# DAV Class 7 Maths Chapter 4 Worksheet 6 Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 4 Worksheet 6 Solutions of Application of Percentage offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 4 WS 6 Solutions

Question 1.
Simplify:
(i) 44 × 5-4
44 × 5-4
= 44 × $$\frac{1}{5^4}=\frac{4^4}{5^4}$$
= $$\frac{256}{625}$$

(ii) 22 × $$\left(\frac{-1}{3}\right)^2$$
22 × $$\left(\frac{-1}{3}\right)^2$$
= 4 × $$\frac{1}{9}$$
= $$\frac{4}{9}$$

(iii) $$\left(\frac{-2}{3}\right)^3 \times\left(\frac{-3}{5}\right)^3$$
$$\left(\frac{-2}{3}\right)^3 \times\left(\frac{-3}{5}\right)^3$$
= $$\left[\frac{-2}{3} \times \frac{-3}{5}\right]^3$$
= $$\left(\frac{2}{5}\right)^3=\frac{8}{125}$$

(iv) $$\left(\frac{1}{2}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2}$$
$$\left(\frac{1}{2}\right)^{-2} \times\left(\frac{2}{5}\right)^{-2}$$
= $$\left(\frac{1}{2} \times \frac{2}{5}\right)^{-2}$$
= $$\left(\frac{1}{5}\right)^{-2}$$
= (5)2
= 25

(v) $$\left(\frac{-5}{6}\right)^4 \div\left(\frac{-7}{6}\right)^4$$
$$\left(\frac{-5}{6}\right)^4 \div\left(\frac{-7}{6}\right)^4$$
= $$\left[\frac{\left(\frac{-5}{6}\right)}{\left(\frac{-7}{6}\right)}\right]^4$$
= $$\left(\frac{5}{6} \times \frac{6}{7}\right)^4=\left(\frac{5}{7}\right)^4$$
= $$\frac{625}{2301}$$

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

(vii) $$\frac{(-64)^3}{(16)^3}$$
$$\frac{(-64)^3}{(16)^3}=\left(\frac{-64}{16}\right)^3$$
= (-4)3
= -64

Question 2.
(i) 32 × (-4)2 = (-12)2x
32 × (-4)2 = (-12)2x
⇒ [3 × (-4)2] = (-12)2x
⇒ (-12)2 = (-12)2x
⇒ 2x = 2
∴ x = 1

(ii) $$\left(\frac{-3}{2}\right)^6 \times\left(\frac{4}{9}\right)^3=\left(\frac{1}{2}\right)^{3 x}$$

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

(iv) $$\left(\frac{9}{4}\right)^3 \times\left(\frac{8}{9}\right)^3$$ = (2)6x
$$\left(\frac{9}{4}\right)^3 \times\left(\frac{8}{9}\right)^3$$ = (2)6x
⇒ $$\left(\frac{9}{4} \times \frac{8}{9}\right)^3$$ = (2)6x
x = $$\frac{1}{2}$$
(v) $$\left(\frac{15}{4}\right)^3 \div\left(\frac{5}{4}\right)^3$$ = 3x
⇒ $$\left[\frac{15}{4} \div \frac{5}{4}\right]^3$$
⇒ $$\left[\frac{15}{4} \times \frac{4}{5}\right]^3$$