900 + 60 + 60 + 4 = 1024

(ii) 352 = (30 + 5)2 = 30(30 + 5) + 5(30 + 5)

= 302 + 30 × 5 + 5 × 30 + 52
= 900 + 150 + 150 + 25 = 1225

40
 Squares and Square Roots
Illustration 5.
For each of the following numbers, find the smallest whole number by which it should be multiplied
so as to get a perfect square number. Also find the square root of the square number so obtained.
(i) 252 (ii) 180
Solution
(i) By prime factorisation we get 2 252
252 = 2 × 2 × 3 × 3 × 7 2 126
It is clear that in order to get
3 63
a perfect square, one more 7
3 21
is required.
So, the given number should be 7 7
multiplied by 7 to make the 1
product a perfect square.
\ 252 × 7 = 1764 is a perfect square.
Thus, 1764 = 2 × 2 × 3 × 3 × 7 × 7
\ 1764 = 2 × 3 × 7 = 42
(ii) By prime factorisation, we get 2 180
180 = 2 × 2 × 3 × 3 × 5 2 90
3 45
It is clear that in order to get a
3 15
perfect square, one more 5 is 5 5
required. 1
So, the given number should be multiplied by 5 makes the product a perfect square.
\ 180 × 5 = 900 is a perfect square.
Thus, 900 = 2 × 2 × 3 × 3 × 5 × 5
\ 900 = 2 × 3 × 5 = 30

Illustration 6.
For each of the following numbers, find the smallest whole number by which it should be divided
so as to get a perfect square. Also find the square root of the square number so obtained.
(i) 252 (ii) 2925
Solution
(i) By prime factorisation, we get
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

252 = 2 × 2 × 3 × 3 × 7
Since the prime factor 7 cannot be paired.
\ The given number should be divided by
2 252
252 2 ´ 2 ´ 3 ´ 3 ´ 7
\ = 2 126
7 7
3 63
=2×2×3×3
3 21
= 36 is a perfect square
7 7
and, 36 = 2 ´ 2 ´ 3 ´ 3
1
=2×3=6

41
Class VIII : Mathematics

(ii) By prime factorisation, we get

3 2925
2925 = 3 × 3 × 5 × 5 × 13
Since the prime factor 13 3 975
cannot be paired. 5 325
\ The given number should be 5 65
divided by 13. 13 13
2925 3 ´ 3 ´ 5 ´ 5 ´ 13 1
\ =
13 13
= 3 × 3 × 5 × 5 = 225 is a perfect square
and, 225 = 3 ´ 3 ´ 5 ´ 5
= 3 × 5 = 15

Illustration 7.
Find the number of digits in the square root of each of the following numbers (without any calculation):
(i) 64 (ii) 144
Solution
We know that if a perfect square is of n-digits, then its square root will have n/2 digits if n is even
(n + 1)
and digits if n is odd.
2
(i) Given number is 64. It is a 2-digit number, i.e., even number of digits.

æ 2ö
\ The number of digits in 64 is çè 1 ÷ø , i.e., 1.

(ii) Given number is 144. It is a 3-digit number, i.e., odd number of digits.

æ 3 + 1ö
\ 144 contains çè 2 ÷ø , i.e., 2-digits.

Illustration 8.
Find the square root of the following decimal numbers.
(i) 2.56 (ii) 51.84
Solution
(i) Here, the number of decimal 1.6
places is already even. So, —
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

1 2.56
mark off periods and proceed –1
as under \ 2.56 = 1.6 26 156
–156
0

(ii) Here, the number of decimal

7.2
—
places are already even. So, 7
51.84
–49
mark off periods and proceed
142 284
as under \ 51.84 = 7.2 –284
0

42
 Squares and Square Roots

Illustration 9.
Find the least number which must be added to each of the following numbers so as to get a perfect
square. Also find the square root of the perfect square so obtained.
(i) 525 (ii) 1750
Solution
(i) We try to find out the square root of 525. 22
—
We observe that 2 525
(22)2 < 525 < (23)2 –4
The required number to be 42 125
added –84
= (23)2 – 525 41
= 529 – 525 = 4
\ Required perfect square number
= 525 + 4 = 529
Clearly, 529 = 23
(ii) We try to find out the square root of 1750
We observe that (41)2 < 1750 < (42)2.
41
The required number to be added — —
= (42)2 – 1750 4 1750
= 1764 – 1750 = 14 –16
\ Required perfect square 81 150
number –81
69
= (1750 + 14) = 1764
Clearly, 1764 = 42

Illustration 10.
There are 500 children in a school. for a P.T. drill they have to stand in such a manner that the
number of rows is equal to number of columns. How many children would be left out in this
arrangement?
Solution
Let us find the square root of 500.
This rows that (22)2 = 484 is less than 500 by 16.
\ 16 students have to go out for others to do the P.T. practice as per condition.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

Illustration 11.
The smallest number in a pythagorean triplet is 12. Find the other two numbers.
Solution
Pythagorean triplets can be obtained by the following formula :
2m, m2 – 1, m2 + 1,
where m is a natural number (m > 1), of these three numbers generally 2m is the smallest.
Here, 2m = 12
So, m=6
m2 – 1 = 36 – 1 = 35
and m2 + 1 = 36 + 1 = 37

43
Class VIII : Mathematics
Illustration 12.
Find the greatest 4-digit number, which is a perfect square. Find its square root.
Solution
The greatest 4-digit number is 9999. First, find the least number which when subtracted from 9999
gives a perfect square.
9 9
9 99 99
– 81
189 189 9
– 17 01
198
The remainder is 198. The required number, which is a perfect square is 9999 – 198 = 9801 and
clearly, 9801 = 99.

Illustration 13.
5929 students are sitting in an auditorium in such a manner that there are as many students in
a row as there are rows in the auditorium. How many rows are there in the auditorium ?
Solution
Let there be 'a' rows in the auditorium
Since the number of students in a row is same as the number of rows in the auditorium.
\ Number of students in a row = a
Þ Number of students in 'a' rows = a × a = a2
It is given that the total number of students in the auditorium = 5929.
\ a2 = 5929
Þ a= 5929
7 5929
Þ a= (7 ´ 7) ´ (11 ´ 11) 847
7
[By prime factorization]
11 121
Þ a = 7 × 11 [Taking one factor in each pair]
Þ a = 77 11
Hence, there are 77 rows in the auditorium.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

44
 Squares and Square Roots

1. The value of 11 + 29 - 15 + 4 - 9

(A) 3 (B) 4 (C) 5 (D) 6

2. Given that 35 = 5.9 , what is the value of 0.35 + 35 ?

(A) 6.49 (B) 0.649 (C) 5.49 (D) 5.09

3. Given that 84.32 = 9.18 , what is the value of (91.8) ?

2

(A) 84320 (B) 8432 (C) 843.2 (D) 84.32

4. What least number must be multiplied to 112, 896 to make it a perfect square
(A) 2 (B) 3 (C) 6 (D) None of these

5. What least number must be subtracted from 1000 to make it a perfect square
(A) 24 (B) 39 (C) 41 (D) 43

6. What least number must be added to 1000 to make it a perfect square.

(A) 24 (B) 39 (C) 41 (D) None of these

7. What least number must be divided to 19, 845 to make it a perfect square.
(A) 5 (B) 7 (C) 9 (D) 3

8. The greatest 5-digit number, which is a perfect square.

(A) 99854 (B) 99855 (C) 99856 (D) 99859

9. The value of 214 + 130 - 88 - 44 + 25 :

(A) 14 (B) 15 (C) 16 (D) 17

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

1 1 1 1 1 1
10. The value of 1 - [1 - 1 + (1 - 1 - 1 )] is
2 2 2 2 2 4

1 1 1 1
(A) (B) (C) (D) 1
2 4 16 5

11. The value of the expression 1 + 23 1 + 24 1 + 25 26 ´ 28 + 1 is equal to :

(A) 24 (B) 25 (C) 26 (D) None of these

12. Sum of the first n odd natural numbers is –

(A) 2n + 1 (B) n2 (C) n2 – 1 (D) 2n2 + 1

45
Class VIII : Mathematics
13. Which of the following is a pythagorean triplet?
(A) (2, 3, 5) (B) (5, 7, 9) (C) (6, 9, 11) (D) (8, 15, 17)

