DAV Class 8 Maths Chapter 3 Brain Teasers Solutions

The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 3 Brain Teasers Solutions of Exponents and Radicals offer comprehensive answers to textbook questions.

DAV Class 8 Maths Ch 3 Brain Teasers Solutions

Question 1A.
Tick (✓) the correct option.

(i) Value of \(\left\{(625)^{\frac{-1}{2}}\right\}^2\) is ________
(a) \(\frac{1}{625}\)
(b) – 25
(c) – 625
(d) 25
Solution:
(a) \(\frac{1}{625}\).

\(\left\{(625)^{-(1 / 2)}\right\}^2=\left\{\frac{1}{(625)^{1 / 2}}\right\}^2\)
= \(\frac{1}{(625)^{(1 / 2) \times 2}}=\frac{1}{625}\)

(ii) If 5^x = 1, value of x is ________
(a ) 5
(b) \(\frac{1}{25}\)
(c) 0
(d) \(\frac{1}{5}\)
Solution:
(c) 0
5^x = 1
⇒ 5^x = 5^x
⇒ x = 0

(iii) Value of (27^1/3 + 64^1/3)² is ________
(a) 7^1/3
(b) 49
(c) 7^1/2
(d) \(\frac{1}{7}\)
Solution:
(b) 49
(27^1/3 + 64^1/3)² = (3^{3 × (1/3)} + 4^{3 × (1/3)})²
= (3 + 4)²
= 7² = 49

(iv) For any two non-zero rational numbers a and b, a⁴ + b⁴ is equal to
(a) (a + b)¹
(b) (a + b)⁰
(c) (a + b)⁴
(d) (a + b)⁸
Solution:
(c) (a + b)⁴
a⁴ + b⁴ = (a + b)⁴

(v) 4.25 × 10^-7 is equal to ________
(a) 0.425000000
(b) 4250000
(c) 0.000000425
(cl) 0.00000425
Solution:
(c) 0.000000425
4.25 × 10^-7 = \(\frac{4.25}{10^7}=\frac{4.25}{10000000}\) = 0.000000425.

Question 1B.
Answer the following questions.

(i) Express \(\left(\frac{13}{21}\right)^{2 / 5}\) as a radical.
Solution:
\(\left(\frac{13}{21}\right)^{2 / 5}=\sqrt[5]{\left(\frac{13}{21}\right)^2}\)

(ii) Find the value of (81)^-3/4
Solution:
(81)^-3/4 = \(\frac{1}{(81)^{3 / 4}}\)
= \(\frac{1}{3^{4 \times(3 / 4)}}\)
= \(\frac{1}{3^3}=\frac{1}{27}\)

(iii) Simplify 5 × 16^3/4
Solution:
5 × 16^3/4 = 5 × (2⁴)^3/4
= 5 × 2³
= 5 × 8 = 40

(iv) Evaluate (0.000064)^5/6
Solution:
(0.000064)^5/6 = (0.2⁶)^5/6
= 0.2^{6 × 5/6}
= (0.2)⁵ = 0.00032

(v) Solve for x if 3^x-1 = \(\frac{1}{27}\)
Solution:
3^x-1 = \(\frac{1}{27}\)
⇒ 3^x-1 = 3^-3
⇒ x – 1 = – 3
⇒ x = – 3 + 1 = – 2.

Question 2.
Simplify: \(\frac{(64)^{-1 / 6} \times(216)^{-1 / 3} \times(81)^{1 / 4}}{(512)^{-1 / 3} \times(16)^{1 / 4} \times(9)^{-1 / 2}}\)
Solution:

Question 3.
Simplify and express the answer with positive exponent: \(\left[\sqrt[3]{x^4 y} \times \frac{1}{\sqrt[3]{x y^7}}\right]^{-4}\)
Solution:

Question 4.
Evaluate:
(i) 3 × (16)^3/4
(ii) 2 × (27)^-2/3
(iii) 2 × 9^3/2 × 9^-1/2
Solution:
(i) 3 × (16)^3/4 = 3 × (2⁴)^3/4
= 3 × 2³
= 3 × 8 = 24

