DAV Class 7 Maths Chapter 4 Worksheet 4 Solutions

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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 = \(\frac{1}{16} \ldots\)
Answer:
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) \(\left(\frac{11}{12}\right)^{\cdots} \div\left(\frac{11}{12}\right)^{22}=\frac{1}{\left(\frac{11}{12}\right)^2}\)
Answer:
DAV Class 7 Maths Chapter 4 Worksheet 4 Solutions 1

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

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

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

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

DAV Class 7 Maths Chapter 4 Worksheet 4 Solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

DAV Class 7 Maths Chapter 4 Worksheet 4 Solutions

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

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

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

(iv) \(\frac{1}{16} \times\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{3(x-2)}\)
Answer:
⇒ \(\left(\frac{1}{2}\right)^4 \times\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^{3 x-6}\)
⇒ \(\left(\frac{1}{2}\right)^{4+2}=\left(\frac{1}{2}\right)^{3 x-6}\)
⇒ \(\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^{3 x-6}\)
⇒ 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
Answer:
(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) \(\left(\frac{12}{13}\right)^6 \div\left(\frac{12}{13}\right)^3=\left(\frac{12}{13}\right)^3\)
Answer:
\(\left(\frac{12}{13}\right)^6 \div\left(\frac{12}{13}\right)^3=\left(\frac{12}{13}\right)^3\)
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 \(\left(\frac{5}{7}\right)^{12}\)
Answer:
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
Answer:
(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) \(\left[\left(\frac{1}{4}\right)^4 \div\left(\frac{1}{4}\right)^3\right]=\frac{1}{4}\)
Answer:
\(\left[\left(\frac{1}{4}\right)^4 \div\left(\frac{1}{4}\right)^3\right]=\frac{1}{4}\)
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
Answer:
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.