14. A perfect square number can never have the digit.....at the units place.
(A) 1 (B) 4 (C) 8 (D) 9

15. Which of the following numbers is a perfect square?

(A) 141 (B) 196 (C) 124 (D) 222

16. 0.9 = ?
(A) 0.3 (B) 0.03 (C) 0.33 (D) None of these

17. What least number must be added to 6072 to make it a perfect square?
(A) 6 (B) 10 (C) 12 (D) 16

57
18. 4 =?
196

1 3 5 9
(A) 2 (B) 2 (C) 2 (D) 2
14 14 14 14

19. 1.0816 = ?
(A) 1.04 (B) 1.286 (C) 0.904 (D) 1.35

20. 0.00059049 is equal to

(A) 0.243 (B) 0.0243 (C) 0.00243 (D) 0.000243

21. What least number must be subtracted from 176 to make it a perfect square?
(A) 16 (B) 10 (C) 7 (D) 4

22. If x * y = x 2 + y 2 , the value of (1 * 2 2 ) (1 * – 2 2 ) is :

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(A) – 7 (B) 0 (C) 2 (D) 9

23. 991 is
(A) Greater than 31 (B) Less than 31 (C) Equal to 31 (D) Equal to 41

24. 169 - 100 =

(A) 2 (B) 3 (C) 2 (D) 3

25. The smallest number by which 12348 must be divided to obtain a perfect square is
(A) 3 (B) 4 (C) 5 (D) 7

46
 Squares and Square Roots

1. Given 5 = 2.236 the value of 45 + 605 - 245 correct to 3 decimal places is :

(A) 15.652 (B) 11.180 (C) 18.652 (D) 16.652

16 n
2. If = then n =
49 49
(A) 4 (B) 7 (C) 16 (D) 28

1
3. Calculate the value of 6
4

1 3 1 1
(A) (B) (C) 1 (D) 2
2 4 2 2

16
4. Calculate the value of 1-
25

9 3 3 1
(A) (B) (C) (D)
5 5 25 25

1
5. 169 - 69 ´ =
52

2 4 1 3
(A) (B) (C) 1 (D) 1
5 5 5 5

6. Given that 4.8 = 2.191 and 48 = 6.928, find the value of 4,800 - 480
(A) 670.89 (B) 437.7 (C) 67.089 (D) 47.37

7. 7, 456 lies between

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(A) 80 and 90 (B) 100 and 110 (C) 800 and 900 (D) 1,000 and 1,100

8. The value of 6.9 ´ 100 lies between

(A) 800 and 900 (B) 200 and 300 (C) 80 and 90 (D) 20 and 30

9. Given that (3.9268)2 = 15.42, find the value of 154200

(A) 3926.8 (B) 392.68 (C) 39.268 (D) 3.9268

10. Given that 6.4 = 2.53, find the value of 640 + 64

(A) 33.30 (B) 27.83 (C) 10.53 (D) 3.33

47
Class VIII : Mathematics

11. Given that 45 = 6.708 and 4.5 = 2.121, find the value of 4,500 + 450
(A) 882.9 (B) 458.7 (C) 88.29 (D) 45.87

12. Which of the following values is not equal to 5 ?

( 5)
2
(A) 52 (B) (C) ( -5 ) 2 (D) - ( 5 )2

13. Given that (5.142)2 = 26.44, find 264400

(A) 5142 (B) 514.2 (C) 51.42 (D) 5.142

14. Given that 64.52 = 8.032, what is the value of (80.32) ?

2

(A) 64,520 (B) 6,452 (C) 645.2 (D) 64.52

15. The value of 2980 correct to two decimal place is

(A) 17.26 (B) 54.59 (C) 58.75 (D) 74.29

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

48
 Squares and Square Roots

Very short answer type questions

1. Which of the following triplets are pythagorean?

(i) (8, 15, 17) (ii) (16, 63, 65) (iii) 12, 16, 20

2. Find the squares of 510 using the identity : (a + b)2 = a2 + 2ab + b2

3. Find the squares of 99 using the identity : (a – b)2 = a2 – 2ab + b2

4. Evaluate:
(i) (38)2 – (37)2 (ii) (75)2 – (74)2 (iii) (141)2 – (140)2

Short answer type questions

5. Write a pythagorean triplet whose smallest number is
(i) 6 (ii) 16

6. Find the square root of each of the following numbers by using the method of prime factorisation.
(i) 5184 (ii) 40000 (iii) 1444

7. Find the value of each of the following, using the column method.
(i) (23)2 (ii) (52)2

8. Find the value of each of the following, using the diagonal method.
(i) (67)2 (ii) (137)2

Long answer type questions

80
9. Find the value of : (i) (ii) 72 × 338
405

10. Find the square root of the following fractions

26 151 324
(i) 23 (ii) 10 (iii)
121 225 361

11. Find the square root of the following decimal numbers.

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

(i) 0.7225 (ii) 150.0625 (iii) 236.144689

12. Find the square root of each of the following correct to three places of decimal.
7
(i) 17 (ii) 1.7 (iii) 2.5 (iv)
8
High order thinking skills (HOTS)
13. Find the least number which must be subtracted from 7581 to obtain a perfect square. Find this
perfect square and its square root.

14. By what smallest number must 180 be multiplied so that it becomes a perfect square? Also, find the
square root of the number obtained.

15. Find the value of 15625 and the use it to find the value of 156.25 + 1.5625 .

49
Class VIII : Mathematics

ANSWERS
CHECK POST-1
1. (i) 1 (ii) 9 (iii) 6 (iv) 6 (v) 4
(vi) 0 (vii) 5 (viii) 9 (ix) 4 (x) 4
2. (i) two (ii) four (iii) No zeros
3. (i) odd (ii) even (iii) even (iv) odd
4. (i) 9 (ii) 6 (iii) 9
5. (i) 0 (ii) 2 (iii) 0
6. (i) 25 (ii) 100

CHECK POST-2
1. (i) No, it is not perfect square (ii) yes, 13 (iii) yes, 6 (iv) No
2. 98 3. (i) 56 (ii) 89 (iii) 30
4. (i) 1.6 (ii) 7.2 (iii) 5.6 5. 3
6. (i) No (ii) No (iii) No 7. 2

EXERCISE-1

Que. 1 2 3 4 5 6 7 8 9 10
Ans. B A B D B A A C B A
Que. 11 12 13 14 15 16 17 18 19 20
Ans. A B D C B D C A A B
Que. 21 22 23 24 25
Ans. C D A D D

EXERCISE-2

Que. 1 2 3 4 5 6 7 8 9 10
Ans. A D D B A D A D B A
Que. 11 12 13 14 15
Ans. C D B B B
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\2. Square and Square Roots

EXERCISE-3
1. (i) Yes (ii) Yes (iii) Yes 2. 260100 3. 9801
4. (i) 75 (ii) 149 (iii) 281
5. (i) (6, 8, 10) (ii) (16, 63, 65) 6. (i) 72 (ii) 200 (iii) 38
7. (i) 529 (ii) 2704 8. (i) 4489 (ii) 18769

4 53 4 18
9. (i) (ii) 156 10. (i) (ii) 3 (iii)
9 11 15 19
11. (i) 0.85 (ii) 12.25 (iii) 15.367
12. (i) 4.123 (ii) 1.304 (iii) 1.581 (iv) 0.935
13. 12, 7569, 87 14. 5,30 15. 13.75

50
 CUBES AND
CHAPTER 3
CUBE ROOTS

1.0 CUBES

2.0 SOME PATTERNS RELATED TO THE CUBE

2.1 Adding Consecutive odd numbers
2.2 Difference of cubes of consecutive numbers

3.0 CUBES AND THEIR PRIME FACTORS

3.1 Column Method

4.0 CUBE ROOTS

4.1 Cube roots through prime factorization method

5.0 CUBE ROOT THROUGH ESTIMATION

EXERCISE-1 (ELEMENTARY)

EXERCISE-2 (SEASONED)

EXERCISE-3 (SUBJECTIVE)
 Cubes and Cube Roots

CUBES AND CUBE ROOTS

1.0 CUBES
Number Cube Number Cube
If we multiply a number by itself three times, then it is the cube 1 1 11 1331
of that number. Cubes in geometry, is a solid figure with 6 2 8 12 1728
faces (which are congruent squares), with 12 edges (which 3 27 13 2197
are equal in length) and with 8 vertices. If a cube has an 4 64 14 2744
3
edge of 1 cm, its volume is = 1 × 1 × 1 = 1 cm . Similarly, 5 125 15 3375
3
if the edge is 2 cm, its volume is = 2 × 2 × 2 = 8 cm . 6 216 16 4096
Here 1, 8 are perfect cubes. But 9 is not a perfect cubes 7 343 17 4913
because there is no natural number which multiplied by 8 512 18 5832
itself three times gives 9. The following table gives the cubes 9 729 19 6859
of the first 20 natural numbers. 10 1000 20 8000

Illustration 1. Evaluate :
3
3 æ 4ö
(i) [4 × (–5)] (ii) ç - ÷
è 5ø
3 3 3
Solution (i) [4 × (–5)] = 4 × (– 5) = 64 × – 125 = – 8000

æ 4ö ( -4 )
3 3
-64
(ii) ç- ÷ = =
è 5ø 53 125

2.0 SOME PATTERNS RELATED TO THE CUBE

2.1 Adding consecutive odd numbers
Observe the following pattern of sum of odd numbers.
3
1 = 1 = 1
3
3 + 5 = 8 = 2
3
7 + 9 + 11 = 27 = 3
3
13 + 15 + 17 + 19 = 64 = 4
3
21 + 23 + 25 + 27 + 29 = 125 = 5
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

3
Illustration 2. How many consecutive odd numbers will be needed to obtain the sum as 10 ?
Solution As we know
3
4 = 13 + 15 + 17 + 19 = 64 (4 consecutive odd no. are needed)
3
5 = 21 + 23 + 25 + 27 + 29 = 125 (5 consecutive odd no. are needed)

3
10 = (10 consecutive odd no. are needed)
2.2 Difference of cubes of consecutive numbers
3 3
2 –1 =1+2×1×3
3 3
3 –2 =1+3×2×3
3 3
4 –3 =1+4×3×3
3 3
(n + 1) – n = 1 + (n + 1) (n) × 3

51
Class VIII : Mathematics

3 3
Illustration 3. Using the above pattern, find the value of 51 – 50 .
3 3
Solution Q (n + 1) – n = 1 + (n + 1) (n) (3)
3 3
\ 51 – 50 = 1 + 51 × 50 × 3 = 7651

3.0 CUBES AND THEIR PRIME FACTORS

Consider the prime factorization of the cubes of the numbers.
Prime factorization of the cube of numbers
3 3 3
4 = 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2 × 2
3 3 3
15 = 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 3 × 5
3 3 3 3
12 = 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2 × 2 × 3

It is observed that each prime factor of a number appears three times in the prime factorization of
its cube. So 64, 3375, 1728 are perfect cubes.
But in the prime factorization of 500
500 = 2 × 2 × 5 × 5 × 5
There are three 5's in the product but only two 2's. So, 500 is not a perfect cube.

