# DAV Class 7 Maths Chapter 4 Brain Teasers Solutions

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

## DAV Class 7 Maths Ch 4 Brain Teasers Solutions

Question 1.
A. Tick (✓) the correct option:
(i) The exponential form of [(22)3] is:
(a) 45
(b) 212
(c) (12)2
(d) 27
(b) 212

[(22)3] = (22)6
= (2)12
= 212
Hence, (b) is the correct answer.

(ii) 2.7 × 10-3 is equal to:
(a) 0.000027
(b) 0.00027
(c) 0.0027
(d) 2.007
(c) 0.0027

2.7 × 10-3
= $$\frac{27}{10} \times \frac{1}{(10)^3}=\frac{27}{10} \times \frac{1}{1000}$$
= $$\frac{27}{10000}$$
= 0.0027
Hence, (c) is the correct answer.

(iii) 9 × 42 is same as:
(a) (12)2
(b) (36)2
(c) (36)3
(d) (18)4
9 × 42 = 3 × 3 × 4 × 4
= (3 × 4) × (3 × 4)
= 12 × 12
= (12)2

(iv) 10 × 1011 is equal to:
(a) (100)4
(b) (10)10
(c) (100)12
(d) (10)12
10 × 1011
= (10)1 × (10)11
= (10)1+11
= (10)12
Hence, (d) is the correct answer.

(v) If $$\left(\frac{3}{5}\right)^{2 x}$$ = 1, then x is equal to:
(a) 2
(b) 0
(c) 1
(d) $$\frac{1}{2}$$
$$\left(\frac{3}{5}\right)^{2 x}$$ = 1
⇒ $$\left(\frac{3}{5}\right)^{2 x}=\left(\frac{3}{5}\right)^0$$
⇒ 2x =1
⇒ x = 0
Hence, (b) is the correct answer.

(i) Simplify $$\left[\left\{\left(\frac{4}{9}\right)^2\right\}^0\right]^5$$
$$\left[\left\{\left(\frac{4}{9}\right)^2\right\}^0\right]^5=\left[\left\{\frac{16}{81}\right\}^0\right]^5$$
= [1]
= 1

(ii) Write (22)3 × 36 in simplified exponential form.
(22)3 × 36
= (2 × 3)6
= (6)6
= 66

(iii) Find the value of x if $$\left[\left(\frac{3}{7}\right)^3\right]^{-2}=\left(\frac{3}{7}\right)^{2 x}$$
$$\left[\left(\frac{3}{7}\right)^3\right]^{-2}=\left(\frac{3}{7}\right)^{2 x}$$
⇒ $$\left(\frac{3}{7}\right)^{-6}=\left(\frac{3}{7}\right)^{2 x}$$
⇒ -6 = 2x
⇒ x = $$\frac{-6}{2}$$
⇒ x = -3

(iv) Simplify 6-2 + $$\left(\frac{3}{2}\right)^{-2}$$
$$\frac{1}{6^2}+\left(\frac{2}{3}\right)^2=\frac{1}{36}+\frac{4}{9}$$
= $$\frac{1+16}{36}$$
= $$\frac{17}{36}$$

(v) Write in usual form 11.2 × (10)-7.
= 11.2 × $$\frac{1}{(10)^7}$$
= $$\frac{112}{10} \times \frac{1}{(10)^7}$$
= $$\frac{112}{(10)^8}$$
= 0.00000112

Question 2.
Write the base and exponent of the following numbers:
(i) $$\left(\frac{1}{9}\right)^{-4}$$
$$\left(\frac{1}{9}\right)^{-4}$$
Base = $$\frac{1}{9}$$
and exponent = -4

(ii) $$\frac{7}{11}$$
$$\frac{7}{11}=\left(\frac{7}{11}\right)^1$$
Base = $$\frac{7}{11}$$
and exponent = 1

(iii) $$\left[\left(\frac{-5}{6}\right)^0\right]^2$$
$$\left[\left(\frac{-5}{6}\right)^0\right]^2=\left(\frac{-5}{6}\right)^0$$
Base = $$\frac{-1}{6}$$,
exponent = 0

(iv) $$\left(\frac{a c}{b}\right)^3$$
$$\left(\frac{a c}{b}\right)^3$$
Base = $$\frac{a c}{b}$$
and exponent = 3

(v) $$\left(\frac{-1}{2} \times \frac{1}{3}\right)^0$$
$$\left(\frac{-1}{2} \times \frac{1}{3}\right)^0=\left(\frac{-1}{6}\right)^0$$
Base = $$\frac{-1}{6}$$
exponent = 0

(vi) $$\left(\frac{-11}{12}\right)^4$$
$$\left(\frac{-11}{12}\right)^4$$
Base = $$\frac{-11}{12}$$
exponent = 4

Question 3.
Express the following as a rational number:
(i) $$\left[\left(\frac{-1}{2}\right)^{-3}\right]^2$$
$$\left[\left(\frac{-1}{2}\right)^{-3}\right]^2=\left(\frac{-1}{2}\right)^{-3 \times 2}$$
= $$\left(\frac{-1}{2}\right)^{-6}=\left(\frac{-2}{1}\right)^6$$
= 64

