DAV Class 7 Maths Chapter 4 Worksheet 5 Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 4 Worksheet 5 Solutions of Application of Percentage offer comprehensive answers to textbook questions.

DAV Class 7 Maths Ch 4 WS 5 Solutions

Question 1.
Simplify:
(i) (2^-1 – 3^-1)^-1 + (6^-1 – 8^-1)^-1
Answer:
(2^-1 – 3^-1)^-1 + (6^-1 – 8^-1)^-1
= \(\left(\frac{1}{2}-\frac{1}{3}\right)^{-1}+\left(\frac{1}{6}-\frac{1}{8}\right)^{-1}\)
= \(\left(\frac{3-2}{6}\right)^{-1}+\left(\frac{4-3}{24}\right)^{-1}\)
= \(\left(\frac{1}{6}\right)^{-1}+\left(\frac{1}{24}\right)^{-1}\)
= 6 + 24
= 30

(ii) (2^-1 – 3^-1)² × \(\left(\frac{-3}{8}\right)^{-1}\)
Answer:
(2^-1 – 3^-1)² × \(\left(\frac{-3}{8}\right)^{-1}\)
= \(\left(\frac{1}{2} \times \frac{1}{3}\right)^2 \times\left(\frac{-8}{3}\right)\)
= \(\left(\frac{1}{6}\right)^2 \times\left(\frac{-8}{3}\right)\)
= \(\frac{1}{36} \times\left(\frac{-8}{3}\right)\)
= \(\frac{-2}{27}\)

(iii) (4^-1 × 3^-1) ÷ 12^-1
Answer:
(4^-1 × 3^-1) ÷ 12^-1
= \(\left(\frac{1}{4} \times \frac{1}{3}\right) \div \frac{1}{12}\)
= \(\frac{1}{12} \div \frac{1}{12}\)
= \(\left(\frac{1}{12}\right)^0\)
= 1

(iv) \(\left(\frac{-1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
Answer:
= (-2)² + (3)² + (4)²
= 4 + 9 + 16
= 29

(v) (2^-1 ÷ 5^-1)² × \(\left(\frac{-5}{8}\right)^{-1}\)
Answer:
= \(\left(\frac{1}{2} \div \frac{1}{5}\right)^2 \times\left(\frac{-8}{5}\right)\)
= \(\left(\frac{1}{2} \times \frac{5}{1}\right)^2 \times\left(\frac{-8}{5}\right)\)
= \(\left(\frac{5}{2}\right)^2 \times\left(\frac{-8}{5}\right)\)
= \(\frac{25}{4} \times\left(\frac{-8}{5}\right)\)
= 5 × (-2)
= -10

(vi) \(\left(\frac{3}{7}\right)^{-2} \div\left(\frac{4}{7}\right)^{-2}\)
Answer:
= \(\left(\frac{7}{3}\right)^2 \div\left(\frac{7}{4}\right)^2\)
= \(\frac{49}{9} \div \frac{49}{16}\)
= \(\frac{49}{9} \times \frac{16}{49}=\frac{16}{9}\)

(vii) \(\left[\left(\frac{2}{5}\right)^{-1} \times\left(\frac{3}{4}\right)^{-1}\right]^{-1}\)
Answer:
= \(\left[\frac{5}{2} \times \frac{4}{3}\right]^{-1}\)
= \(\left[\frac{20}{6}\right]^1\)
= \(\frac{6}{20}=\frac{3}{10}\)

(viii) [4⁵ ÷ 4⁸] × 64
Answer:
= 4^5-8 × 64
= 4^-3 × 64
= \(\frac{1}{4^3}\) × 64
= \(\frac{1}{64}\) × 64
= 1

(ix) \(\left(\frac{4}{5}\right)^{-3} \times\left(\frac{4}{5}\right)^2 \times\left(\frac{4}{5}\right)^3\)
Answer:
= \(\left(\frac{4}{5}\right)^{-3+2+3}\)
= \(\left(\frac{4}{5}\right)^2\)
= \(\frac{16}{25}\)