3.1 Column Method

For finding the cube by column method use following formula :
3 3 2 2 3
(a + b) = a + 3a b + 3ab + b
In this method, first find the 2 digits of a number.
3
Example : 24 , so here the two digits are 2 and 4.
a = 2 and b = 4 by using the column method :
Column-I Column-II Column-III Column-IV
a3 3×a2×b 3×a×b2 b3
23 = 8 3 × 22 × 4 = 48 3 × 2 × 42 = 96
43 = 64
+5 +10 +6
13 58 102 64
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

13 8 2 4
3
\ 24 = 13824

Illustration 4. Is 11025 a perfect cube? If not, find the smallest natural number by which 11025 must
be multiplied so that product is a perfect cube.
Solution 11025 = 3 × 3 × 5 × 5 × 7 × 7

The prime factors 3, 5, 7 do not appear in a group of three. Therefore, 11025 is not a
perfect cube. To make it a cube, we need one more 3, 5 and 7. In that case
11025 × 3 × 5 × 7 = 3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 7 = 1157625
Here the smallest natural number by which 11025 should be multiplied to make it a
perfect cube is 105.

52
 Cubes and Cube Roots

1. Find the volume of a cube with edge 2.2 cm.

2. What is the smallest number by which 8575 may be multiplied so that the product is a perfect cube?
3. What is the smallest number by which 8575 may be divided so that the quotient is a perfect cube?

4.0 CUBE ROOTS

Finding the cube root is the opposite operation of cubing. The symbol for cube is 3 i.e., the same as

square root but with 3 written in the stroke as shown.

Consider the following

Statement Inference Statement Inference

3
13 = 1 1 =1 6 3 = 216 3
216 = 6
3 3 3
23 = 8 8= 2 =2 7 3 = 343 3
343 = 7
3 3 3 3 3 3
3 = 27 27 = 3 = 3 8 = 512 512 = 8
3 3 3
4 = 64 64 = 4 3 = 4 9 = 729 729 = 9
3 3

5 3 = 125 3
125 = 5 10 3 = 1000 3
1000 = 10

4.1 Cube roots through prime factorization method

Consider 74088. We find its cube root by prime factorization.
3 3 3
74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7 = 2 × 3 × 7

3
74088 = 3
23 ´ 33 ´ 7 3 = 3
(2 ´ 3 ´ 7)3 = 3
(42)3 = 42

Step-1 : To find the cube root of a perfect cube; find its prime factors and make them into
group of 3.
Step-2 : Pick one factor from each group and multiply them.
Step-3 : The product will be the cube root of the given number.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

Illustration 5. Find the cube root of 5832 by prime factorization method.

Solution 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 we can
3 3 3 3 3
5832 = 2 × 3 × 3 = (2 × 3 × 3) = (18)

3
5832 = 3
18 3 = 18

1. Find the cube root of 0.003375.

2. Find the cube root of 512.

53
Class VIII : Mathematics

5.0 CUBE ROOT THROUGH ESTIMATION

If you know that the given number is a cube number then following method can be used.
Step-1 : Take any cube number say 857375 and start making groups of three digits. Starting
from the right most digit of the number.
857 375

Second group First group

We get 375 and 857 as two groups of three digits each.

Step-2 : First group i.e., 375 will give you the one's (or unit's) digit of the required cube root.
The number 375 ends with 5. We know that 5 comes at the unit's place of a number
only when it's cube root ends in 5.
So, we get 5 at the unit's place of the cube root.
Step-3 : Now take another group, i.e., 857.
3 3
We know that 9 = 729 and 10 = 1000. Also, 729 < 857 < 1000. We take the one's place,
of the smallest number 729 as the ten's place of the required cube root.
So, we get 3
857375 = 95.

Illustration 6. Find the cube root of 17576 through estimation.

Solution The given number is 17576.
Step-1 : From groups of three starting from the right most digit of 17576. 17 576. In
this case one group i.e., 576 has three digits whereas 17 has only two
digits.
Step-2 : Take 576. The digit 6 is at its one's place.
We take the one's place of the required cube root as 6.
Step-3 : Take the other group, i.e., 17.
Cube of 2 is 8 and cube of 3 is 27. 17 lies between 8 and 27.
The smaller number among 2 and 3 is 2.
The one's place of 2 is 2 itself. Take 2 as ten's place of the cube root of
17576.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

Thus, 3 17576 = 26.

• Only 4 natural numbers, less than 100 are perfect cubes and less than 1000 only 9 numbers are
perfect cubes.
• The square of a negative integer is positive but the cube of a negative integer is negative.
• Sum of first n cubes is the nth triangle number squared.
2
3 3 3 3 2 é n(n + 1) ù
1 + 2 +3 + ... + n =(1+ 2 +.....+n) = ê .
ë 2 úû

54
 Cubes and Cube Roots

SOME WORKED OUT ILLUSTRATIONS

Illustration 1.
Which of the following numbers are not perfect cube?
(i) 216 (ii) 128
Solution
(i) Resolving 216 into prime factors, we find that
2 216
216 = 2 × 2 × 2 × 3 × 3 × 3 2 108
Clearly, the prime factors of 2 54
216 can be grouped into triples 3 27
3 9
of equal factors and no factor is
3 3
left over.
1
\ 216 is a perfect cube
(ii) Resolving 128 into prime factors, we find that 2 128
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 2 64
2 32
Now, if we try to group together
2 16
triples of equal factors, we are 2 8
2 4
lef t with a single fact or, 2.
2 2
\ 128 is not a perfect cube. 1
Illustration 2.
Find the smallest number by which each of the following numbers must be multiplied to obtain a
perfect cube.
(i) 243 (ii) 256
Solution 3 243
(i) Writing 243 as a product of 3 81
3 27
prime we have
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

3 9
243 = 3 × 3 × 3 × 3 × 3
3 3
Clearly, to make it a perfect 1
cube, it must be multiply by 3.
2 256
(ii) Writing 256 as a product of 2 128
2 64
prime factors, we have 2 32
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 2 16
2 8
Clearly, to make it a perfect 2 4
cube, it must be multiply by 2. 2 2
1
55
Class VIII : Mathematics

Illustration 3.
Find the smallest number by which each of the following numbers must be divided to obtain a perfect
cube.
(i) 81 (ii) 128
Solution
3 81
(i) Writing 81 as a product of
3 27
pr im e f a c t o r s , w e ha ve
3 9
81 = 3 × 3 × 3 × 3
3 3
Clearly, to make it a perfect 2 128
1
cube, it must be divided by 3. 2 64
2 32
(ii) Writing 128 as a product of 2 16
pr im e f a c t o r s , w e ha ve 2 8
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 2 4
Clearly to make it a perfect 2 2
cube, it must be divided by 2. 1

Illustration 4.
Parikshit makes a cuboid of plasticine of side 5 cm, 2 cm, 5 cm. How many such cuboids will be
need to form a cube ?
Solution
Volume of the cuboid = 5 × 2 × 5 = 2 × 5 × 5
To make it a cube, we require 2 × 2 × 5, i.e., 20 such cuboids.

Illustration 5.
Find the cube root of each of the following numbers by prime factorisation method :
(i) 64 (ii) 512 (iii) 10648
Solution

(i) Re so lv i ng 64 i nto pr im e 2 64
2 32
factors, we get 2 16
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

64 = 2 × 2 × 2 × 2 × 2 × 2 2 8
2 4
\ 3
64 = (2 × 2) = 4 2 2
1 2 512
2 256
(ii) Resolving 512 into prime
2 128
factors, we get 2 64
2 32
512 = 2 × 2 × 2 × 2 × 2 × 2
2 16
×2×2×2 2 8
2 4
\ 3
512 = (2 × 2 × 2) = 8
2 2
1

56
 Cubes and Cube Roots

(iii) Resolving 10648 into prime 2 10648

2 5324
factors, we get
2 2662
10648 = 2 × 2 × 2
11 1331
× 11 × 11 × 11 11 121
\ 11 11
3
10648 = (2 × 11) = 22
1
Illustration 6.
State true or false
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25
(iv) There is no perfect cube which ends with 8.
Solution
(i) False
(ii) True
2 3
(iii) False, as 15 = 225 and 15 = 3375
3 3
(iv) False, as 8 = 2 , 1728 = 12 , etc.

Illustration 7.
Find the smallest number by which each of the following numbers must be multiplied to obtain a
perfect cube.
(i) 243 (ii) 675
Solution
(i) 3 243
3 81
3 27 =3×3×3×3×3
3 9
3 3
1
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

The prime factor 3 does not appear in groups of three so, 243 is not a perfect cube. To make
it a perfect cube, we need one more 3.

