w NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers Exercise 12.1
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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)$(\frac {1}{2})^{-5}$
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.
NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1
Answer
(i) Exercise 12.1 NCERT Solutions for Exponents Class 8 Maths
= (-4)5 / (-4)8
= (-4)5-8
=1/ (-4)3 (ii)
Chapter 12  Exercise 12.1 NCERT Solutions for Exponents Class 8 Maths
=1/23x2
=1/26

(iv )
Exercise 12.1 chapter 12 NCERT Solutions for Exponents Class 8 Maths
=(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 5m   ÷  5-3 = 55
Answer
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. $\left ( \frac {5}{8} \right )^{-7} \times \left ( \frac {8}{5} \right )^{-4}$
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.
Ncert Solutions for Exponents Class 8 Maths Chapter 12 Exercise 12.1 pdf download

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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