w NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers Exercise 12.1 # NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1

In this page we have NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1 for Exponent and Power.This exercise has questions about Exponents and power, Powers with Negative Exponents and Laws of exponents. Hope you like them and do not forget to like , social share and comment at the end of the page.

## NCERT Solutions for Class 8 Maths Chapter 12 Ex 12.1

Question 1
Evaluate
(i) $3^{-2}$
(ii) $(-4)^{-2}$
(iii)$(\frac {1}{2})^{-5}$
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. (i) = (-4)5 / (-4)8
= (-4)5-8
=1/ (-4)3 (ii) =1/23x2
=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. $\left \{ \left (\frac {-2}{3 }\right )^{-2} \right \}^2$
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. $(5^{-1} \times 2^{-1}) \times 6^{-1}$
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 5m   ÷  5-3 = 55
5m   ÷   5-3 = 55
5m × (1/5-3)  = 55
5m+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. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}$
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
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 