SEBA Class 10 Maths Revision Exercise R-3 Solutions: Indices and Power | Revision Chapter
Get free and reliable SEBA Class 10 Maths Revision Exercise R-3 Solutions: Indices and Power from the new SCERT textbook 2026. Here, we have solved all the questions and MCQs from this Revision Exercise R-3.
See More:
- SEBA Class 10 Maths Chapter R-1 Solutions: Square and Square Root
- SEBA Class 10 Maths Chapter R-2 Solutions: Cube and Cube Root
- SEBA Class 10 Maths Chapter R-3 Solutions: Indices and Power
- SEBA Class 10 Maths Chapter R-4 Solutions: Factorisation
- SEBA Class 10 Maths Chapter R-5 Solutions: Congruence of Triangles
1. Find the value of:
(i) $11^3$
$\mathbf{Sol^n.}$
$11^3$
$= 11 \times 11 \times 11$
$= 121 \times 11$
$= 1331$
(ii) $2 \times 10^3$
$\mathbf{Sol^n.}$
$2 \times 10^3 $
$= 2 \times 1000$
$= 2000$
(iii) $(\frac{1}{2})^{-5}$
$\mathbf{Sol^n.}$
Using the law $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$, we get
$(\frac{1}{2})^{-5} = (\frac{2}{1})^5$
$= 2^5$
$= 32$
(iv) $(-4)^{-2}$
$\mathbf{Sol^n.}$
Using the law $a^{-m} = \frac{1}{a^m}$, we have
$(-4)^{-2} = \frac{1}{(-4)^2}$
$= \frac{1}{16}$
2. Express the following numbers in terms of powers of their prime factors.
(i) 729
$\mathbf{Sol^n.}$
By prime factorisation, we get:
$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$
$= 3^6$
(ii) 3125
$\mathbf{Sol^n.}$
$3125 = 5 \times 5 \times 5 \times 5 \times 5$
$= 5^5$
(iii) 3600
$\mathbf{Sol^n.}$
$3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$
$= 2^4 \times 3^2 \times 5^2$
(iv) $108 \times 192$
$\mathbf{Sol^n.}$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$
$192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1$
Therefore:
$108 \times 192 = (2^2 \times 3^3) \times (2^6 \times 3^1)$
$= 2^{2+6} \times 3^{3+1}$
$= 2^8 \times 3^4$
3. Simplify:
(i) $(-3)^2 \times (-5)^2$
$\mathbf{Sol^n.}$
$(-3)^2 \times (-5)^2 $
$= [(-3) \times (-5)]^2$
$= (15)^2$
$= 225$
(ii) $(2^3 \times 2)^4$
$\mathbf{Sol^n.}$
$(2^3 \times 2)^4 $
$= (2^{3+1})^4$
$= (2^4)^4$
$= 2^{4 \times 4}$
$= 2^{16}$
$= 65536$
(iii) $2^0 \times 3^0 \times 4^0$
$\mathbf{Sol^n.}$
We know that $a^0 = 1$ for any non-zero integer $a$.
$2^0 \times 3^0 \times 4^0 $
$= 1 \times 1 \times 1$
$= 1$
(iv) $(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}$
$\mathbf{Sol^n.}$
Using the law $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$:
$(\frac{5}{8})^{-7} = (\frac{8}{5})^7$
Therefore:
$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4} $
$= (\frac{8}{5})^7 \times (\frac{8}{5})^{-4}$
$= (\frac{8}{5})^{7 + (-4)}$
$= (\frac{8}{5})^3$
$= \frac{8^3}{5^3}$
$= \frac{512}{125}$
4. Compare the following numbers:
(i) $2^8$, $8^2$
$\mathbf{Sol^n.}$
$2^8 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$
$8^2 = 8 \times 8 = 64$
Since $256 > 64$,
$\therefore 2^8 > 8^2$
(ii) $2.7 \times 10^{12}$, $1.5 \times 10^8$
$\mathbf{Sol^n.}$
Comparing the powers of $10$, we get
In $2.7 \times 10^{12}$, the exponent of $10$ is $12$.
In $1.5 \times 10^8$, the exponent of $10$ is $8$.