3 67 5
(ii)
3 2 25
3 75
=3×3×3×5×5
5 25
5 5
1
The prime factor 5 does not appear in group of three so 675 is not perfect cube. To make it a
perfect cube we need one more 5

57
Class VIII : Mathematics

Illustration 8.
Show that 0.001728 is the cube of a rational number
Solution
We have,
1728 2´2´2´2´2´2´3´3´3
0.001728 = =
1000000 2´2´2´2´2´2´5´5´5´5´5´5
3 3
(2 ´ 2 ´ 3)3 æ 12 ö æ 3 ö
Þ 0.001728 = 3 = ç ÷ =ç ÷
(2 ´ 2 ´ 5 ´ 5) è 100 ø è 25 ø

3
Hence, 0.001728 is the cube of the rational number .
25

Illustration 9.
Three numbers are in the ratio of 2 : 3 : 4. The sum of their cubes is 33957. Find the numbers.
Solution
Let the number be 2x, 3x and 4x. Then,
3 3 3
(2x) + (3x) + (4x) = 33957
3 3 3
Þ 8x + 27x + 64x = 33957
3
Þ 99x = 33957

3 33957
Þ x = = 343
99
3
Þ x =7×7×7
[Resolving 343 into prime factors]

Þ x= 3
7´7´7 = 7
Hence, the numbers are :
2x = 2 × 7 = 14, 3x = 3 × 7 = 21 and 4x = 4 × 7 = 28

Illustration 10.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

Find the cube root of – 226981 through estimation.

Solution

We have, 3
–226981 = – 3 226981
Consider the number 226981
Since unit's digit of 226981 is 1. Therefore, units digit of its cube root is also 1
Now, Number obtained by striking out units, tens and hundreds digits = 226
3 3
Q 6 < 226 < 7
\ Tens digit of the cube root of 226981 is 6.
3
Hence, 226981 = 61

\ 3
–226981 = – 61

58
 Cubes and Cube Roots

Illustration 11.
Is 1188 a perfect cube? If not by which smallest natural number should 1188 be divided so that the
quotient is a perfect cube?
Solution
1188 = 2 × 2 × 3 × 3 × 3 × 11
The primes 2 and 11 do not appear in groups of three so 1188 is not a perfect cube. So if we divided 1188
by 2 × 2 × 11 = 44.
Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is
44.

1188 3
= 27 = 3
44
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

59
Class VIII : Mathematics

æ 27 ö
1. Write ç ÷ in index form.
è 125 ø

3 3 3
æ 3ö æ5ö æ4ö
(A) ç ÷ (B) ç ÷ (C) ç ÷ (D) None of these
è5ø è 3ø è5ø

3
2. The value of (3.1) is
(A) 27.971 (B) 29.791 (C) 29.971 (D) 27.197

3. Number of digits in the cube of a two digit number may be :

(A) 4 (B) 5 (C) 6 (D) All of these

4. Which of the following numbers is a perfect cube?

(A) 1525 (B) 1728 (C) 1458 (D) 3993

5. Which of the following numbers is not a perfect cube ?

(A) 2197 (B) 512 (C) 2916 (D) 343

6. Which of the following numbers are cubes of rational numbers?

27 125
, , 0.001331, 0.04
64 128

27 125
(A) ,0.001331 (B) (C) 0.04 (D) None of these
64 128
7. The largest four digit number which is a perfect cube is :
(A) 8000 (B) 9261 (C) 9999 (D) None of these

8. The smallest natural number by which 25 must be multiplied to get a perfect cube is
(A) 5 (B) 20 (C) 25 (D) None of these
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

9. The smallest natural number by which 1296 be divided to get a perfect cube is
(A) 16 (B) 6 (C) 60 (D) None of these

10. Cube root of the quotient of two negative integers is

(A) Positive (B) Zero (C) Negative (D) None of these

11. The value of 3

-a 3 ´ 3 - b3 is
(A) a (B) b (C) ab (D) None of these

12. 3
5832 = ?
(A) 22 (B) 18 (C) 16 (D) 14

60
 Cubes and Cube Roots

3
-1728
13. Evaluate :
2744
6 6 3 12
(A) – (B) – (C) – (D) –
11 7 4 17

14. Which of the following numbers are the cube of a negative whole number?
(– 64, – 2197, – 1056, – 3888)
(A) – 64, – 2197 (B) – 1056, – 3888 (C) – 64, – 1056 (D) – 2197, – 3888

1/3
15. The value of [64 × (– 2744)] is
(A) 56 (B) – 56 (C) 65 (D) – 65

16. What is the least number by which 8640 is divided, the quotient as a complete cube number?
(A) 6 (B) 7 (C) 5 (D) 8

17. If the volume of a cube is 512 cu. metres. What is the length of one side of the cube?
(A) 6.8 m (B) 7.8 m (C) 8 m (D) 9 m
1
18. The value of (3 3 + 4 3 + 5 3 )3 is equal to :
(A) 6 (B) 12 (C) 36 (D) 216

19. How many such numbers are there which are equal to their cube but not equal to their square?
(A) 2 (B) 1 (C) 0 (D) 3

20. The cube of a number is 8 times the cube of another number. If the sum of the cubes of numbers
is 243, find the difference of the number:
(A) 3 (B) 4 (C) 6 (D) None of these

3
21. Calculate the value of (–0.4) .
(A) 0.640 (B) 0.064 (C) –0.064 (D) –0.640

3 3
22. 3 – (–0.6) = .........
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

(A) 27.216 (B) 26.784 (C) –26.784 (D) –27.216

3
16 æ 1 ö
23. ´ ç -1 ÷ = ..........
9 è 2ø

8 8
(A) –12 (B) –6 (C) - (D)
3 9

24. Which of the following is the cube of an integer ?

(A) –200 (B) 9 (C) 512 (D) 1,024

25. 3
0.125 + 3 = .......
(A) 8 (B) 3.5 (C) 2 (D) 0.35

61
Class VIII : Mathematics

192
1. Calculate the value of 3 -
81

5 4 3 13
(A) - (B) - (C) (D)
3 3 2 9

2. 3
216 is ............
(A) less than 6 (B) greater than 6 (C) equal to 6 (D) equal to 9

3. 3+ 3
-343 = ........
(A) 16 (B) 11 (C) –4 (D) –7

17
4. 3 -4 = ..........
27

11 5 2 5
(A) (B) (C) - (D) -
9 9 3 3

3
5. Given that 512 = 83 and 3.375 = 1.53, find the value of 512 ´ 3 3.375
(A) 12 (B) 9.5 (C) 8 (D) 1.5

3
6. Given that x = –6, find the value of x
(A) 216 (B) 18 (C) –18 (D) –216

-3
æ bö
( b)
3
7. If x = a3/2, y = çç ÷÷ ,z= , xyz = 8000, then a is
è aø

(A) 10 (B) 15 (C) 20 (D) 25

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

8. If 3
5 = 1.710, then 3
320 is
(A) 5.829 (B) 6.829 (C) 5.839 (D) 6.840

16 4 3
9. If 3 16 = ´ y , then y =
27 3

2 2 4
(A) (B) 7 (C) (D)
3 9 9

10. 13 + 23 + 33 + ......93 is equal to

(A) 729 (B) 2025 (C) 1331 (D) None of these

62
 Cubes and Cube Roots

11. If the cube root of 1728 is 12, then cube root of 1.728 is :
(A) 1.2 (B) 0.12 (C) 1.02 (D) 0.012

12. There are two number such that the sum of the number is 50 and their difference is 14. Find
difference of their cube ?
(A) 26936 (B) 23966 (C) 23696 (D) 32696

13. Find the value of A if 3

500.A = 10
(A) 3 (B) 2 (C) 4 (D) 5

14. Which one of the following is the cube of an odd number.

(A) 64 (B) 343 (C) 8000 (D) 1728

15. Calculate the value of 3

-64 + 9 2
(A) –4 (B) 3 (C) 5 (D) 77
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

63
Class VIII : Mathematics

Very short answer type questions

1. Find the cubes of :

(i) 13 (ii) 1.3 (iii) 0.4

2. Which of the following numbers are perfect cubes?

(i) 4096 (ii) 392

3. Find the value of each of the following using the short-cut or column method :
(i) (25)3 (ii) (47)3 (iii) (84)3

4. Find the cube roots of the following by successive subtraction of numbers :

(i) 64 (ii) 512

5. Find the cube roots of the following by prime factorization :

(i) 5832 (ii) 1728

6. Find the cubes of :

(i) 0.6 (ii) – 3.1

Short answer type questions

7. Find the cube root of the following :

(i) – 27000 (ii) – 0.000001

8. Evaluate :
3 3
æ4ö æ 10 ö
(i) ç ÷ (ii) ç ÷
è7ø è 11 ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

9. Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

10. Evaluate : 3
8 ´ 125

11. Find the cube roots of the following rational numbers :

4913 -512
(i) (ii)
3375 343

64
 Cubes and Cube Roots

Long answer type questions

12. Find the smallest number which when multiplied with 3600 will make the product a perfect cube.
Further, find the cube root of the product.

13. Evaluate : 3
8 ´ 17 ´ 17 ´ 17

14. By what smallest number 3645 be multiplied so that the product becomes a perfect cube?

15. By what smallest number 29160 be divided so that the quotient becomes a perfect cube?

16. By what smallest number should we divide 9000 so that the quotient becomes a perfect cube. Find
the cube root of the quotient.