(ii) $$\left(\frac{1}{3^2}\right)^{-1}$$ × 9-2
$$\left(\frac{1}{3^2}\right)^{-1}$$ × 9-2
= 32 × 9-2
= 9 × 9-2
= 91-2
= 9-1
= $$\frac{1}{9}$$

(iii) $$\left(\frac{16}{25}\right)^{-1} \div\left(\frac{4}{5}\right)^{-3}$$
$$\left(\frac{16}{25}\right)^{-1} \div\left(\frac{4}{5}\right)^{-3}=\left(\frac{25}{16}\right) \div\left(\frac{5}{4}\right)^3$$
= $$\left(\frac{5}{4}\right)^2 \div\left(\frac{5}{4}\right)^3=\left(\frac{5}{4}\right)^{2-3}$$
= $$\left(\frac{5}{4}\right)^{-1}=\frac{4}{5}$$

(iv) $$\left(\frac{2}{3}\right)^{-1}-\left(\frac{3}{2}\right)^{-2}$$
$$\left(\frac{2}{3}\right)^{-1}-\left(\frac{3}{2}\right)^{-2}=\frac{3}{2}-\left(\frac{2}{3}\right)^2$$
= $$\frac{3}{2}-\frac{4}{9}=\frac{27-8}{18}$$
= $$\frac{19}{18}$$

(v) $$\left[\left(\frac{5}{6}\right)^0+\left(\frac{3}{4}\right)^0\right] \div\left(\frac{2}{3}\right)^0$$
$$\left[\left(\frac{5}{6}\right)^0+\left(\frac{3}{4}\right)^0\right] \div\left(\frac{2}{3}\right)^0$$
= [1 + 1] ÷ 1
= 2 ÷ 1
= 2

(vi) $$\frac{\left(\frac{-3}{5}\right)^7}{\left(\frac{-3}{5}\right)^7}$$
$$\frac{\left(\frac{-3}{5}\right)^7}{\left(\frac{-3}{5}\right)^7}=\left(\frac{-3}{5}\right)^{7-7}$$
= $$\left(\frac{-3}{5}\right)^0$$
= 1

Question 4.
Express the following in the exponential form:
(i) (2.5)2 ÷ $$\left(\frac{1}{2.5}\right)^2$$
(2.5)2 ÷ $$\left(\frac{1}{2.5}\right)^2$$
= (2.5)2 ÷ (2.5)-2
= (2.5)4-2
= (2.5)2

(ii) (52 × 55) ÷ 510
(52 × 55) ÷ 510
= 52+5 ÷ 510
= 57 ÷ 510
= 57-10
= 5-3

Question 5.
Find the reciprocals of:
(i) $$\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3$$
$$\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3$$
= $$\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3$$
∴ Reciprocal of $$\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3$$
= $$\frac{1}{\left(\frac{2}{3}\right)^{-5}}=\left(\frac{2}{3}\right)^5$$

(ii) $$\left(\frac{3}{4}\right)^{-2} \times\left(\frac{3}{4}\right)^3$$
$$\left(\frac{3}{4}\right)^{-2} \times\left(\frac{3}{4}\right)^3$$
= $$\left(\frac{3}{4}\right)^{-2+3}=\frac{3}{4}$$
∴ Reciprocal of $$\frac{3}{4}$$
= $$\frac{1}{3 / 4}=\frac{4}{3}$$

(iii) $$\left[\left(\frac{7}{8}\right)^{-3}\right]^2$$
$$\left[\left(\frac{7}{8}\right)^{-3}\right]^2=\left(\frac{7}{8}\right)^{-3 \times 2}$$
= $$\left(\frac{7}{8}\right)^{-6}$$
∴ Reciprocal of $$\left(\frac{7}{8}\right)^{-6}$$
= $$\frac{1}{\left(\frac{7}{8}\right)^{-6}}=\left(\frac{7}{8}\right)^6$$

(iv) $$\left(\frac{3}{5} \times \frac{-2}{9}\right)^{-2}$$
$$\left(\frac{3}{5} \times \frac{-2}{9}\right)^{-2}$$
= $$\left(\frac{-6}{45}\right)^{-2}$$
∴ Reciprocal of $$\left(\frac{-6}{45}\right)^{-2}$$
= $$\frac{1}{\left(\frac{-6}{45}\right)^{-2}}=\left(\frac{-6}{45}\right)^2$$

(v) $$\frac{\left(\frac{-3}{2}\right)^{-1} \times\left(\frac{2}{3}\right)^2}{\left(\frac{2}{3}\right)^{-1} \div\left(\frac{3}{2}\right)}$$

Question 6.
Express (153)-16 as a single exponent of 15.
Solution:
(153)16 = 153 × (-16)
= (15)-48

Question 7.
By what number should (7)-2 be multiplied so that the product may be equal to (545T1?
Let x be the required number.
x x (7)-2 = (343)-1
⇒ x = (343)-1 -(7)-2
⇒ x = (73)-1 + (7)-2
⇒ x = 7-3 + 7-2
⇒ x = 7-3+2
= 7-1 = $$\frac{1}{7}$$
Hence the required number is $$\frac{1}{7}$$.