(x) \(\frac{\left(\frac{-1}{2}\right)^{-3}}{\left(\frac{-1}{2}\right)^{-4}}-\frac{\left(\frac{-1}{2}\right)^{-5}}{\left(\frac{-1}{2}\right)^{-6}}\)
Answer:
= \(\left(\frac{-1}{2}\right)^{-3+4}-\left(\frac{-1}{2}\right)^{-5+6}\)
= \(\left(\frac{-1}{2}\right)-\left(\frac{-1}{2}\right)\)
= \(\frac{-1}{2}+\frac{1}{2}\)
= 0

Question 2.
By what number should we multiply (2^-5) so that the product may be equal to (2^-1)?
Answer:
Let x be multiplied to (2^-5)
∴ x × 2^-5 = 2^-1
⇒ x = 2^-1 + 2^-5
⇒ x = 2^-1-(-5)
⇒ x = 2^-1+5
⇒ x = 2⁴ = 16

Question 3.
By what number should \(\left(\frac{1}{5}\right)^{-3}\) be multiplied so that the product may be equal to \(\left(\frac{1}{5}\right)^{-3}\)?
Answer:
Let x be the required number
∴ x × \(\left(\frac{-1}{5}\right)^{-1}=\left(\frac{1}{5}\right)^{-3}\)
⇒ x = \(\left(\frac{1}{5}\right)^{-3} \div\left(\frac{-1}{5}\right)^{-1}\)
⇒ x = 5³ ÷ (-5)
⇒ x = 125 ÷ (- 5)
⇒ x = – 25.

Question 4.
By what number should (24)^-1 be divided so that the quotient may be equal to (4)^-1 ?
Answer:
Let the required number be x.
∴ (24)^-1 ÷ x = 4^-1
⇒ (24)^-1 = 4^-1 × x
⇒ \(\frac{1}{24}=\frac{1}{4}\) × x
∴ x = 4 × \(\frac{1}{24}\)
x = \(\frac{1}{6}\)

Question 5.
By what number should be divided so that the quotient may be equal to (
Answer:
Let the required number be x.

Question 6.
Find the value of x so that:
(i) \(\left(\frac{3}{4}\right)^{-9} \times\left(\frac{3}{4}\right)^{-7}=\left(\frac{3}{4}\right)^{4 x}\)
Answer:
\(\left(\frac{3}{4}\right)^{-9} \times\left(\frac{3}{4}\right)^{-7}=\left(\frac{3}{4}\right)^{4 x}\)
⇒ \(\left(\frac{3}{4}\right)^{-9-7}=\left(\frac{3}{4}\right)^{4 x}\)
⇒ \(\left(\frac{3}{4}\right)^{-16}=\left(\frac{3}{4}\right)^{4 x}\)
⇒ 4x = -1615
∴ x = -4

(ii) \(\left(\frac{2}{9}\right)^{-6} \times\left(\frac{2}{9}\right)^3=\left(\frac{2}{9}\right)^{2 x-1}\)
Answer:
\(\left(\frac{2}{9}\right)^{-6} \times\left(\frac{2}{9}\right)^3=\left(\frac{2}{9}\right)^{2 x-1}\)
⇒ \(\left(\frac{2}{9}\right)^{-6+3}=\left(\frac{2}{9}\right)^{2 x-1}\)
⇒ \(\left(\frac{2}{9}\right)^{-3}=\left(\frac{2}{9}\right)^{2 x-1}\)
⇒ 2x – 1 = -3
⇒ 2x = 1 – 3
⇒ 2x = -2
∴ x = -1