High order thinking skills (HOTS)

17. Find the cube roots of the :

2744 473
(i) 729 × 216 (ii) (iii) 4
35937 2197

18. Solve

3
216 3
-125
(i) (ii)
2197 512

19. Evaluate :
(i) 3
-1331 (ii) 3
64 ´ 729

20. Show that 3 216 ´ (-343) = 3

216 × 3
-343
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

65
Class VIII : Mathematics

ANSWERS
CHECK POST-1

1. 10.648 cm3 2. 5 3. 25

CHECK POST-2
1. 0.15 2. 8

EXERCISE-1

Que. 1 2 3 4 5 6 7 8 9 10
Ans. A B D B C A B A B A
Que. 11 12 13 14 15 16 17 18 19 20
Ans. C B B A B C C A B A
Que. 21 22 23 24 25
Ans. C A B C B

EXERCISE-2

Que. 1 2 3 4 5 6 7 8 9 10
Ans. B C C D A D C D B B
Que. 11 12 13 14 15
Ans. A A B B C

EXERCISE-3
1. (i) 2197 (ii) 2.197 (iii) 0.064 2. (i) Yes (ii) No

3. (i) 15625 (ii) 103823, (iii) 592704 4. (i) 4, (ii) 8

5. (i) 18, (ii) 12 6. (i) 0.216 (ii) – 29.791 JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\3.Cubes and Cube Roots\Theory

64 1000
7. (i) – 30 (ii) –0.01 8. (i) (ii) 9. 25 10. 10
343 1331

17 8
11. (i) (ii) - 12.60, 60 13. 34 14.25 15.5
15 7

14 8 6 5
16.9, 10 17. (i) 54 (ii) (iii) 1 18. (i) (ii) –
33 13 13 8

19. (i) – 11 (ii) 36

66
 EXPONENTS AND
CHAPTER
4
POWERS

1.0 INTRODUCTION

2.0 POWERS WITH NEGATIVE EXPONENTS

3.0 LAWS OF EXPONENTS

4.0 WRITING EXPANDED FORM OF A DECIMAL NUMBER USING

EXPONENTIAL NOTATION

5.0 SCIENTIFIC NOTATION

6.0 COMPARING VERY LARGE AND VERY SMALL NUMBERS

EXERCISE-1 (ELEMENTARY)

EXERCISE-2 (SEASONED)

EXERCISE-3 (SUBJECTIVE)
 Exponents and Power

EXPONENTS AND POWER

1.0 INTRODUCTION
4
An expression like 2 × 2 × 2 × 2 can be written as the power 2 .

4
The base is the number
that is multiplied 2 The exponent tells how many
time the base is used as a factor.

The number that is expressed using

an exponent is called a power.

The table below shows how to write and read powers.

Powers Words Repeated factors

21 2 to the first power 2
22 2 to the second power or 2 squared 2×2
23 2 to the third power or 2 cubed 2×2×2
24 2 to the fourth power 2 × 2 × 2× 2
? ? ?
th 2 × 2 × 2 × ... 2
2n 2 to the n power n tim e s

4 th
Similarily 3 = 3 × 3 × 3 × 3 and is read as 3 to the 4 power or 3 to the power 4

2.0 POWERS WITH NEGATIVE EXPONENTS

–n
How can you evaluate an expression of the form a ?
Definition : In general for non-zero integer a,
1
a-n = where n is a positive integer..
an

Illustration 1. Find the value of

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

-3
–3 1 æ -6 ö
(i) 4 (ii) (iii) ç ÷
7 -2 è 7 ø
–3 1 1
Solution (i) 4 = 3
=
4 64
1 2
(ii) -2 = 7 = 7 × 7 = 49
7
-3
æ -6 ö 1 1 73
(iii) ç ÷ = 3
= =
è 7 ø æ -6 ö (-6)3 (-6)3
ç ÷ 73
è 7 ø
343 -343
= =
-216 216

67
Class VIII : Mathematics
Illustration 2. Write the reciprocal of
2 4
3 æ6ö æ -3 ö
(i) 4 (ii) ç ÷ (iii) ç ÷
è5ø è 2 ø
3 1 –3
Solution (i) Reciprocal of 4 = 3 or 4
4
1
2 2 -2
æ6ö 2 1 52 æ 5 ö æ6ö
(ii) Reciprocal of ç ÷ = æ 6 ö = 2 = 2 = ç ÷ or ç ÷
è5ø ç ÷ 6 6 è6ø è5ø
è5ø
2
5
4 4 –4
æ -3 ö 1 1 (2)4 æ 2 ö æ -3 ö
(iii) Reciprocal of ç ÷ = = = =ç ÷ or ç ÷
è 2 ø æ -3 ö
4
(–3)4 (–3) 4
è -3 ø è 2 ø
ç 2 ÷ (2)4
è ø

ìïæ 1 ö -2 æ 1 ö -3 üï æ 1 ö -2
1. Simplify : íç 4 ÷ - ç 2 ÷ ý ¸ ç 4 ÷
ïîè ø è ø ïþ è ø

2. Rewrite the following using positive exponents. Assume that no denominators are equal to 0.
–6 –9 3 –4 2
(i) b (ii) c d (iii) 7x y

3.0 LAWS OF EXPONENTS

You have studied the laws of exponents, with exponents as positive integer. These laws also hold
true for negative exponents.
Let us see if these hold true for integral exponents i.e. exponents which can be negative also.
m n m+n
(i) x ×x =x
Example :

–2 –3 1 1 1 1 1 –5
2 ×2 = 2 × 3 = 2 3 = 3+ 2 = =2
2 2 2 ´2 2 25
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

(–2)+ (–3) –5
So, 2 =2
m n m–n
(ii) x ¸x =x
Example :

–1 –2 1 1 1 1 1 4´4 4 1
4 ¸4 = ¸ 2 = ¸ = × = =4=4
4 4 4 4´4 4 1 1
[(–1) – (–2)] (–1 + 2)
So, 4 =4
m n m× n
(iii) (x ) = x
Example :
-3 3
æ 1 ö æ8ö
–1 –3
(8 ) = ç 1 ÷ = ç ÷ = 83
è8 ø è1ø

68
 Exponents and Power
m m m
(iv) x × y = (xy)
Example :
1 1 1 1
2
–2
×3 =
–2
2
´ 2 = 2 2
= = (2 × 3)
–2
2 3 2 ´3 (2 ´ 3)2
–2 –2 –2
So, 2 × 3 = (2 × 3)

m
xm æ x ö
(v) =ç ÷
ym è y ø
Example :
2 -2
3 -2 5 2 æ 5 ö æ3ö
-2
= 2 =ç ÷ =ç ÷
5 3 è3ø è5ø

–2
3 –2 æ3ö
So, =ç ÷
5 -2 è5ø

Illustration 3. Simplify and write the answer in an exponential form

7 11 3 –8
(i) (3 ¸ 3 ) × 3

2 2
ìïæ 6 ö 3 üï ìïæ 6 ö -4 üï
(ii) íç ÷ ý ´ íç ÷ ý
îïè 7 ø þï ïîè 7 ø þï
7 11 3 –8
Solution (i) (3 ¸ 3 ) × 3

7–11 3 –8 –4 3 –8 –12 –8 –12 + (–8) –20 1

= (3 ) × 3 = (3 ) × 3 =3 ×3 =3 =3 =
3 20

2 2
ìïæ 6 ö 3 üï ìïæ 6 ö -4 üï 3´ 2 -4´ 2
æ6ö æ6ö
(ii) íç ÷ ý ´ íç ÷ ý = ç ÷ × ç ÷
îïè 7 ø þï ïîè 7 ø þï è7ø è7ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

6 -8 6 +(-8) -2 2
æ6ö æ6ö æ6ö æ6ö æ7ö
= ç ÷ ´ç ÷ =ç ÷ =ç ÷ =ç ÷
è7ø è7ø è7ø è7ø è6ø

-5 x 3
æ -11 ö æ -11 ö æ -11 ö
Illustration 4. Find the value of x : ç ÷ ´ç ÷ =ç ÷
è 7 ø è 7 ø è 7 ø

-5 x 3 (-5)+ x 3
æ -11 ö æ -11 ö æ -11 ö æ -11 ö æ -11 ö
Solution ç ÷ ´ç ÷ =ç ÷ Þ ç ÷ =ç ÷
è 7 ø è 7 ø è 7 ø è 7 ø è 7 ø
Since, the base is the same on both sides of the expression,their exponents should also
be the same.
Þ –5 + x = 3 Þ x = 3 + 5 \ x=8

69
Class VIII : Mathematics

-4 2
æ -2 ö æ -5 ö
Illustration 5 Evaluate : ç ÷ × ç ÷
è 7 ø è 7 ø
Solution We have,

-4 2 4 2
æ -2 ö æ -5 ö æ 7 ö æ -5 ö
Þ ç ÷ × ç ÷ = ç ÷ × ç ÷
è 7 ø è 7 ø è -2 ø è 7 ø

4 2
æ -7 ö æ -5 ö é æ 7 ö æ -7 ö ù
= ç ÷ × ç ÷ êQ ç -2 ÷ = ç 2 ÷ ú
è 2 ø è 7 ø ë è ø è øû

