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

(iii) Value of (271/3 + 641/3)2 is ________
(a) 71/3
(b) 49
(c) 71/2
(d) $$\frac{1}{7}$$
Solution:
(b) 49
(271/3 + 641/3)2 = (33 × (1/3) + 43 × (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.

(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 × 163/4
Solution:
5 × 163/4 = 5 × (24)3/4
= 5 × 23
= 5 × 8 = 40

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

(v) Solve for x if 3x-1 = $$\frac{1}{27}$$
Solution:
3x-1 = $$\frac{1}{27}$$
⇒ 3x-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 × 93/2 × 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 × 33 × -2/3
= 2 × 3-2
= 2 × $$\frac{1}{3^2}$$
= $$2 \times \frac{1}{9}=\frac{2}{9}$$

(iii) 2 × 93/2 × 9-1/2 = 2 × $$9^{\frac{3}{2}-\frac{1}{2}}$$
= 2 × 92/2
= 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) 2x + 2x + 2x = 192
(ii) 8255 = (32)x
(iii) 22x + 2 = 42x – 1
Solution:
(i) 2x + 2x + 2x = 192
⇒ 3 × 2x = 192
⇒ 2x = $$\frac{192}{3}$$
⇒ 2x = 64
⇒ 2x = 26
∴ x = 6

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

(iii) 22x + 2 = 42x – 1
⇒ 22x + 2 = 22(2x -1)
⇒ 2 (x + 1 ) = 2 (2x – 1)
⇒ x + 1 = 2x – 1
⇒ 2x – x = 1 + 1
⇒ x = 2.

Question 7.
If 4x – 4x – 1 = 24, then find the value of x.
Solution:
4x – 4x – 1 = 24
⇒ 4x – $$\frac{4^x}{4}$$ = 24
⇒ $$\frac{4 \times 4^x-4^x}{4}$$ = 24
⇒ 3 × 4x = 4 × 24
⇒ 3 × 4x = 96
⇒ 4x = $$\frac{96}{3}$$
⇒ (22)x = 32
⇒ 22x = (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 3x – 1 = 9 and 4y + 2 = 64, find the value of $$\frac{y}{x}-\frac{x}{y}$$.
Solution:
∵ 3x – 1 = 9
⇒ 3x – 1 = 32
⇒ x – 1 = 2
⇒ x = 3.
Again, 4y + 2 = 64
⇒ 4y + 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 9x × 32 × (3-x/2)-2 = $$\frac{1}{27}$$, find x.
Solution:
We have 9x × 32 × (3-x/2)-2 = $$\frac{1}{27}$$
⇒ 32x × 32 × 3(- x/2) × (- 2) = 3-3
⇒ 32x × 32 × 3x = 3– 3
⇒ 32x + 2 + x = 3– 3
⇒ 33x + 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

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 = a3 – 10
= a– 7 = $$\frac{1}{a^7}$$

(iii) 32 × 3– 5 × 36 = 32 – 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 = (25 – 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 5m ÷ 5– 3= 55.
Solution:
5m ÷ 5– 3 = 55
⇒ 5m – (- 3) = 55
⇒ 5m + 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
⇒ 52 × 1/2 = 5

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

(ii) 2x – 3 = 1
⇒ 2x – 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 52x + 1 + 25 = 125, find the value of x.
Solution:
52x + 1 + 25 = 125
⇒ 52x + 1 6 + 52 = 53
⇒ 52x + 1 – 2 = 53
⇒ 52x – 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 34x = $$\frac{1}{81}$$ 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