(iii) \(\left(\frac{5}{7}\right)^2 \div\left(\frac{5}{7}\right)^{3 x+1}=\left(\frac{5}{7}\right)^4\)
Answer:
⇒ \(\left(\frac{5}{7}\right)^2 \div\left(\frac{5}{7}\right)^{3 x+1}=\left(\frac{5}{7}\right)^4\)
⇒ \(\left(\frac{5}{7}\right)^{2-(3 x+1)}=\left(\frac{5}{7}\right)^4\)
⇒ 2 – (3x + 1) = 4
⇒ 2 – 3x – 1 = 4
⇒ 1 – 3x = 4
⇒ -3x = 4 – 1
⇒ – 3x = 3
∴ x = -1

(iv) \(\left(\frac{-6}{11}\right)^x \div\left[\left(\frac{-6}{11}\right)^{-2}\right]^{-1}=\left[\left(\frac{-6}{11}\right)^2\right]^{-3}\)
Answer:
\(\left(\frac{-6}{11}\right)^x \div\left[\left(\frac{-6}{11}\right)^{-2}\right]^{-1}=\left[\left(\frac{-6}{11}\right)^2\right]^{-3}\)
⇒ \(\left(\frac{-6}{11}\right)^x \div\left(\frac{-6}{11}\right)^2=\left(\frac{-6}{11}\right)^{-6}\)
⇒ \(\left(\frac{-6}{11}\right)^{x-2}=\left(\frac{-6}{11}\right)^{-6}\)
⇒ x – 2 = – 6
⇒ x = -6 + 2
⇒ x = -4

Question 7.
If \(\frac{p}{q}=\left(\frac{2}{3}\right)^2 \times\left(\frac{1}{3}\right)^{-4}\), find the value of \(\left(\frac{p}{q}\right)^{-2}\).
Answer:
\(\frac{p}{q}=\left(\frac{2}{3}\right)^2 \times\left(\frac{1}{3}\right)^{-4}\)
\(\frac{p}{q}=\frac{2^2}{3^2}\) × 3⁴
\(\frac{p}{q}\) = 2² × 3^4-2
\(\frac{p}{q}\) = 2² × 3²
\(\frac{p}{q}\) = 4 × 9 = 36

\(\left(\frac{p}{q}\right)^{-2}\) = (36)^-2
\(\left(\frac{p}{q}\right)^{-2}=\left(\frac{1}{36}\right)^2=\frac{1}{1296}\)

Question 8.
If a = \(\left(\frac{3}{5}\right)^{-2} \div\left(\frac{7}{5}\right)^0\), find the value of a^-3
Answer:
a = \(\left(\frac{3}{5}\right)^{-2} \div\left(\frac{7}{5}\right)^0\)
a = \(\left(\frac{5}{3}\right)^2\) ÷
a = \(\frac{25}{9}\) ÷ 1
= \(\frac{25}{9}\)

a^-3 = \(\left(\frac{25}{9}\right)^{-3}=\left(\frac{9}{25}\right)^3=\frac{729}{15625}\)

Question 9.
Simplify:
\(\left[\left(\frac{2}{3}\right)^2\right]^3 \times\left(\frac{2}{3}\right)^{-4}\) × 3^-1 × \(\frac{1}{6}\)
Answer:
\(\left[\left(\frac{2}{3}\right)^2\right]^3 \times\left(\frac{2}{3}\right)^{-4}\) × 3^-1 × \(\frac{1}{6}\)
= \(\left(\frac{2}{3}\right)^{2 \times 3} \times\left(\frac{2}{3}\right)^{-4}\) × 3^-1 × \(\frac{1}{6}\)
= \(\left(\frac{2}{3}\right)^6 \times\left(\frac{2}{3}\right)^{-4} \times \frac{1}{3} \times \frac{1}{6}\)
= \(\left(\frac{2}{3}\right)^{6-4} \times \frac{1}{18}\)
= \(\left(\frac{2}{3}\right)^2 \times \frac{1}{18}\)
= \(\frac{4}{9} \times \frac{1}{18}=\frac{2}{81}\)

