NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1

NCERT Solutions for Class 8 Maths Chapter 12 Ex 12.1

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 2 Exercise 2.6 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 12.1 as pdf
2. This chapter 12 has total 2 Exercise 12.1 and 12.2. 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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<a href="https://physicscatalyst.com/class8/ncert_solutions_class8_maths_exponents_1.php">NCERT Solutions for Class 8 Maths Chapter 12 Exponents Exercise 12.1</a> Also Read

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