(ii) 2 × (27)^-2/3 = 2 × (3³)^-2/3
= 2 × 3^{3 × -2/3}
= 2 × 3^-2
= 2 × \(\frac{1}{3^2}\)
= \(2 \times \frac{1}{9}=\frac{2}{9}\)

(iii) 2 × 9^3/2 × 9^-1/2 = 2 × \(9^{\frac{3}{2}-\frac{1}{2}}\)
= 2 × 9^2/2
= 2 × 9 = 18

Question 5.
Find the value of: [(5)² + (12)²]^1/2
Solution:
[(5)² + (12)²]^1/2 = [25 + 144]^1/2
= [169]^1/2 = (13²)^1/2
= \(13^{2 \times \frac{1}{2}}\) = 13

Question 6.
Find the value of x, if
(i) 2^x + 2^x + 2^x = 192
(ii) 8²⁵⁵ = (32)^x
(iii) 2^{2x + 2} = 4^{2x – 1}
Solution:
(i) 2^x + 2^x + 2^x = 192
⇒ 3 × 2^x = 192
⇒ 2^x = \(\frac{192}{3}\)
⇒ 2^x = 64
⇒ 2^x = 2⁶
∴ x = 6

(ii) 8²⁵⁵ = (32)^x
⇒ (2³)²⁵⁵ = (2⁵)^x
⇒ 2^{3 × 255} = 2^5x
⇒ 2⁷⁶⁵ = 2^5x
∴ 5x = 765
⇒ x = \(\frac{765}{5}\) = 153

(iii) 2^{2x + 2} = 4^{2x – 1}
⇒ 2^{2x + 2} = 2^{2(2x -1)}
⇒ 2 (x + 1 ) = 2 (2x – 1)
⇒ x + 1 = 2x – 1
⇒ 2x – x = 1 + 1
⇒ x = 2.

Question 7.
If 4^x – 4^{x – 1} = 24, then find the value of x.
Solution:
4^x – 4^{x – 1} = 24
⇒ 4^x – \(\frac{4^x}{4}\) = 24
⇒ \(\frac{4 \times 4^x-4^x}{4}\) = 24
⇒ 3 × 4^x = 4 × 24
⇒ 3 × 4^x = 96
⇒ 4^x = \(\frac{96}{3}\)
⇒ (2²)^x = 32
⇒ 2^2x = (2)⁵
⇒ 2x = 5
∴ x = \(\frac{5}{2}\)

DAV Class 8 Maths Chapter 3 HOTS

Question 1.
Evaluate: (6^{– 1} – 8^{– 1})^{– 1} + (2^{– 1} – 3^{– 1})^{– 1}
Solution:
(6^{– 1} – 8^{– 1}) + (2^{– 1} – 3^{– 1})^{– 1} = \(\left(\frac{1}{6}-\frac{1}{8}\right)^{-1}+\left(\frac{1}{2}-\frac{1}{3}\right)^{-1}\)
= \(\left(\frac{4-3}{24}\right)^{-1}+\left(\frac{3-2}{6}\right)^{-1}=\left(\frac{1}{24}\right)^{-1}+\left(\frac{1}{6}\right)^{-1}\)
= 24 + 6 = 30

Question 2.
If 3^{x – 1} = 9 and 4^{y + 2} = 64, find the value of \(\frac{y}{x}-\frac{x}{y}\).
Solution:
∵ 3^{x – 1} = 9
⇒ 3^{x – 1} = 3²
⇒ x – 1 = 2
⇒ x = 3.
Again, 4^{y + 2} = 64
⇒ 4^{y + 2} = 4³
⇒ y + 2 = 3
⇒ y = 3 – 2 = 1
Thus, \(\frac{y}{x}-\frac{x}{y}=\frac{1}{3}-\frac{3}{1}=\frac{1-9}{3}=\frac{-8}{3}\).