Since $12 > 8$,
$\therefore 2.7 \times 10^{12} > 1.5 \times 10^8$
5. Express the following with the help of positive power.
(i) $2^{-3} \times (-7)^{-3}$
$\mathbf{Sol^n.}$
Using the law $a^m \times b^m = (ab)^m$, we get
$2^{-3} \times (-7)^{-3} $
$= [2 \times (-7)]^{-3}$
$= (-14)^{-3}$
$= \frac{1}{(-14)^3}$ [Since, $a^{-m} = \frac{1}{a^m}$]
(ii) $(-3)^{-4} \times (\frac{5}{3})^{-4}$
$\mathbf{Sol^n.}$
Using the law $a^m \times b^m = (ab)^m$:
$(-3)^{-4} \times (\frac{5}{3})^{-4} $
$= \left[(-3) \times \frac{5}{3}\right]^{-4}$
$= (-5)^{-4}$
$= \frac{1}{(-5)^4}$ [$a^{-m} = \frac{1}{a^m}$]
$= \frac{1}{5^4}$
6. Express the following numbers in standard form.
(i) 3,430,000
$\mathbf{Sol^n.}$
$3430000 $
$= 3.43 \times 10^6$
(ii) 70,040,000,000
$\mathbf{Sol^n.}$
$70040000000 $
$= 7.004 \times 10^{10}$
(iii) 0.00000015
$\mathbf{Sol^n.}$
$0.00000015 $
$= 1.5 \times 10^{-7}$
(iv) 0.00001436
$\mathbf{Sol^n.}$
$0.00001436 $
$= 1.436 \times 10^{-5}$
7. Express the following in general form.
(i) $1.0001 \times 10^9$
$\mathbf{Sol^n.}$
$1.0001 \times 10^9 $
$= \frac{10001}{10000} \times 10^9$
$= \frac{10001}{10^4} \times 10^9$
$= 10001 \times 10^{9-4}$
$= 10001 \times 10^5$
$= 1,000,100,000$
(ii) $3.02 \times 10^{-6}$
$\mathbf{Sol^n.}$
$3.02 \times 10^{-6} $
$= \frac{3.02}{10^6}$
$= \frac{3.02}{1000000}$
$= 0.00000302$
8. Find the value of $m$ such that $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$\mathbf{Sol^n.}$
Given:
$(-3)^{m+1} \times (-3)^5 = (-3)^7$
$\Rightarrow (-3)^{(m+1) + 5} = (-3)^7$
$\Rightarrow (-3)^{m+6} = (-3)^7$
∴ $m + 6 = 7$
$\Rightarrow m = 7 – 6$
$\therefore m = 1$
9. Find the correct option of the following :
(a) The value of $3^{-3}$ is:
(i) $3^3$
(ii) $3^{\frac{1}{3}}$
(iii) $\frac{1}{3^3}$
(iv) $3 \times 3$
$\mathbf{Sol^n.}$
Using $a^{-m} = \frac{1}{a^m}$:
$3^{-3} = \frac{1}{3^3}$
So, the correct option is (iii) $\frac{1}{3^3}$
(b) The value of $(\frac{2}{3})^{-2}$ is :
(i) $(2 \times 3)^{-2}$
(ii) $\frac{1}{(2 \times 3)^2}$
(iii) $(\frac{3}{2})^{-2}$
(iv) $(\frac{3}{2})^2$
$\mathbf{Sol^n.}$
Using $(\frac{a}{b})^{-m} = (\frac{b}{a})^m$:
$(\frac{2}{3})^{-2} = (\frac{3}{2})^2$
So, the correct option is (iv) $(\frac{3}{2})^2$
(c) The value of $(-\frac{2}{3})^4$ is:
(i) $\frac{8}{12}$
(ii) $\frac{16}{81}$
(iii) $-\frac{16}{81}$
(iv) $-\frac{8}{12}$
$\mathbf{Sol^n.}$
$(-\frac{2}{3})^4 = \frac{(-2)^4}{3^4} = \frac{16}{81}$
So, the correct option is (ii) $\frac{16}{81}$
(d) The standard form of 0.000064 is :
(i) $64 \times 10^4$
(ii) $64 \times 10^{-4}$
(iii) $6.4 \times 10^5$
(iv) $6.4 \times 10^{-5}$
$\mathbf{Sol^n.}$
$0.000064 = \frac{6.4}{10^5} = 6.4 \times 10^{-5}$
So, the correct option is (iv) $6.4 \times 10^{-5}$
(e) The value of $2.03 \times 10^{-5}$ is:
(i) 0.203
(ii) 0.0000203
(iii) 203000
(iv) 0.00203
$\mathbf{Sol^n.}$
$2.03 \times 10^{-5} $
$= \frac{2.03}{10^5} $
$= \frac{2.03}{100000} $
$= 0.0000203$
So, the correct option is (ii) 0.0000203