(-7)4 (-5)2 7 4 ´ (-5)2 4 4

= 4 × 2 = 4 2 [Q (–7) = 7 ]
2 7 2 ´7

7 2 ´ (-5)2 49 ´ (-5) ´ (-5) 1225

= 4 = =
2 16 16

-1 -1 -1
æ1ö æ1ö æ1ö
Illustration 6 Simplify : ç ÷ +ç ÷ +ç ÷
è 2ø è 3ø è4ø

-1 -1 -1 1 1 1
æ1ö æ1ö æ1ö æ 2ö æ 3ö æ 4 ö
Solution ç ÷ +ç ÷ +ç ÷ = ç ÷ +ç ÷ +ç ÷
è 2ø è 3ø è4ø è1ø è1ø è1ø
=2+3+4=9

–1 –1
Illustration 7 By what number should (–24) be divided so that the quotient may be 3 ?
Solution Let the required number be x. Then,
–1 –1
(–24) ¸ x=3

1
(-24)-1 –1 1 é 1ù
Q a -1 = ú
Þ =3 Þ -24 = ê
x x 3 ë aû

1 1 3 -3 1
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

Þ = Þ 3 = –24x Þ x = Þ x= =–
-24x 3 -24 24 8

–1 –1 –1 –1 1 x
Illustration 8 Find x, if (4 + 8 ) × (3 –9 )¸ =5
12

æ1 1ö æ1 1ö 12 x
Solution ç + ÷ × ç - ÷× =5
è4 8ø è3 9ø 1

æ 2 + 1 ö æ 3 - 1 ö 12 3 2 12
Þ ç ÷´ç ÷´ = 5x Þ ´ ´ = 5x
è 8 ø è 9 ø 1 8 9 1
x 0 x
Þ1=5 Þ 5 =5 Þ x=0

70
 Exponents and Power

-5 m -2
æ 8 ö æ 8 ö æ 8ö
Illustration 9 Find m, if ç ÷ ´ç ÷ = ç ÷
è 11 ø è 11 ø è 11 ø

-5 + m -2
æ 8 ö æ 8 ö
Solution = ç ÷ =ç ÷
è 11 ø è 11 ø
Þ –5+m=–2
Þ m=–2 +5
Þ m=3

-10 -10 -10

x æ4ö æ 18 ö æxö
1. Find the value of , if ç ÷ ´ç ÷ =ç ÷
y è9ø è 7 ø èyø
9
2. The diameter of the sun is approximately 20 metres. A model of the sun has a diameter of
–1
20 meters. How many models would fit across the diameter of the sun?

–1 2
3. A framed window is 3 yards wide. The side of the house is 3 yards wide. How many framed
windows could fit across the side of the house?
-1
ìïæ 3 ö -1 æ 1 ö -1 üï
4. Evaluate íç ÷ - ç ÷ ý
îïè 4 ø è 4 ø þï

-3 -2
æ -3 ö æ9ö
5. By what number should ç ÷ be divided so that the quotient is ç ÷ ?
è 2 ø è4ø

x x–1
6. If 4 – 4 = 24, then find the value of x.

2 -4
æ 3ö æ 2ö –2
7. If x = ç ÷ ´ ç ÷ , find the value of x .
è 2ø è 3ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

4.0 WRITING EXPANDED FORM OF A DECIMAL NUMBER USING

EXPONENTIAL NOTATION
Look at the expanded form of 96829.653.

1 1 1
= 9 × 10000 + 6 × 1000 + 8 × 100 + 2 × 10 + 9 × 1 + 6 × +5× +3× .
10 100 1000
We can express this expansion in exponential notation using exponents of 10. Therefore,
4 3 2 1 0 –1 –2 –3
96829.653=9 × 10 +6 × 10 +8 × 10 + 2 × 10 + 9 × 10 +6 × 10 +5 × 10 +3 × 10 .
We observe that, the exponents of 10 start from the highest value and go on decreasing by 1 at
each step, from left to right.

71
Class VIII : Mathematics

Illustration 10. Expand the following number using exponents : 1256 ·249.

2 4 9
Solution 1256 · 249 = 1 × 1000 + 2 × 100 + 5 × 10 + 6 + + +
10 100 1000
–1 –2 –3
= 1 × 1000 + 2 × 100 + 5 × 10 + 6 + 2× 10 + 4 × 10 + 9 × 10

5.0 SCIENTIFIC NOTATION

The sun is located at a distance of 300,000,000,000,000,000,000 m from the centre of our galaxy.
The average size of an atom is about 0.00000003 centimeter across.
The length of these numbers in standard notation makes them awkward to work with. Scientific
notation is a shorthand way of writing such numbers.
To express any number in scientific notation, write it as the product of a power of ten and a number
greater than or equal to 1 but less than 10.
20
In scientific notation, the sun is located at a distance of 3.0 × 10 m from the centre of our galaxy and the
–8
size of each atom is 3.0 × 10 centimeters across.

6.0 COMPARING VERY LARGE AND VERY SMALL NUMBERS

To compare two numbers written in scientific notation, first compare the powers of ten. The number
with the greater power of ten is greater. If the powers of ten are the same, compare the values
between one and ten.
13 9 13 9
2.7 × 10 > 2.7 × 10 10 > 10
22 22
3.98 × 10 > 2.52 × 10 3.98 > 2.52

–3
Illustration 11.The thickness of a sheet of paper is 1.6 × 10 cm and the thickness of a human hair
–3
is 5 × 10 cm. Compare the two.

Thickness of hair 5 ´ 10 -3
=
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

Solution
Thickness of paper 1.6 ´ 10 -3

5 ´ 10 -3 ´ 103 5
= = = 3.125
1.6 1.6
The hair is approximately three times thicker than the paper.

72
 Exponents and Power

1. Write the following numbers in the standard form :

(i) 0.00925 (ii) 457000000 (iii) 0.32458

11
2. ASTRONOMY The distance between the Sun and the Moon is 1.49984 × 10 m and the distance
8
between the Moon and the Earth is 3.84 × 10 m. During Lunar eclipse, the Earth comes between
the Sun and the Moon. Find out the distance between the Sun and the Earth at this time.

3. Simplify and write the answer in an exponential form :

æ 1 ö é æ 3 ö 5 æ 3 ö 2 ù æ 343 ö 3
(i) ç ´ 7 -3 ÷ (ii) êç ÷ ´ ç ÷ ú ¸ ç ÷
è 125 ø ëê è 7 ø è 7 ø ûú è 27 ø

4. Write the following numbers in the standard from :

(i) 0.00001145 (ii) 239.8956 (iii) 0.3002457

5. The distance between the Sun and Saturn is 1,433,500,000,000 m. The distance between Saturn
and Uranus is 1,439,000,000,000 m. The distance between the Sun and the Earth is
149,600,000,000 m.
(i) Arrange the distance in an ascending order.
(ii) Subtract the distance between the Sun and the Earth from the distance between the Sun
and the Saturn

0
• 0 ¹1
• A number less than 1 will have a negative exponent when written in scientific notation.
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

73
Class VIII : Mathematics

SOME WORKED OUT ILLUSTRATIONS

Illustration 1
Evaluate.
(i) 3–2 (ii) (–4)–2
Solution
1 1 1
(i) 3–2 = 2
= =
3 3´3 9
1 1 1 1
(ii) (–4)–2 = 2
= 2 2
= =
(-4) (-1) ´ (4) 1 ´ 16 16

Illustration 2
Simplify and express the result in power notation with positive exponent.
2 4
æ 1 ö æ5ö
(i) (–4) ¸ (– 4)
5 8
(ii) ç 3 ÷ (iii) (–3) × ç ÷
4
è2 ø è 3ø
Solution

(-4)5 1 1 1 1 1
(i) (–4) ¸ (– 4) =
5
8
= 8
8-5 = 3
= 3 3 = =- 3
(-4) (-4) (-4) (-1) (4) -1 ´ 4 3
4
2
æ 1 ö 1 1
(ii) ç 3 ÷ = 3´ 2 = 6
è2 ø 2 2
4
æ5ö 54
(iii) (–3) × ç ÷
4
= (–1) × 3 ×
4 4
è 3ø 34
= 1 × 34–4 × 54 = 30 × 54 = 1 × 54 = 54

Illustration 3
Find the value of :
(i) (30 + 4–1) × 22 (ii) (2–1 × 4–1) ¸ 2–2
Solution

æ 1ö 4 +1 5
(i) (30 + 4–1) × 22 = ç 1 + ÷ ´ 4 = ´4 = ´4 = 5
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

è 4ø 4 4

æ1 1ö 1 1 1 1 4 1
(ii) (2–1 × 4–1) ¸ 2–2 = ç ´ ÷ ¸ 2 = ¸ = ´ =
è2 4ø 2 8 4 8 1 2

Illustration 4

8 -1 ´ 5 3
Evaluate :
2-4
Solution

8 -1 ´ 5 3 1 3 4 1
-4 = ´ 5 ´ 2 = ´ 125 ´ 16 = 125 × 2 = 250
2 8 8

74
 Exponents and Power

Illustration 5
m –3 5
Find the value of m for which 5 ¸ 5 = 5 .
Solution
m –3 5
We have, 5 ¸ 5 = 5
1
Þ
m
5 ¸ = 55 Þ 5m × 53 = 55
53
m+3 5
Þ 5 =5
m n
Þ m + 3 = 5 [Q if a = a then m = n]
Þ m=5–3=2