Question 10.
Find the reciprocals of
(i) \(\left(\frac{1}{2}\right)^{-2} \div\left(\frac{2}{3}\right)^{-3}\)
Answer:
\(\left(\frac{1}{2}\right)^{-2} \div\left(\frac{2}{3}\right)^{-3}\) = (2)² ÷ \(\left(\frac{3}{2}\right)^3\)
= 8 ÷ \(\frac{27}{8}\)
= 8 × \(\frac{8}{27}\)
= \(\frac{64}{27}\)
∴ Reciprocal of \(\frac{64}{27}=\frac{27}{64}\)

(ii) \(\left(\frac{2}{5}\right)^3 \times\left(\frac{5}{4}\right)^2\)
Answer:
\(\left(\frac{2}{5}\right)^3 \times\left(\frac{5}{4}\right)^2\)
= \(\frac{8}{125} \times \frac{25}{16}\)
= \(\frac{1}{10}\)

Question 11.
Express the following as a rational number with positive exponent:
(i) \(\left(\frac{3}{2}\right)^{-4}\)
Answer:
\(\left(\frac{3}{2}\right)^{-4}=\left(\frac{2}{3}\right)^4\)

(ii) (3^-3)²
Answer:
(3^-3)² = 3^-6
= \(\frac{1}{3^6}\)

(iii) 7² × 7^-3
Answer:
7² × 7^-3 = 7^2-3
= 7^-1
= \(\frac{1}{7}\)

(iv) \(\left[\left(\frac{5}{8}\right)^{-2}\right]^\beta\)
Answer:
\(\left[\left(\frac{5}{8}\right)^{-2}\right]^\beta=\left(\frac{5}{8}\right)^{-2 \times 3}\)
= \(\left(\frac{5}{8}\right)^{-6}=\left(\frac{8}{5}\right)^6\)

(v) \(\left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^5\)
Answer:
\(\left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^5\)
= \(\left(\frac{2}{5}\right)^{-3+5}=\left(\frac{2}{5}\right)^2\)

(vi) (8³ ÷ 8⁵) × 8^-4
Answer:
(8³ ÷ 8⁵) × 8^-4
= 8^3-5 × 8^-4
= 8^-2 × 8^-4
= 8^-2-4
= 8^-6
= \(\left(\frac{1}{8}\right)^6\)

Question 12.
Express the following as a rational number with negative exponent:
(i) \(\left(\frac{1}{7}\right)^5\)
Answer:
\(\left(\frac{1}{7}\right)^5\)
= (7)^-5

(ii) (3²)⁹
Answer:
(3²)⁹ = 3^2×9
= 3¹⁸
= \(\left(\frac{1}{3}\right)^{-18}\)

(iii) 5³ × 5²
Answer:
5³ × 5² = 5³⁺²
= 3⁵
= \(\left(\frac{1}{3}\right)^{-5}\)

(iv) \(\left[\left(\frac{-8}{9}\right)^3\right]^2\)
Answer:
\(\left[\left(\frac{-8}{9}\right)^3\right]^2=\left(\frac{-8}{9}\right)^{3 \times 2}\)
= \(\left(\frac{-8}{9}\right)^6=\left(\frac{-9}{8}\right)^{-6}\)

(v) \(\left(\frac{5}{7}\right)^3 \div\left(\frac{5}{7}\right)^2\)
Answer:
\(\left(\frac{5}{7}\right)^3 \div\left(\frac{5}{7}\right)^2=\left(\frac{5}{7}\right)^{3-2}\)
= \(\left(\frac{5}{7}\right)^1=\left(\frac{7}{5}\right)^{-1}\)

(vi) (2⁶ ÷ 2⁵) × 2²
Answer:
(2⁶ ÷ 2⁵) × 2²
= 2^6-5 × 2²
= 2¹ × 2²
= 2¹⁺²
= 2³
= \(\left(\frac{1}{2}\right)^{-3}\)