# DAV Class 7 Maths Chapter 4 Worksheet 4 Solutions

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

## DAV Class 7 Maths Ch 4 WS 4 Solutions

Question 1.
Fill in the blanks:
(i) 162 ÷ 162 = $$\frac{1}{16} \ldots$$
162 ÷ 162 = $$\frac{1}{16} \ldots$$
⇒ 1615-19 = $$\frac{1}{16} \ldots$$
⇒ 16-4 = $$\frac{1}{16} \ldots$$
⇒ $$\frac{1}{16^4}=\frac{1}{16^{\prime \prime}}$$
∴ (……….) = 4

(ii) $$\left(\frac{11}{12}\right)^{\cdots} \div\left(\frac{11}{12}\right)^{22}=\frac{1}{\left(\frac{11}{12}\right)^2}$$

(iii) $$\left(\frac{1}{3^2}\right)^{\cdots}=\left(\frac{1}{3^6}\right)$$
$$\left(\frac{1}{3^2}\right)^{\cdots}=\frac{1}{(3)^6}$$
⇒ $$\left(\frac{1}{3^2}\right)^{\cdots}=\left(\frac{1}{3^2}\right)^3$$
∴ (……….) = 3

(iv) $$\left[\left(\frac{-1}{2}\right)^4\right]^2=\left(\frac{-1}{2}\right)^{\cdots}$$
$$\left[\left(\frac{-1}{2}\right)^4\right]^2=\left[\left(\frac{-1}{2}\right)\right]^*$$
⇒ $$\left(\frac{-1}{2}\right)^{4 \times 2}=\left(\frac{-1}{2}\right)^{\cdots}$$
⇒ $$\left(\frac{-1}{2}\right)^8=\left(\frac{-1}{2}\right)^{\prime}$$
∴ (……….) = 8

(v) $$\left(\frac{1}{12}\right)^5 \div\left(\frac{1}{12}\right)^5=\left(\frac{1}{12}\right)^5$$
$$\left(\frac{1}{12}\right)^5 \div\left(\frac{1}{12}\right)^5=\left(\frac{1}{12}\right)^{\cdots}$$
⇒ $$\left(\frac{1}{12}\right)^{5-5}=\left(\frac{1}{12}\right)^3$$
⇒ $$\left(\frac{1}{12}\right)^0=\left(\frac{1}{12}\right)^{\cdots}$$
∴ (……….) = 0

(vi) $$\left[\left(\frac{-1}{7}\right)^{o p}\right]^0=\left(\frac{-1}{7}\right)^{\cdots}$$
$$\left[\left(\frac{-1}{7}\right)^0\right]^p=\left[\frac{-1}{7}\right]$$
⇒ $$\left(\frac{-1}{7}\right)^{0 \times 7}=\left(\frac{-1}{7}\right)^{\cdots}$$
⇒ $$\left(\frac{-1}{7}\right)^0=\left(\frac{-1}{7}\right)^{\cdots}$$
∴ (……….) = 0

Question 2.
Simplify and express the result in exponential form:
(i) $$\left[\left(\frac{9}{2}\right)^0\right]^5$$
$$\left[\left(\frac{9}{2}\right)^{0 p}\right]^0=\left(\frac{9}{2}\right)^{0 \times 5}$$
= $$\left(\frac{9}{2}\right)^0$$

(ii) $$\left[\left(\frac{-5}{17}\right)^6\right]^3$$
$$\left[\left(\frac{-5}{17}\right)^6\right]^3=\left(\frac{-5}{17}\right)^{6 \times 3}$$
= $$\left(\frac{-5}{17}\right)^{18}$$

(iii) $$\left[\left(\frac{4}{7}\right)^4\right]^4$$
$$\left[\left(\frac{4}{7}\right)^4\right]^4=\left(\frac{4}{7}\right)^{4 \times 4}$$
= $$\left(\frac{4}{7}\right)^{16}$$

(iv) $$\left[\left(\frac{-2}{5}\right)^3 \times\left(\frac{-2}{5}\right)^2\right]^4$$
$$\left[\left(\frac{-2}{5}\right)^3 \times\left(\frac{-2}{5}\right)^2\right]^4=\left[\left(\frac{-2}{5}\right)^{3+2}\right]^4$$
= $$\left[\left(\frac{-2}{5}\right)^5\right]^4=\left(\frac{-2}{5}\right)^{5 \times 4}$$
= $$\left(\frac{-2}{5}\right)^{20}$$