Illustration 6
-1
ìïæ 1 ö -1 æ 1 ö -1 üï
Evaluate íç ÷ - ç ÷ ý
ïîè 3 ø è 4 ø ïþ

Solution
-1
ìïæ 1 ö -1 æ 1 ö -1 üï 1 1
íç ÷ - ç ÷ ý = (31 – 41)–1 =(–1)–1 = 1
= = -1
ïîè 3 ø è 4 ø ïþ (-1) -1

Illustration 7

25 ´ t -4
Simplify : (t ¹ 0)
5-3 ´ 10 ´ t -8
Solution

25 ´ t -4 25 ´ 5 3 -4 + 8 5 ´ 53 4 625 4
= ´t = ´t = t
5-3 ´ 10 ´ t -8 10 2 2

Illustration 8
Express the following numbers in standard form:
(i) 0.0000000000085 (ii) 0.00000000000942
Solution
85
(i) 0.0000000000085 =
10000000000000
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

85 8.5 ´ 10
= 13
=
10 1013
= 8.5 × 10 × 10–13 = 8.5 × 10–12

(ii) 0.00000000000942
942
=
100000000000000
942 9.42 ´ 100
= 14
=
10 1014
= 9.42 × 102 × 10–14 = 9.42 × 10–12

75
Class VIII : Mathematics

Illustration 9
Express the following numbers in usual form.
(i) 3.02 × 10–6 (ii) 4.5 × 104 (iii) 3 × 10–8
Solution

302 1 302
(i) 3.02 × 10–6 = ´ 6 = = 0.00000302
100 10 100000000

45
(ii) 4.5 × 104 = ´ 10 4 = 45 × 103 = 45000
10

3 3
(iii) 3 × 10–8 = 8
= = 0.00000003
10 100000000

Illustration 10
Express the number appearing in the following statements in standard form :

1
(i) 1 micron is equal to m
1000000
(ii) Charge of an electron is 0.00000000000000000016 coulomb.
(iii) Size of a bacteria is 0.0000005 m.
Solution

1
(i) 1 micron = m = 1 × 10–6 m
1000000
(ii) Charge of an electron is

16
= 0.00000000000000000016 =
100000000000000000000

16 1.6 ´ 10 1.6
= 20
= 20
= = 1.6 × 10–19
10 10 1019

5 5
(iii) Size of a bacteria = 0.0000005 m = m = 7 m = 5 × 10–7 m
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

10000000 10

Illustration 11
In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016
mm. What is the total thickness of the stack?
Solution
Thickness of one book = 20 mm
\ Thickness of 5 books = 5 × 20 mm = 100 mm
Thickness of 1 sheet paper = 0.016 mm
Thickness of 5 sheets paper = 5 × 0.016 mm = 0.08 mm
Total thickness = 100 mm + 0.08 mm = 100.08 mm = 1.0008 × 102 mm

76
 Exponents and Power

-3
æ 2ö
1. The value of ç ÷ is
è5ø
8 25 125 2
(A) - (B) (C) (D) -
125 4 8 5

2. The value of (–3)–4 is

1 1
(A) 12 (B) 81 (C) - (D)
12 81

3. The value of (–2)–5 is

-1 1
(A) –32 (B) (C) 32 (D)
32 32

4. (2–5 ÷ 2–2) = ?
1 -1 1 1
(A) (B) (C) - (D)
128 128 8 8

-1
æ -3 ö
5. ç ÷ =?
è 2 ø
2 2 3
(A) (B) - (C) (D) None of these
3 3 2

0
æ 5ö
6. çè ÷ø = ?
6
6
(A) (B) 0 (C) 1 (D) None of these
5

30 + 20 + 11
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

7. is equal to
52 ¸ 5

3 3
(A) 0 (B) (C) (D) 15
5 125

4
ìïæ 1 ö 2 üï
8. íç ÷ ý = ?
îïè 3 ø þï

6 8 16 24
æ1ö æ1ö æ1ö æ1ö
(A) ç ÷ (B) ç ÷ (C) ç ÷ (D) ç ÷
è 3ø è 3ø è 3ø è 3ø

77
Class VIII : Mathematics

3 8
æ -1 ö æ -1 ö
9. ç ÷ ¸ç ÷ = ?
è 5 ø è 5 ø
5 11 5
æ 1ö æ -1 ö æ1ö
(A) ç - ÷ (B) ç ÷ (C) (–5)5 (D) ç ÷
è 5ø è 5 ø è5ø

10. (6–1 – 8–1)–1 = ?

1 1
(A) - (B) –2 (C) (D) 24
2 24

11. (5–1 × 3–1)–1 = ?

1 -1
(A) (B) (C) 15 (D) –15
15 15

2
æ -2 ö
12. ç ÷ = ?
è 3 ø
4 -2 4 -4
(A) (B) (C) (D)
3 9 9 9

-2 -2 -2
æ1ö æ1ö æ1ö
13. ç ÷ +ç ÷ +ç ÷ = ?
è 2ø è 3ø è4ø
61 144
(A) (B) 29 (C) (D) None of these
144 61

-1
ìï -1 æ 3 ö -1 üï
14. í6 + ç ÷ ý = ?
ïî è 2 ø ïþ

2 5 6
(A) (B) (C) (D) None of these
3 6 5

-1
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

ìïæ 3 ö -1 æ 1 ö -1 üï
15. íç ÷ - ç ÷ ý = ?
îïè 4 ø è 4 ø þï

3 -3 8 -8
(A) (B) (C) (D)
8 8 3 3

-1
éì 2 ü -2 ù
ê ïíæ - 1 ö ïý ú = ?
16. ê ïçè 2 ÷ø ï ú
ëî þ û

1 -1
(A) (B) 16 (C) (D) –16
16 16

78
 Exponents and Power

-3
æ 2ö
17. (32 – 22) × ç ÷ = ?
è3ø
45 8 8 135
(A) (B) (C) (D)
8 45 135 8

ìïæ 1 ö-3 æ 1 ö-3 üï æ 1 ö -3

18. íç ÷ - ç ÷ ý ¸ ç ÷ = ?
ïîè 3 ø è 2 ø ïþ è 4 ø

19 64 27
(A) (B) (C) (D) None of these
64 19 16

-5 11 8x
æ5ö æ5ö æ5ö
19. If ç ÷ ´ç ÷ =ç ÷ , then x = ?
è 3ø è3ø è 3ø
-1 -3 3 4
(A) (B) (C) (D)
2 4 4 3

20. By what number should (–8)–1 be multiplied to get 10–1 ?

4 -5 -4
(A) (B) (C) (D) None of these
5 4 5

a 1
21. If 3 = and 5 b = 125 , then the value of (a+b) is :
9
(A) 1 (B) 2 (C) 0 (D)

22. Simplify (a–1 + b–1)–1 =

a+ b ab a b
(A) (B) (C) (D)
ab a+b a+b a+b

x -3 2x - 6
æ 3ö æ 5ö
23. If ç ÷ =ç ÷ , then x is equal to:
è 5ø è 3ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

(A) 3 (B) 0 (C) 1 (D) 2

20
24. The value of 22 is :
(A) 4 (B) 8 (C) 16 (D) 2

6 12 3x
25. Solve for x : æç 2 ö÷ ´ æç 3 ö÷ æ 3ö
=ç ÷
è 3 ø è 2ø è 2ø

1 1
(A) 3 (B) 2 (C) (D)
3 2

79
Class VIII : Mathematics

-2 2
æ4ö æ1ö
1. If x = ç ÷ ¸ ç ÷ , then the value of x–1 ?
è5ø è4ø

1 1
(A) 5 (B) (C) (D) 25
5 25

–1 –1 –1 –1
2. The value of (3 +4 ) ÷5 is

7 60 7 7
(A) (B) (C) (D)
10 7 5 15

ìïæ 1 ö -2 æ 1 ö -3 üï æ 1 ö -3
3. íç ÷ + ç ÷ ý ¸ ç ÷ = ?
ïîè 3 ø è 2 ø ïþ è 4 ø

17 27 64 16
(A) (B) (C) (D)
64 16 19 25

-4 3x 5
æ 7 ö æ 7 ö æ 7 ö
4. The value of x for which ç ÷ ´ç ÷ = ç ÷ , is
è 12 ø è 12 ø è 12 ø
(A) –1 (B) 1 (C) 2 (D) 3

3x–1
5. If (2 + 10) ÷ 7 = 6, then x is equal to
(A) –2 (B) 0 (C) 1 (D) 2

6. 3670000 in standard form is

4 5 6
(A) 367 × 10 (B) 36.7 × 10 (C) 3.67 × 10 (D) None of these

7. 0.0000463 in standard form is

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

–7 –5 –9 –6
(A) 463 × 10 (B) 4.63 × 10 (C) 4.63 × 10 (D) 46.3 × 10

4
8. 0.000367 × 10 in usual form is
(A) 3.67 (B) 36.7 (C) 0.367 (D) 0.0367

2x–y x+y 2 2
9. if 2 = 32 & 2 = 16, then x + y = ?
(A) 9 (B) 10 (C) 11 (D) 13

M n Mn
10. If a .a = a , then M(n – 2) + n(M – 2) = ?
1
(A) 1 (B) –1 (C) 0 (D)
2

80
 Exponents and Power

11. If x y = 121 and x, y are natural numbers, the x - y is :

(A) 6 (B) 11 (C) 1 (D) 3

–100
12. The multiplicative inverse of 2 is
10 100
(A) 2 (B) 2 (C) 2 (D) 100

( )
0
13. The value of 2-1 + 3-1 is :

