S.chand publication New Learning Composite mathematics solution of class 7 Chapter 4 Powers and Exponents Exercise 4B

 Exercise 4B

Q1 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 1

(a) 2⁵ x 2³

Sol : 2⁸

(b) 4⁷ x 4⁵

Sol : 4¹²

(c) 9⁸ x 9⁷

Sol : 9¹⁵

(d) 10⁵ x 10³ x 10

Sol : 10⁹

(e) $\left(\frac{1}{2}\right)^7 \times \left(1/2\right)^3$

Sol :

$\left(\frac{1}{2}\right)^{10}$

(f) $\left(-\frac{7}{9}\right)^{11} \times \left(-\frac{7}{9}\right)^6$

Sol :

$=\left(-\frac{7}{9}\right)^{17}$

(g) $\left(\frac{5}{13}\right)^9 \times \left(\frac{5}{13}\right)^{11}$

Sol :

$=\left(\frac{5}{13}\right)^{20}$

(h) $\left(-\frac{4}{15}\right)^{8} \times \left(-\frac{4}{15}\right)^{5} \times \left(-\frac{4}{15}\right)^9$

Sol :

$=\left(-\frac{4}{15}\right)^{22}$

(i) p⁷ x p⁸ x p⁹ x p⁶

Sol :

p³⁰

Q2 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 2

Divide and write the quotient as one power.

(a) 3⁶ ÷ 3⁴

Sol :

=3^(6-4)

=3²

(b) 7¹¹ ÷ 7⁵

Sol :

=7^(11-5)

=7⁶

(c) (-11)⁷ ÷ (-11)⁴

Sol :

=-11^7-4

=(-11)³

(d) x¹⁰ ÷ x⁴

Sol :

=x^(10-4)

=x⁶

(e) $(\frac{1}{3})^7 \div (\frac{1}{3})^5$
Sol :

$=\left(\frac{1}{3}\right)^{7-5}=\left(\frac{1}{3}\right)^2$

(f) $\left(-\frac{8}{17}\right)^{20} \div \left(-\frac{8}{17}\right)^{15}$
Sol :

$=\left(-\frac{8}{17}\right)^{20-15}=\left(-\frac{8}{17}\right)^{5}$

(g) $\left(\frac{3}{19}\right)^8 \div \left(\frac{3}{19}\right)$
Sol :

$=\left(\frac{3}{19}\right)^7$

(h) (-23)⁷÷(-23)⁷

Sol :

=(-23)⁰

(i) $\frac{p^5}{p}$

Sol :

=p⁴

(j) $\frac{(-x)^{17}}{(-x)^5}$

Sol :

=(-x)¹²

Q3 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 3

Simplify, leaving each answer in exponential form.

(a) (3⁴)²

Sol : 3⁸

(b) (6⁵)⁷

Sol : 6³⁵

(c) (17²)³

Sol : 17⁶

(d) $[\left(\frac{7}{3}\right)^2]^5$

Sol : $\left(\frac{7}{3}\right)^{10}$

(e) $[\left(-\frac{7}{8}\right)^6]^5$
Sol : $\left(-\frac{7}{8}\right)^{30}$

(f) (m⁹)³

Sol : m²⁷

(g) (t⁴)⁴

Sol : t¹⁶

(h) (n¹⁰)⁹

Sol : n⁹⁰

Q4 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 4

Simplify:

(a) (xy)²

Sol : x²y²

(b) (7a²)³

Sol : 343a⁶

(c) (-2x⁵y²)⁷

Sol : -128x³⁵y¹⁴

(d) (-3p³q⁴)⁴

Sol : 81p¹²q¹⁶

Q5 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 5

Simplify:

(a) (8³)² ÷ 8⁴

Sol : 8⁶÷ 8⁴

= 8²=64

(b) (2²)³ x 5²

Sol :

= 2⁶×5²

=64×25=1600

(c) (3a²)³ ÷ (-3a⁶)

Sol :

=(27a⁶)÷(-3a⁶)

$=\frac{27a^6}{-3a^6}$

=-9

(d) (-2x²y³)⁴ x (1/2 x⁴y⁷)⁴

Sol :

$=16x^8y^{12} \times \frac{1}{16}x^{16}y^{28}$

=x²⁴y⁴⁰

(e) (19⁵)⁰

Sol :

=(19)⁰

=1

(f) (5⁰ – 7⁰) x 35⁰
Sol :
=(1-1)1
=0

Q6 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 6

Find the value of x in each case.

(a) x⁶ = 64

Sol :

x⁶ = 2⁶

∴x=2

(b) (-3)^x-1 = -243

Sol :

(-3)^x-1 =(-3)⁵

x-1=5

x=5+1=6

(c) $\left(\frac{3}{4}\right)^4 \times \left(\frac{3}{4}\right)^5 = \left(\frac{3}{4}\right)^{5x - 1}$

Sol :

$\left(\frac{3}{4}\right)^9=\left(\frac{3}{4}\right)^{5x-1}$

9=5x-1

5x=9+1

$x=\frac{10}{2}=2$

(d) 5⁹÷ (5)^{3 – x} = 5¹⁰

Sol :

5^9-3+x=5¹⁰

5^6+x=5¹⁰

6+x=10

∴x=10-6=4

Q7 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 7

Simplify the following:

(a) $\frac{(3^3)^2 \times 5^2}{9^2 \times 5}$
Sol :

$=\frac{3^6 \times 5}{9^2}=\frac{729 \times 5}{81}$

=45

(b) $\frac{3^8 \times a^6}{9^3 \times a^3}$

Sol :

$=\frac{3^8 \times a^3}{9^3}$
$=\frac{6561 \times a^3}{729}$
=9a³

(c) (5⁰ + 4⁰ + 3⁰) x 3⁴

Sol :

=(1+1+1).81

=3×81=243

Q8 Ex 4B Class 7 Chapter 4 Powers and Exponent Schand New Learning Composite mathematics

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Question 8

If [3m² ÷ (3m)² ]^m  = 1/81 then the value of m is

(a) -3

(b) -6

(c) -3

(d) 4

Sol :

[3m² ÷ (3m)² ]^m  = $\frac{1}{81}$

$\left[\frac{3m^2}{9m^2}\right]^m=\frac{1}{81}$

$\left[\frac{1}{3}\right]^m=\frac{1}{81}$

$\left[\frac{1}{3}\right]^m=\left[\frac{1}{3}\right]^4$

m=4

Ans: d