(v) $$\left(\frac{3}{5}\right)^3 \div\left(\frac{3}{5}\right)^8$$
$$\left(\frac{3}{5}\right)^3 \div\left(\frac{3}{5}\right)^8=\left(\frac{3}{5}\right)^{3-8}$$
= $$\left(\frac{3}{5}\right)^{-5}$$
= $$\left[\frac{5}{3}\right]^5$$

(vi) $$\left[\left(\frac{12}{5}\right)^0 \times\left(\frac{12}{5}\right)\right]^5$$
$$\left[\left(\frac{12}{5}\right)^0 \times\left(\frac{12}{5}\right)\right]^p=\left[\left(\frac{12}{5}\right)^{0+1}\right]^5$$
= $$\left[\left(\frac{12}{5}\right)^1\right]^p$$
= $$\left(\frac{12}{5}\right)^5$$

Question 3.
Evaluate:
(i) $$\left(\frac{-3}{4}\right)^3 \div\left(\frac{-3}{4}\right)^5$$
$$\left(\frac{-3}{4}\right)^3 \div\left(\frac{-3}{4}\right)^5=\left(\frac{-3}{4}\right)^{3-5}$$
= $$\left(\frac{-3}{4}\right)^{-2}=\left(\frac{-4}{3}\right)^2$$
= $$\frac{16}{9}$$

(ii) $$\left(\frac{1}{5^2}\right)^2 \times \frac{1}{5}$$
$$\left(\frac{1}{5^2}\right)^2 \times \frac{1}{5}=\frac{1}{5^{2 \times 2}} \times \frac{1}{5}$$
= $$\frac{1}{5^4} \times \frac{1}{5}=\frac{1}{5^{4+1}}=\frac{1}{5^5}$$
= $$\frac{1}{3125}$$

(iii) $$\left[\left(\frac{-5}{6}\right)^2\right]^2 \div\left(\frac{-5}{6}\right)^2$$
$$\left[\left(\frac{-5}{6}\right)^2\right]^2 \div\left(\frac{-5}{6}\right)^2=\left(\frac{-5}{6}\right)^{2 \times 2} \div\left(\frac{-5}{6}\right)^2$$
= $$\left(\frac{-5}{6}\right)^4 \div\left(\frac{-5}{6}\right)^2=\left(\frac{-5}{6}\right)^{4-2}=\left(\frac{-5}{6}\right)^2$$
= $$\frac{25}{36}$$

(iv) $$\left(\frac{2}{3}\right)^2 \div\left[\left(\frac{2}{3}\right)^2\right]^p$$
$$\left(\frac{2}{3}\right)^2 \div\left[\left(\frac{2}{3}\right)^2\right]^p=\left(\frac{2}{3}\right)^2 \div\left(\frac{2}{3}\right)^0$$
= $$\left(\frac{2}{3}\right)^{2-0}=\left(\frac{2}{3}\right)^2$$
= $$\frac{4}{9}$$

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

Question 4.
(i) $$\left(\frac{5}{7}\right)^{3 \times 2-6}$$
$$\left(\frac{5}{7}\right)^{3 \times 2-6}=\left(\frac{5}{7}\right)^{6-6}=\left(\frac{5}{7}\right)^0$$
= 1

(ii) (30 + 40) × 50
(30 + 40) × 50
= (1 + 1) × 1
= 2 × 1
= 2

(iii) 10 × 20 + 30 × 40 + 50 × 60
10 × 20 + 30 × 40 + 50 × 60
= 1 × 1 + 1 × 1 + 1 × 1
= 1 + 1 + 1
= 3

(iv) $$\left(\frac{-5}{9}\right)^{9-3 \times 2-3}$$
$$\left(\frac{-5}{9}\right)^{9-3 \times 2-3}=\left(\frac{-5}{9}\right)^{9-6-3}$$
= $$\left(\frac{-5}{9}\right)^{9-9}=\left(\frac{-5}{9}\right)^0$$
= 1

Question 5.
Find the value of x so that:
(i) $$\left(\frac{3}{4}\right)^{2 x+1}=\left[\left(\frac{3}{4}\right)^3\right]^3$$
⇒ $$\left(\frac{3}{4}\right)^{2 x+1}=\left(\frac{3}{4}\right)^{3 \times 3}$$
⇒ $$\left(\frac{3}{4}\right)^{2 x+1}=\left(\frac{3}{4}\right)^9$$
⇒ 2x + 1 = 9
⇒ 2x = 9 – 1
⇒ 2x = 8
x = 4