DAV Class 8 Maths Chapter 3 Enrichment Questions

Question 1.
If 9^x × 3² × (3^-x/2)^-2 = \(\frac{1}{27}\), find x.
Solution:
We have 9^x × 3² × (3^-x/2)^-2 = \(\frac{1}{27}\)
⇒ 3^2x × 3² × 3^{(- x/2) × (- 2)} = 3^-3
⇒ 3^2x × 3² × 3^x = 3^{– 3}
⇒ 3^{2x + 2 + x} = 3^{– 3}
⇒ 3^{3x + 2} = 3^{– 3}
⇒ 3x + 2 = – 3
⇒ 3x = – 3 – 2 = – 5
∴ x = \(\frac{- 5}{3}\)

Question 2.
By what number should \(\left(\frac{-3}{2}\right)^{-3}\) be divided so that the quotient may be \(\left(\frac{-8}{27}\right)^{-2}\).
Solution:
Let the number be x such that

Additional Questions:

Question 1.
Simplify and write in exponential form
(i) (- 3)^{– 3} × (- 3)^{– 4}
(ii) a³ × a^{– 10}
(iii) 3² × 3^{– 5} × 3⁶
Solution:
(i) (- 3)^{– 3} × (- 3)^{– 4} = (- 3)^{– 3 – 4}
= (- 3)^{– 7} = – \(\frac{1}{3^7}\)

(ii) a³ × a^{– 10} = a^{3 – 10}
= a^{– 7} = \(\frac{1}{a^7}\)

(iii) 3² × 3^{– 5} × 3⁶ = 3^{2 – 5 + 6}
= 3³ = 27

Question 2.
Simplify and write the answer in the exponential form
(i) (2⁵ + 2⁸)⁵ × 2^{– 5}
(ii) (- 4)^{– 3} × (5)^{– 3} × (- 5)^{– 3}
(iii) \(\frac{1}{8}\) × (3)^{– 3}
(iv) (- 3)⁴ × \(\left(\frac{5}{3}\right)^4\)
Solution:
(i) (2⁵ + 2⁸)⁵ × 2^{– 5} = (2^{5 – 8})⁵ × 2^{– 5}
= (2^{– 3})⁵ × 2^{– 5}
= 2^{– 15} × 2^{– 5}
= 2^{– 15 – 5} = 2^{– 20} = \(\frac{1}{20}\)

(ii) (- 4)^{– 3} × (5)^{– 3} × (- 5)^{– 3} = \(\left(\frac{1}{-4}\right)^3 \times\left(\frac{1}{5}\right)^3 \times\left(\frac{1}{-5}\right)^3=-\frac{1}{(4)^3} \times \frac{1}{5^3} \times\left(\frac{-1}{5^3}\right)\)

= \(\frac{1}{4^3} \times \frac{1}{5^{3+3}}=\frac{1}{4^3 \times 5^6}\)

= \(\frac{1}{\left(2^2\right)^3 \times 5^6}=\frac{1}{2^6 \times 5^6}\)

= \(\frac{1}{(2 \times 5)^6}=\frac{1}{10^6}\)

(iii) \(\frac{1}{8}\) × (3)^{– 3} = \(\frac{1}{8} \times \frac{1}{(3)^3}\)

= \(\frac{1}{(2)^3} \times \frac{1}{(3)^3}=\frac{1}{(2 \times 3)^3}\)

= \(\frac{1}{(6)^3}=\left(\frac{1}{6}\right)^3\)

(iv) (- 3)⁴ × \(\left(\frac{5}{3}\right)^4\)

= (- 3)⁴ × \(\frac{(5)^4}{(3)^4}\)

= – (3)^{4 – 4} × (5)⁴
= – (3)⁰ × (5)⁴
= – (5)⁴ = – 5⁴

Question 3.
Find ‘k’ so that (- 3)^{k + 1} × (- 3)⁵ = (- 3)⁷
Solution:
(- 3)^{k + 1} × (- 3)⁵ = (- 3)⁷
⇒ (- 3)^{k + 1 + 5} = (- 3)⁷
⇒ (- 3)^{k + 6} = (- 3)⁷
Equating the powers on bothsides,
k + 6 = 7
∴ k = 7 – 6 = 1
Hence, k = 1.

