NCERT Solutions for Class 8 Maths Chapter 10 Exercise 10.1

NCERT Solutions for Class 8 Maths Chapter 10 Ex 10.1

Question 1
Evaluate
(i) $3^{-2}$
(ii) $(-4)^{-2}$
(iii)$(\frac {1}{2})^{-5}$
Answer
As we know that
$b^{-n}= \frac {1}{b^n}$
(i) $3^{-2}=\frac {1}{3^2} =\frac {1}{9}$
(ii) $(-4)^{-2}=\frac {1}{(-4)^2}= \frac {1}{16}$
(iii) $(\frac {1}{2})^{-5}=\frac {1}{2^{-5}}=2^5= 32$

Question 2
Simplify and express the result in power notation with positive exponent.

Answer
(i)
= (-4)⁵ / (-4)⁸
= (-4)^5-8
=1/ (-4)³ (ii)

=1/2^3x2
=1/2⁶

(iv )

=(3^-7/ 3^-10) × 3^-5
=3^-7+10 × 3^-5
=3³ × 3^-5
=3^-2
=1/3²
(v) 2 ^{- 3} × ( - 7) ^{- 3}
= (2× -7)^-3
= (-14)^-3
=1/ (-14)³

Question 3
Find the value of

1. (3⁰ + 4 ^{- 1}) × 2²
2. (2 ^{- 1} × 4 ^{- 1})  ÷ 2 ^{- 2}
3. $(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$
4. (3 ^{- 1} + 4 ^{- 1} + 5 ^{- 1})⁰
5. $\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$

Answer
1.(3⁰ + 4 ^{- 1}) × 2² = (1+1/4) × 4
= 4+1=5
2. (2 ^{- 1} × 4 ^{- 1})  ÷ 2 ^{- 2}
= [(1/2) × (1/4)] ÷ (1/4)
= (1/8)  ×  (4)  =1/2
3.
$(\frac {1}{2})^{-2} + (\frac {1}{3})^{-2} + (\frac {1}{4})^{-2}$ = 2² + 3²  + 4²
=4+9+16=29
4. (3 ^{- 1} + 4 ^{- 1} + 5 ^{- 1})⁰
=1   as  a⁰=1
5.$\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$
$=\left (\frac {-2}{3 }\right )^{-4}$
$=\frac {81}{16}$

Question 4
Evaluate
i. $ \frac {8^{-1} \times 5^3}{2^{-4}}$
ii. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
Answer
i. $ \frac {8^{-1} \times 5^3}{2^{-4}}$
$=\frac {2^4 \times 5^3}{8} = 250$
ii. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
$=( \frac {1}{5} \times \frac {1}{2} ) \times \frac {1}{6}$
$=\frac {1}{60}$


Question 5
Find the value of m for which 5^m   ÷  5^-3 = 5⁵
Answer
5^m   ÷   5^-3 = 5⁵
5^m × (1/5^-3)  = 5⁵
5^m+3 =5⁵
So  m+3=5
m=2

Question 6
Evaluate
i. $\left \{ \left ( \frac {1}{3} \right )^{-1} - \left ( \frac {1}{4} \right )^{-1} \right \}^{-1}$
ii. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}$
Answer
i. $\left \{ \left ( \frac {1}{3} \right )^{-1} - \left ( \frac {1}{4} \right )^{-1} \right \}^{-1}$
$={3^2 - 4^1}^{-1} = -1$ ii. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}$
$=\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {5}{8} \right )^{4}$
$=\left ( \frac {5}{8} \right )^{-3}= \frac {512}{125}$

Question 7
Simplify.

Summary

1. Class 8 Maths chapter 10 Exercise 10.1 Solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
   Download Exponents ex 10.1 as pdf
2. This chapter 10 has total 2 Exercise 10.1 and 10.2. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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