(ii) $$\left(\frac{2}{5}\right)^3 \times\left(\frac{2}{5}\right)^6=\left(\frac{2}{5}\right)^{3 x}$$
⇒ $$\left(\frac{2}{5}\right)^{3+6}=\left(\frac{2}{5}\right)^{3 x}$$
⇒ $$\left(\frac{2}{5}\right)^9=\left(\frac{2}{5}\right)^{3 x}$$
3x = 9
x = 9 ÷ 3
= 3

(iii) $$\left(\frac{-1}{5}\right)^{20} \div\left(\frac{-1}{5}\right)^{15}=\left(\frac{-1}{5}\right)^{5 x}$$
$$\left(\frac{-1}{5}\right)^{20} \div\left(\frac{-1}{5}\right)^{15}=\left(\frac{-1}{5}\right)^{5 x}$$
⇒ $$\left(\frac{-1}{5}\right)^{20-15}=\left(\frac{-1}{5}\right)^{5 x}$$
⇒ $$\left(\frac{-1}{5}\right)^5=\left(\frac{-1}{5}\right)^{5 x}$$
⇒ 5x = 5
x = 5 ÷ 5 = 1
x = 1

(iv) $$\frac{1}{16} \times\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{3(x-2)}$$
⇒ $$\left(\frac{1}{2}\right)^4 \times\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{3 x-6}$$
⇒ $$\left(\frac{1}{2}\right)^{4+2}=\left(\frac{1}{2}\right)^{3 x-6}$$
⇒ $$\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^{3 x-6}$$
⇒ 3x – 6 = 6
⇒ 3x = 6 + 6 = 12
⇒ x = 12 ÷ 3 = 4
x = 4

Question 6.
Which of the following statements are true?
(i) (0.6)8 ÷ (0.6)7 = (0.6)2
(0.6)8 ÷ (0.6)7 = (0.6)2
L.H.S (0.6)8-7 = 0.6 and R.H.S = (0.6)2
∴ L.H.S ≠ R.H.S
Hence it is False.

(ii) $$\left(\frac{12}{13}\right)^6 \div\left(\frac{12}{13}\right)^3=\left(\frac{12}{13}\right)^3$$
$$\left(\frac{12}{13}\right)^6 \div\left(\frac{12}{13}\right)^3=\left(\frac{12}{13}\right)^3$$
L.H.S = $$\left(\frac{12}{13}\right)^{6-3}=\left(\frac{12}{13}\right)^3$$
R.H.S = $$\left(\frac{12}{13}\right)^3$$
L.H.S = R.H.S
Hence it is true.

(iii) The reciprocal of $$\left(\frac{7}{5}\right)^{12}$$ is $$\left(\frac{5}{7}\right)^{12}$$
The reciprocal of $$\left(\frac{7}{5}\right)^{12}$$ is $$\left(\frac{5}{7}\right)^{12}$$
L.H.S Reciprocal of $$\left(\frac{7}{5}\right)^{12}$$
= $$\frac{1}{\left(\frac{7}{5}\right)^{12}}=\left(\frac{5}{7}\right)^{12}$$
R.H.S = $$\left(\frac{5}{7}\right)^{12}$$
L.H.S = R.H.S
Hence it is true.

(iv) (5 + 5)5 = 55 + 55
(5 + 5)5 = 55 + 55
L.H.S = (5 + 5)5 = 105
= 100000

R.H.S = 55 + 55
= 3125 + 3125
= 6250
L.H.S ≠ R.H.S
Hence it is false.

(v) $$\left[\left(\frac{1}{4}\right)^4 \div\left(\frac{1}{4}\right)^3\right]=\frac{1}{4}$$
$$\left[\left(\frac{1}{4}\right)^4 \div\left(\frac{1}{4}\right)^3\right]=\frac{1}{4}$$
L.H.S = $$\left(\frac{1}{4}\right)^4 \div\left(\frac{1}{4}\right)^3=\left(\frac{1}{4}\right)^{4-3}$$
= $$\frac{1}{4}$$

R.H.S = $$\frac{1}{4}$$
L.H.S = R.H.S
Hence it is true.

(vi) $$\left(\frac{1}{7} \times \frac{1}{7^2}\right) \div \frac{1}{7^3}$$ = 1
L.H.S = $$\left(\frac{1}{7} \times \frac{1}{7^2}\right) \div \frac{1}{7^3}$$
= $$\frac{1}{7^{1+2}} \div \frac{1}{7^3}$$
= $$\frac{1}{7^3} \div \frac{1}{7^3}$$
= 1 and R.H.S = 1
L.H.S = R.H.S
Hence it is true.