Question 4.
Find the value of x:

(i) (2^{– 1} × 4^{– 1}) + 2^{– 2}
(ii) \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
(iii) \(\left[\left(\frac{-2}{3}\right)^{-2}\right]^2\)
(iv) (3^{– 1} + 4^{– 1} + 5^{– 1})⁰
Solution:
(i) (2^{– 1} × 4^{– 1}) + 2^{– 2} = (2^{– 1} × (2)^{– 2}) + 2^{– 2}
= 2^{– 1 – 2} + 2^{– 2}
=2^{– 3} + 2^{– 2}
= 2^{– 3 – (- 2)}
= 2^{– 3 + 2}
= 2^{– 1} = \(\frac{1}{2}\)

(ii) \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\) = (2)² + (3)² + (4)²
= 4 + 9 + 16 = 29

(iii) \(\left[\left(\frac{-2}{3}\right)^{-2}\right]^2=\left(\frac{-2}{3}\right)^{-2 \times 2}\)
= \(\left(\frac{-2}{3}\right)^{-1}=-\frac{3}{2}\)

(iv) (3^{– 1} + 4^{– 1} + 5^{– 1})⁰ = 1.

Question 5.
Evaluate:
(i) \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
(ii) (5^{– 1} × 2^{– 1}) × 6^{– 1}
Solution:
(i) \(\frac{8^{-1} \times 5^3}{2^{-4}}=\frac{\frac{1}{8} \times 125}{2^{-4}}\)
= \(\frac{125}{8} \times 2^4=\frac{125}{8} \times 16\)
= 125 × 2 = 250.

(ii) (5^{– 1} × 2^{– 1}) × 6^{– 1}
= \(\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}\)
= \(\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}\)

Question 6.
Find the value of m for which 5^m ÷ 5^{– 3}= 5⁵.
Solution:
5^m ÷ 5^{– 3} = 5⁵
⇒ 5^{m – (- 3)} = 5⁵
⇒ 5^{m + 3} = 5⁵
⇒ m + 3 = 5
⇒ m = 5 – 3
∴ m = 2

Question 7.
Evaluate :
(i) \(\left[\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right]^{-1}\)
(ii) \(\left(\frac{5}{8}\right)^{-2} \times\left(\frac{8}{5}\right)^{-3}\)
Solution:
(i) \(\left[\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right]^{-1}\)
= (3 – 4)^{– 1}
= (- 1)^{– 1}
= \(\frac{1}{- 1}\) = – 1

(ii) \(\left(\frac{5}{8}\right)^{-2} \times\left(\frac{8}{5}\right)^{-3}=\left(\frac{8}{5}\right)^2 \times\left(\frac{8}{5}\right)^{-3}\)
= \(\left(\frac{8}{5}\right)^{2-3}=\left(\frac{8}{5}\right)^{-1}=\frac{5}{8}\)

Question 8.
Simplify :
(i) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\) (t ≠ 0)

(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution:
(i) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\)

(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}=\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}}\)
= 3^{– 5 + 5} × 2^{– 5 + 5} × 5^{– 5 + 3 + 7 + 5}
= 3⁰ × 2⁰ × 5¹⁰ = 5¹⁰

Question 9.
Evaluate :
(i) (0.000064)^5/6
(ii) [{(625)^-1/2}^-1/4]²
Solution:
(0.000064)^5/6 = [(0.2)⁶]^5/6
= (0.2)⁵ = 0.00032

(ii) [{(625)^-1/2}^-1/4]² = \(\left[\left\{25^{2 \times \frac{-1}{2}}\right\}^{\frac{-1}{4}}\right]^2=\left[(25)^{-1 \times \frac{-1}{4}}\right]^2\)
= (25)^{1/4 × 2}
= (25)^1/2
= (5²)^1/2
⇒ 5^{2 × 1/2} = 5

Question 10.
Solve each of the following exponential equations:
(i) 7^x = 343
(ii) 2^{x – 3} = 1
Solution:
(i) 7^x = 343
⇒ 7^x = (7)³
⇒ x = 3

(ii) 2^{x – 3} = 1
⇒ 2^{x – 3} = (2)⁰
⇒ x – 3 = 0
⇒ x = 3

Question 11.
Evaluate:
\(\sqrt[3]{100} \times \sqrt[3]{270}\)
Solution:
\(\sqrt[3]{100} \times \sqrt[3]{270}=\sqrt[3]{100 \times 270}\)
[∵ \(\sqrt[3]{a} \times \sqrt[3]{b}=\sqrt[3]{a b}\)]
= \(\sqrt[3]{27000}=\sqrt[3]{30^3}\) = 30

