Exponents and powers class 8 worksheets with answers

Short Answer type

Question 1
Evaluate

1. 2^-2
2. (-2)^-2
3. (3/2)^-5

Answer

As we know that
b^-n= 1/bⁿ
(i) 2^-2 = 1/ 2² =1/4
(ii) (-2)^-2 = 1/(-2)²= 1/4
(iii) (3/2)^-5 =3-⁵/ 2^-5 =2⁵/3⁵ = 32/243

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Question 2
Simplify and express the result in power notation with positive exponent.
(i) (-2)⁵ ÷ (-2)⁴
(ii) (1/2)² × (2/5)²
(iii) (-5)² × (3/5)

Answer

(i) (-2)⁵ ÷ (-2)⁴
= (-2)⁵ / (-2)⁴
= (-2)^5-4
=-2
(ii) (1/2)² × (2/5)²
= (1/4) X (4/25)
= 1/25
(iii) (-5)² × (3/5)
=25 × (3/5) =15

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Question 3
Find the value of.
(i)(4⁰ + 4^-1) × 2²
(ii)(3^-1 × 9^-1) ÷ 3^-2
(iii)(11^-1 + 12^-1 + 13^-1)⁰

Answer

(i)(4⁰ + 4^-1) × 2² = (1+1/4) × 4
= 4+1=5
(ii)(3^-1 × 9^-1) ÷ 3^-2 = [(1/3) × (1/9)] ÷ (1/9)
=1/3
(iii)(11^-1 + 12^-1 + 13^-1)⁰
=1 as a⁰=1

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Question 4
Find the value of x here
$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$

Answer

$( \frac {11}{9})^3 \times (\frac {9}{11})^6 = (\frac {11}{9})^{2x-1}$
$(\frac {11}{9})^{3-6} = (\frac {11}{9})^{2x-1}$
-3=2x-1
Or
x=-1

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Question 5
Find the value of m for which 2^m ÷ 2^-4 = 4⁵

Answer

2^m ÷ 2^-4 = 4⁵
2^m × (1/2^-4) = 2¹⁰
2^m+4 =2¹⁰
So m+4=10
m=6

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Question 6
Express the following numbers in standard form.
(i) 0.0000000015
(ii) 0.00000001425
(iii) 102000000000000000

Answer

(i)0.0000000015
=1.5 ×10^-9
(ii)0.00000001425
=1.425×10^-8
(iii)102000000000000000
=1.02×10¹⁷

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Question 7
Express the following numbers in usual form.
(i) 34.02 x 10^-5
(ii) 9.5 x 10⁵
(iii) 9 x 10^-4
(iv) 2.0001 x 10⁸

Answer

(i)34.02 x 10^-5 = .0003402
(ii)9.5 x 10⁵
=950000
(iii) 9 x 10^-4
=.0009
(iv)2.0001 x 10⁸
=200010000

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Question 8
Solve for the variables
(i) $\frac {1}{y^{-5}} = 32$
(ii) $ (\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
(iii) $ \frac {3^x .3}{3^{2x+1}} = 27$

Answer

(i) $\frac {1}{y^{-5}} = 32$
$y^5 = 32$
$y^5 = 2^5$
Therefore y=2
(ii) $ (\frac {1}{7})^3 \div (\frac {1}{7})^8 = 7^n$
$(\frac {1}{7})^3 \times 7^8 = 7^n$
$7^5 = 7^n$
Hence n=5
(iii) $ \frac {3^x .3}{3^{2x+1}} = 27$
$\frac {3^{x+1}}{3^{2x+1}} = 27$
$3^{x + 1 - 2x -1} =27$
$3^{-x}= 27$
$3^{-x}=3^3$
Hence x =-3

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Question 9
Find the Multiplicative inverse of
(i) $3^{25}$
(ii) $4^{-3}$
(iii) $(\frac {2}{3})^{-2}$

Answer

(i) $3^{-25}$
(ii) $4^{3}$
(iii) $(\frac {2}{3})^{2}$

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Question 10
Simplify the following
(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$

Answer

(i) $1 + \left [ \left ( \frac{1}{3} \right )^{-3} - \left ( \frac{1}{2} \right )^{-3} \right ] \div 38$
=$1 + (3^3 - 2^3) \div 38$
=$1 + 19 \div 38$
=$\frac {3}{2}$

(ii) $\frac {1}{1 + p^{a-b}} + \frac {1}{1 + p^{b-a}}$
=$\frac {p^b}{p^b + p^a} + \frac {p^a}{p^a + p^b}$
=$\frac {p^b + p^a}{p^b + p^a}$
=1

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True and false

Question 11
True and False
(i) Very small numbers can be expressed in standard form using negative exponents.
(ii) $a^p \times b^q= (ab)^{pq}$
(iii) $.00567 = 5.67 \times 10^{-3}$
(iv) $\frac {1}{(8)^{-3}}= 2^9$
(v) $4^0 =4$
(vi)$ (-1)^3=1$
(vii) The multiplicative inverse of $(–2)^{–2}$ is $(2)^2$.

Answer

(i)True
(ii) False
(iii) True
(iv) True as $\frac {1}{(8)^{-3}}=8^3=(2^3)^3= 2^9$
(v) False
(vi) False
(vii) True as $(–2)^{–2}= \frac {1}{4}$

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Fill in the blanks

Question 12
Fill in the blank
(i) The multiplicative inverse of $(3)^{–2}$ is _____
(ii) $9^7 \times 3^{-2}$=_____
(iii) .0000067 in exponent form _____
(iv) Very large numbers can be expressed in standard form by using_________ exponents
(v) $5^0$ =____
(vi)$ (-1)^100$=____
(vii) The value of $[4^{–1} +3^{–1} + 6^{–2}]^{–1}$ is ____.

Answer

(i)$(3)^{2}$
(ii) $3^{11}$
(iii) $6.7 \times 10^6$
(iv) positive
(v) 1
(vi) 1
(vii) 18/11

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Match the column

Question 13

Answer

(p) -> (ii)
(q) -> (iii)
(r) -> (iv)
(s) -> (i)

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Multiple choice questions

Question 14
If a be any non-zero integer, then $a^{–1}$ is equal to
(a) a
(b) -a
(c) 1/a
(d) -1/a
Question 15
The multiplicative inverse of $(–7)^{–2} \div (90)^{–1}$ is
(a) $-(7)^2 \times (90)^{–1}$
(b) $-(7)^{-}2 \times (90)^{–1}$
(c) $-(7)^2 \div (90)^{–1}$
(d) $-(7)^2 \times (90)^{1}$
Question 16
If $5^{3x–1} \div 25 = 125$, Then the value x is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

(13) (c)
(14) $(–7)^{–2} \div (90)^{–1}= - (1/7)^2 \times 90^1$. So Multiplication inverse is $-(7)^2 \times (90)^{–1}$. Hence Option (a)
(15) $5^{3x–1} \div 25 = 125$
$5^{3x–1} \times 5^2=5^3$
$5^{3x-3}=5^3$
Hence x=2
Option(b) is correct

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Summary

This exponents and powers class 8 worksheets with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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Also Read

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• Notes
  • Exponent and Power Class 8 Notes
• Ncert Solutions
  • NCERT Solutions for Class 8 Maths Chapter 10 Exponents and Power Exercise 10.1
  • NCERT solutions for class 8 maths chapter 10 exercise 10.2
• Assignments
  • Exponent and Power Class 8 Important questions

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