5 6
(A) (B) 1 (C) (D) None of these
6 5

0 0 0 0
14. The value of (6 – 7 ) × (6 + 7 ) is :
(A) 0 (B) 1 (C) 2 (D) None of these

5
15. 3 with negative exponent can be written as :

-5 -5
–5 æ 1ö æ 1ö
(A) (–3) (B) ç - ÷ (C) ç ÷ (D) None of these
è 3ø è 3ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

81
Class VIII : Mathematics

Very short answer type questions

1. Evaluate :
-5
æ1ö
(i) 4–3 (ii) ç ÷
è 2ø

2. Express as rational number :

-1 -3
æ -3 ö æ -5 ö
(i) ç ÷ (ii) ç ÷
è 4 ø è 7 ø

3. Express as a power of a rational number with positive exponent :

4
é æ 4 ö -2 ù
(i) 86 × 8–5 (ii) êç ÷ ú
ëè 5 ø û

4. Express as a power of a rational number with negative exponent :

5
æ3ö
2 éæ -7 ö 2 ù
(i) ç ÷ (ii) êç ÷ ú
è4ø ëè 3 ø û

5. Express each of the following with positive indices :

-1 -2 7
(i) (ii) (iii) -5 / 6 (iv) (x–3)4
x2 x 5
x

6. Simplify and express in the exponential form :

3 7 4 3
æ5ö æ5ö æ -2 ö æ -2 ö æ -2 ö
(i) ç ÷ ´ ç ÷ (ii) ç ÷ ´ ç ÷ ´ ç ÷
è 2ø è 2ø è 3 ø è 3 ø è 3 ø
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

6 4
æ4ö æ4ö
(iii) ç ÷ ¸ ç ÷
è9ø è9ø

Short answer type questions

7. Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

8. Express the following numbers in usual form.

(i) 3.02 × 10–6 (ii) 4.5 × 104 (iii) 3 × 10–8

82
 Exponents and Power

9. Evaluate :
2 2 6 -4
æ5ö æ5ö æ5ö æ5ö
(i) ç ÷ ´ ç ÷ (ii) ç ÷ ´ ç ÷
è 3ø è3ø è6ø è6ø

10. Find the value of :

(i) (30 + 4–1) × 22 (ii) (2–1 × 4–1) ¸ 2–2
-2 -2 -2
æ1ö æ1ö æ1ö
(iii) ç ÷ +ç ÷ +ç ÷
è 2ø è 3ø è4ø

11. Simplify :
-3 4 -5 -3
æ 2ö æ 2ö æ -3 ö æ -3 ö
(i) ç - ÷ ´ç- ÷ (ii) ç ÷ ¸ç ÷
è 5ø è 5ø è 4 ø è 4 ø

Long answer type questions

12. Simplify :
0
3
æ 3 ö æ 15 ö
3 ææ 4ö 8ö
(i) ç ÷ ´ ç ÷ (ii) ç ç ÷ ÷
è5ø è 2 ø èè 7ø ø

éæ -2 ö3 æ -2 ö ù æ -2 ö 2
(iii) êç ÷ ´ ç ÷ ú ¸ ç ÷
ëè 3 ø è 3 ø û è 3 ø

13. (a) Evaluate : [(5–1× 3–1)–1 ¸ 6–1]

-3 -2
æ -2 ö æ 4 ö
(b) By what number should ç ÷ be divided so that the quotient may be ç ÷ ?
è 3 ø è 27 ø
14. Evaluate :
-2 1
(ii) é
2 ù2
(i) (512) 9
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

êë(216)3 úû

-1 3
(iii) 10 ¸8 3 (iv) (16) 4
1 -1
(v) 27 3 ´ 16 4

15. Evaluate :
3 3
(i) ( 0.04 ) 2 (ii) ( 6.25 ) 2
-2 2
(iii) ( 0.03125 ) 5 (iv) ( 0.008 ) 3

83
Class VIII : Mathematics

High order thinking skills (HOTS)

16. (a) INSECTS Fire ants are 2–3 inches long. The ants are in a line across a porch that is 2 6 inches
long. how many fire ants are there?

(b) MEASUREMENT The diameter of a grain of sand is about 2–9 metres. The diameter of
a grain of powder is about 2–17 metres. How many grains of powder could fit across one
grain of sand ?

17. (a) ASTRONOMY The star Betelgeuse, in the constellation of Orion is approximately 3.36 ×
1015 miles from Earth. This is approximately 1.24 × 106 times as far as Pluto's minimum
distance from Earth. What is Pluto's approximate minimum distance from Earth? Write your
answer in scientific notation.
(b) For what positive values of x will x18 be greater than x20?

18. (a) For what negative values of x will x20 be greater than x18 ?
(b) For what negative values of x will x18 be equal to x20 ?

19. BIOLOGY Escherichia coli is a type of bacterium that is sometimes found in swimming pools.
Each E. coli bacterium has a mass of 2 × 10–12 gram. The number of bacteria increase so that,
after 30 hours, one bacterium has been replaced by a population of 4.8 × 108 bacteria.
(i) Suppose a pool begins with a population of only 1 bacterium. What would be the mass of
the population after 30 hours?
(ii) A small paper clip has a mass of about 1 gram. The paper clip has how many times the mass
of the 4.8 × 108 E. coli bacteria?

20. PHYSICAL SCIENCE The speed of light is about 2 × 105 miles per second.
(i) On average, it takes light about 500 seconds to travel from the sun to Earth. What is the
average distance from Earth to the sun? Write your answer in scientific notation.
(ii) The star Alpha Centauri is approximately 2.5 × 1013 miles from Earth. How many seconds
does it take light to travel between Alpha Centauri and Earth?
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

(iii) Use your answer to Part (ii) to estimate how many years it takes for light to travel between
Alpha Centauri and Earth.

84
 Exponents and Power

ANSWERS
CHECK POST-1

1 1 d3 7y 2
1. 2. (i) (ii) (iii)
2 b6 c9 x4

CHECK POST-2
8 -3
1. 2. 2010 models 3. 27 windows 4.
7 8
12
-3 5 æ 2ö
5. 6. 7. ç ÷
2 2 è 3ø

CHECK POST-3

1. (i) 9.25 × 10–3 (ii) 4.57 × 108 (iii) 3.2458 × 10–1

2. 1.496 × 1011 m

16
1 æ 3ö
3. (i) (ii) ç ÷
35 3 è 7ø

4. (i) 1.145 × 10–5 (ii) 2.398956 × 102 (iii) 3.002457 × 10–1

5. (i) 0.1496 × 1012 < 1.4335 × 1012 < 1.439 × 1012 (ii) 1.2839 × 1012 m

EXERCISE-1

Q ue. 1 2 3 4 5 6 7 8 9 10
Ans. C D B D B C B B C D
Q ue. 11 12 13 14 15 16 17 18 19 20
Ans. C C B C B A D A C C
Q ue. 21 22 23 24 25
JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

Ans. A B A A B

EXERCISE-2

Que. 1 2 3 4 5 6 7 8 9 10
Ans. C B A D D C B A B C
Que. 11 12 13 14 15
Ans. D C B A C

85
Class VIII : Mathematics
EXERCISE-3
Very short answer type questions

1 4 343
1. (i) (ii) 32 2. (i) - (ii) -
64 3 125

8 -2 -10
æ 5ö æ 4ö æ -3 ö
3. (i) 8 (ii) ç ÷ 4. (i) ç ÷ (ii) ç ÷
è 4ø è 3ø è 7ø

10 8 2
1 1 1 æ 5ö æ -2 ö æ 4ö
5. (i) (ii) (iii) 7x5/6 (iv) 6. (i) ç ÷ (ii) ç ÷ (iii) ç ÷
x1/ 2 x 2/5 x12 è 2ø è 3ø è 9ø

Short answer type questions

7. (i) 8.5 × 10–12 (ii) 9.42 × 10–12 8. (i) 0.00000302 (ii) 45000 (iii) 0.00000003

625 25 1
9. (i) (ii) 10. (i) 5 (ii) (iii) 29
81 36 2

-2 16
11. (i) (ii)
5 9

Long answer type questions

729 4 -2
12. (i) (ii) 1 (iii) 13. (a) 90 (b)
8 9 27

1 3
14. (i) (ii) 6 (iii) 20 (iv) 8 (v) 15. (i) 0.008 (ii) 15.625 (iii) 4 (iv) 0.04
4 2

High order thinking skills (HOTS)

JPR\Comp.251\Allen-Junior wing (2020-21)\maths\VIII\Unit-1\4.Exponents and Powers

16. (a) 29 (b) 28

p
17. (a) 2.7 × 109 miles (b) x = , where p, q ¹ 0 and p < q
q

p p
18. (a) x = - , where p, q ¹ 0 and p > q (b) x = - where p = q and q ¹ 0
q q

19. (i) 9.6 × 10–4 (ii) 9.6 × 10–4

20. (i) 108 (ii) 1.25 × 108 sec (iii) 3.96

86