Question 12.
Evaluate: \(\sqrt[3]{(-1728) \times(-2197)}\)
Solution:
\(\sqrt[3]{(-1728) \times(-2197)}=\sqrt[3]{(-12)^3} \times \sqrt[3]{(-13)^3}\)
[∵ \(\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}\)]
= – 12 × – 13 = 156

Question 13.
Find the value of (2^{– 1} × 5^{– 1})^{– 1} ÷ 5^{– 1}.
Solution:
(2^{– 1} × 5^{– 1})^{– 1} ÷ 5^{– 1}
⇒ (10^{– 1})^{– 1} ÷ 5^{– 1}
= 10 ÷ \(\frac{1}{5}\)
= 10 × 5 = 50

Question 14.
Evaluate: (0.03125)^-2/5
Solution:
(0.03125)^-2/5 = (0.5 × 0.5 × 0.5 × 0.5 × 0.5)^-2/5
= (0.5)^{5 × – 2/5}
= (0.5)^{– 2}
= \(\frac{1}{(0.5)^2}=\frac{1}{0.25}=\frac{100}{25}\) = 4

Question 15.
If 5^{2x + 1} + 25 = 125, find the value of x.
Solution:
5^{2x + 1} + 25 = 125
⇒ 5^{2x + 1} 6 + 5² = 5³
⇒ 5^{2x + 1 – 2} = 5³
⇒ 5^{2x – 1} = 5³
Equating the power on bothsides,
2x – 1 = 3
⇒ 2x = 3 + 1
⇒ 2x = 4
∴ x = 2.

Question 16.
Evaluate: [{(625)^-1/2}^-1/4]².
Solution:

Question 17.
Find the value of: 4 × 81^-1/2 \(\left(81^{\frac{1}{2}}+81^{\frac{3}{2}}\right)\).
Solution:
4 × 81^-1/2 \(\left(81^{\frac{1}{2}}+81^{\frac{3}{2}}\right)\) = \(=4 \times 9^{2 \times\left(-\frac{1}{2}\right)}\left(9^{2 \times \frac{1}{2}}+9^{2 \times \frac{3}{2}}\right)\)

= 4 × 9^{– 1} (9 + 9³)
= 4 × 9^{– 1} × 9 (1 + 9²)
= 4 × 9^{– 1 + 1} (1 + 81)
= 4 × 9⁰ (82)
= 4 × 1 × 82 = 328.

Question 18.
Solve for x : \(\left(\frac{8}{3}\right)^{x+1} \times\left(\frac{8}{3}\right)^{x+5}=\left(\frac{8}{3}\right)^x \times \frac{64}{9}\)
Solution:

Equating the powers on both sides,
x + 6 = 2
∴ x = 2 – 6 = – 4.

Multiple Choice Questions:

Question 1.
If 3^4x = \(\frac{1}{81}\) then the value of x is
(a) 1
(b) – 1
(c) 2
(d) \(\frac{1}{2}\)
Solution:
(b) – 1

Question 2.
The value of (x^{– 2} + x^{– 3} + x^{– 4} + x-5)0 is
(a) 0
(b) – 1
(c) – 14
(d) 1
Solution:
(d) 1

Question 3.
The value of (27)^6/5 ÷ (27)^1/5 is
(a) 3
(b) 9
(c) 27
(d) 36
Solution:
(c) 27

Question 4.
The value of \(\left[\left(\frac{36}{25}\right)^{3 / 2}\right]^{5 / 3}\) is ________
(a) \(\left(\frac{6}{5}\right)^4\)
(b) \(\left(\frac{6}{5}\right)^5\)
(c) \(\left(\frac{6}{5}\right)^3\)
(d) \(\left(\frac{6}{5}\right)^2\)
Solution:
(c) \(\left(\frac{6}{5}\right)^3\)

Question 5.
The value of (343)^-2/3 is ________
(a) \(\frac{1}{49}\)
(b) \(\frac{1}{7}\)
(c) 7
(d) 49
Solution:
(a) \(\frac{1}{49}\)

Question 6.
The vaiue of (125)^2/3 is ________
(a) 5
(b) \(\frac{1}{5}\)
(c) 25
(d) \(\frac{1}{25}\)
Solution:
(c) 25