# 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
(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 × $$\left(\frac{-3}{8}\right)^{-1}$$
(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
(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) $$\left(\frac{-1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}$$
= (-2)2 + (3)2 + (4)2
= 4 + 9 + 16
= 29

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

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

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

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

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

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

Question 2.
By what number should we multiply (2-5) so that the product may be equal to (2-1)?
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 $$\left(\frac{1}{5}\right)^{-3}$$?
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 ?
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 (
Let the required number be x.

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

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

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

(iv) $$\left(\frac{-6}{11}\right)^x \div\left[\left(\frac{-6}{11}\right)^{-2}\right]^{-1}=\left[\left(\frac{-6}{11}\right)^2\right]^{-3}$$
$$\left(\frac{-6}{11}\right)^x \div\left[\left(\frac{-6}{11}\right)^{-2}\right]^{-1}=\left[\left(\frac{-6}{11}\right)^2\right]^{-3}$$
⇒ $$\left(\frac{-6}{11}\right)^x \div\left(\frac{-6}{11}\right)^2=\left(\frac{-6}{11}\right)^{-6}$$
⇒ $$\left(\frac{-6}{11}\right)^{x-2}=\left(\frac{-6}{11}\right)^{-6}$$
⇒ 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 $$\left(\frac{p}{q}\right)^{-2}$$.
$$\frac{p}{q}=\left(\frac{2}{3}\right)^2 \times\left(\frac{1}{3}\right)^{-4}$$
$$\frac{p}{q}=\frac{2^2}{3^2}$$ × 34
$$\frac{p}{q}$$ = 22 × 34-2
$$\frac{p}{q}$$ = 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
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 × $$\frac{1}{6}$$
$$\left[\left(\frac{2}{3}\right)^2\right]^3 \times\left(\frac{2}{3}\right)^{-4}$$ × 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) $$\left(\frac{1}{2}\right)^{-2} \div\left(\frac{2}{3}\right)^{-3}$$
$$\left(\frac{1}{2}\right)^{-2} \div\left(\frac{2}{3}\right)^{-3}$$ = (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) $$\left(\frac{2}{5}\right)^3 \times\left(\frac{5}{4}\right)^2$$
$$\left(\frac{2}{5}\right)^3 \times\left(\frac{5}{4}\right)^2$$
= $$\frac{8}{125} \times \frac{25}{16}$$
= $$\frac{1}{10}$$

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

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

(iii) 72 × 7-3
72 × 7-3 = 72-3
= 7-1
= $$\frac{1}{7}$$

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

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

(vi) (83 ÷ 85) × 8-4
(83 ÷ 85) × 8-4
= 83-5 × 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) $$\left(\frac{1}{7}\right)^5$$
$$\left(\frac{1}{7}\right)^5$$
= (7)-5

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

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

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

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