S.chand publication New Learning Composite mathematics solution of class 8 Chapter 2 Exponents Exercise 2A

 Exercise 2A

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Q1 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 1

Evaluate

(a) 9⁰

Sol : 1

(b) $\left(\frac{1}{3}\right)^0$

Sol : 1

(c) $\left(-\frac{64}{21}\right)^0$

Sol : 1

(d) (395.008)⁰

Sol : 1

(e) $\left(\frac{3xy}{ab}-z\right)^0$

Sol : 1

(f) $\frac{1}{a^0 + b^0 + c^0}$

Sol : $=\frac{1}{3}$

(g) (8⁰ – 2⁰)x 25⁰

Sol : 

=(1-1)×1

=0×1=0

(h) (35⁰ + 2⁰ + 3⁰) ÷ 3

Sol : 1

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Q2 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 2

Simplifying giving the answer as a fraction:
(a) 6^-1

Sol : $=\frac{1}{6}$

(b) 10^-1

Sol : $=\frac{1}{10}$

(c) 4^-3

Sol : $=\frac{1}{4^3}=\frac{1}{64}$

(d) 2^-7

Sol : 

$=\frac{1}{2^7}=\frac{1}{128}$

(e) 3^-3

Sol : 

$=\frac{1}{3^3}=\frac{1}{27}$

(f) p^-1

Sol : 

$=\frac{1}{p}$

(g) (x^-2)²

Sol : 

$=\frac{1}{x^4}$

(h) (10a)^-2

Sol : 

$=\frac{1}{100a^2}$

(i) (x+y)^-2

Sol : 

$=\frac{1}{(x+y)^2}$

(j) (m⁵)^-3

Sol : 

$=\frac{1}{m^{15}}$

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Q3 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 3

Find the reciprocal of the following.

(a) (3)^-2

Sol : 

$=\frac{1}{3^2}=\frac{1}{9}$

(b) (-2)^-5

Sol : 

$=\frac{1}{(-2)^5}=-\frac{1}{32}$

(c) $\left(\frac{2}{5}\right)^{-3}$
Sol : 

$=\left(\frac{5}{2}\right)^3=\frac{125}{8}$

(d) $\left(\frac{2}{3}\right)^{-4}$

Sol : 

$=\left(\frac{3}{2}\right)^4=\frac{81}{16}$

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Q4 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 4

Simplify and express the result in power notation with positive exponent.
(a) (-2)^-4 x (-2)^-6

Sol : $=\frac{1}{-2^4}\times \frac{1}{-2^6}=\frac{1}{2^{10}}$

(b) 3⁷ x 3^-9 x 3⁶

Sol : $=\frac{1}{3^7}\times \frac{1}{3^9}\times \frac{1}{3^{6}}$

=3⁴

(c) $3^7\times \left(\frac{3}{5}\right)^{-7}$

Sol : $=3^7 \times \left(\frac{5}{3}\right)^7=3^7 \times  \frac{5^7}{3^{7}}$

=5⁷

(d) 2^-7 x [(2)^-3]^-4

Sol : $=\frac{1}{2^7}\times 2^{12}$

=2⁵

(f) (4^-8 ÷ 4^-10) x 4^-5

Sol : $=\frac{4^{-8}}{4^{-10}}\times 4^{-5}$ $=\frac{1}{4^2}\times 4^{-5}=4^{-3}=\frac{1}{4^3}$

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Q5 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 5

Evaluate:
(a) (8⁰ + 5^-1)×$\left(\frac{1}{5}\right)^{-2}$

Sol :

$=\left(1+\frac{1}{5}\right)\times (5^{-1})^{-2}$

$=\frac{5+1}{5}\times 5^2$

$=\frac{6}{5}\times 5^2$

=6×5=30

(b) (30⁰+ 15⁰)×$\left(\frac{1}{2}\right)^{-3}$

Sol :

=(1+1)×(2^-1)^-3=2×2³

=2⁴=16

(c) (7 ^-1+ 8^-1 + 9^-1)⁰÷$\left(-\frac{1}{9}\right)^{-2}$

Sol :

$=\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)^0 \div \left(\frac{9}{-1}\right)^2$
$=1\div \frac{81}{1}=\frac{1}{81}$

(d) $\left[\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{6}\right)^{-2}\right]$

Sol :

=3²+4²+6²

=9+16+36

=25+36=61

(e) $\left\{\left(-\frac{4}{5}\right)^{-2}\right\}^{-2}$

Sol :

$=\left\{\left(-\frac{5}{4}\right)\right\}^{-2}=\left(-\frac{5}{4}\right)^{-4}$

$=\left(\frac{4}{5}\right)^4=\frac{256}{625}$

(f) $9^{-1}\times \frac{4^3}{3^{-4}}$

Sol :

$=\frac{(3^2)^{-1}\times 4^3}{3^{-4}}=\frac{3^{-2}\times 4^3}{3^{-4}}$

$=\frac{4^3}{3^{-2}}$

=4³×3²=576

(g) $(5^{-1}\times 4^{-1})\times \left(\frac{3}{2}\right)^{-2}$

Sol :

$=\left(\frac{1}{5}\times \frac{1}{4}\right)\times \left(\frac{2}{3}\right)^2$

$=\frac{1}{20}\times \frac{2^2}{3^2}$

$=\frac{1}{20}\times \frac{4}{9}=\frac{1}{45}$

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Q6 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 6

Evaluate:

(a) $\left[\left(\frac{1}{4}\right)^{-1}-\left(\frac{1}{5}\right)^{-1}\right]+1$
Sol :

=(4-5)+1

=-1+1=0

(b) $\left[\left(\frac{11}{12}\right)^{-10} \div \left(\frac{11}{12}\right)^{-12}\right]\times 1\frac{23}{121}$

Sol :

$=\left[\left(\frac{12}{11}\right)^{10}\div \left(\frac{12}{11}\right)^{12}\right]\times \frac{144}{121}$

$=\left(\frac{12}{11}\right)^{-2}\times \left(\frac{12}{11}\right)^{2}$

$=\frac{(11)^2}{(12)^2}\times \frac{(12)^2}{(11)^2}$

=1

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Q7 | Ex-2A |Class 8 |Exponents | S.Chand | New Learning | Composite maths | myhelper

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Question 7

Find the value of m for which, 7^2m ÷ 7^-4 = 7^-6

Sol :

$=\frac{7^{2m}}{7^{-4}}=$7^-6

⇒7^2m=7^-6×7^-4=7^-10

⇒(7²)^m=(7²)^-5

⇒m=-5