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
Answer:
4⁴ × 5^-4
= 4⁴ × \(\frac{1}{5^4}=\frac{4^4}{5^4}\)
= \(\frac{256}{625}\)

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

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

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

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

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

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

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

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

(iii) \(\left(\frac{4}{5}\right)^{-2} \div\left(\frac{-4}{5}\right)^{-2}\) = (1)^3x
Answer:
\(\left(\frac{4}{5}\right)^{-2} \div\left(\frac{-4}{5}\right)^{-2}\) = (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
Answer:
\(\left(\frac{9}{4}\right)^3 \times\left(\frac{8}{9}\right)^3\) = (2)^6x
⇒ \(\left(\frac{9}{4} \times \frac{8}{9}\right)^3\) = (2)^6x
⇒ (2)³ = (2)^6x
⇒ 6x = 3
x = \(\frac{1}{2}\)

(v) \(\left(\frac{15}{4}\right)^3 \div\left(\frac{5}{4}\right)^3\) = 3^x
Answer:
⇒ \(\left[\frac{15}{4} \div \frac{5}{4}\right]^3\)
⇒ \(\left[\frac{15}{4} \times \frac{4}{5}\right]^3\)
= (3) = 3^x
∴ x = 3