Question 8.
By what number should $$\left(\frac{-3}{4}\right)^5$$ be multiplied so that the product may be equal to $$\left(\frac{-64}{27}\right)^{-1}$$?
Let the required number be x

Hence the required number is $$\frac{16}{9}$$

Question 9.
By what number should (- 512)-1 be divided so that the quotient may be equal to (8)-2?
Solution
Let the required number be x
(- 512)-1 + x = (8)-2
⇒ (-512)-1 + (8)-2 = x
⇒ [(-8)-3]-1 + (8)-2 = x
⇒ (-8)-3 + (8)-2 = x
-(8)-3+2 = x
-(8)-1 = x

Question 10.
By what number should $$\left(\frac{144}{225}\right)^{-1}$$ be divided so that the quotient may be equal to $$\left(\frac{12}{15}\right)^{-4}$$?
Let the required number be x

Question 11.
Write the following numbers in usual form:
(i) 5.3 × 105
5.3 × 105 = 5.3 × 100000 = 530000

(ii) 2.9 × 10-10
2.9 × 10-10 = 2.9 × $$\frac{1}{10000000000}$$
= 0.00000000029

(iii) 4.6 × 10-12
4.6 × 10-12 = 4.6 × $$\frac{1}{1000000000000}$$
= 0.0000000000046

(iv) 1.08 × 107
1.08 × 107 = 1.08 × 10000000
= 10800000

(v) 3.09 × 1011
3.09 × 1011 = 3.09 × 100000000000
= 309000000000

(vi) 6.00005 × 109
6.00005 × 109 = 6.00005 × 100000000
= 6000050000.

Question 12.
Write the following numbers in the form k × 10n where 1 ≤ k < 10 and n is an integer:
(i) 762850
762850 = 7.62850 × 105

(ii) 2500000
2500000 = 2.5 × 106

(iii) 0.09
0.09 = 9.0 × 10-2

(iv) 0.0000076
0.0000076 = 7.6 × 10-6

(v) 0.0000008
0.00000008 = 8.0 × 10-8

(vi) 4592000000000
459200000000 = 4.592 × 1011

(vii) $$\frac{315 \times 10^5}{0.7 \times 10^3}$$
$$\frac{315 \times 10^5}{0.7 \times 10^3}$$
= 450 × 102
= 45000 = 4.5 × 104

(viii) $$\frac{4.4 \times 10^{-7}}{44 \times 10^{-5}}$$
$$\frac{4.4 \times 10^{-7}}{44 \times 10^{-5}}=\frac{1}{10} \times \frac{10^{-7}}{10^{-5}}$$
= 0.1 × 10-2
= 0.1 × $$\frac{1}{100}$$
= 0.001
= 1.0 × 10-3

(ix) (7 × 102) × (8 × 103)
(7 × 102) × (8 × 103)
= 7 × 8 × 102+3
= 56 × 105
= 5.6 × 106

Question 13.
Simplify the following:
(i) $$\left(\frac{7}{8}\right)^{-3} \times\left(\frac{9}{5}\right)^0 \times 8^{-2} \times\left(\frac{1}{7}\right)^{-1}$$

(ii) $$\left\{\left[\left(\frac{4}{5}\right)^3\right]^2 \div\left(\frac{4}{5}\right)^2\right\} \times\left(\frac{1}{4}\right)^{-2} \times 4^{-1}$$

(iii) $$\left(\frac{-4}{5}\right)^3 \times 5^2 \times\left(\frac{-1}{2}\right)^5 \times\left(\frac{1}{2}\right)^{-3}$$

(iv) $$\left(\frac{-4}{5}\right)^3 \times 5^2 \times\left(\frac{-1}{2}\right)^5 \times\left(\frac{1}{2}\right)^{-3}$$

Question 14.
Express the numbers appearing in the following statements in the form k × 10n where 1 ≤ k < n and n is an integer.
(i) The mean distance of the moon from the earth is 384,400,000 metres.
384,400,000 metres = 3.844 × 108 metres

(ii) The distance travelled by a ray of light in one year is 9,460,500,000,000,000 m.
9,460,500,000,000,000 m = 9.4605 × 1015 m

(iii) The number of red blood cells per cubic mm of human blood is approximately 5.5 millions.
5.5 millions = 5500000 = 5.5 × 106

Question 15.
Find the value of x so that
(i) $$\left(\frac{-7}{11}\right)^{-3} \times\left(\frac{-7}{11}\right)^{5 x}=\left[\left(\frac{-7}{11}\right)^{-2}\right]^{-1}$$
$$\left(\frac{-7}{11}\right)^{-3} \times\left(\frac{-7}{11}\right)^{5 x}=\left[\left(\frac{-7}{11}\right)^{-2}\right]^{-1}$$
⇒ $$\left(\frac{-7}{11}\right)^{-3+5 x}=\left(\frac{-7}{11}\right)^2$$
⇒ -3 + 5x = 2
⇒ 5x = 2 + 3
⇒ 5x = 5
⇒ x = 1

(ii) $$\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left[\left(\frac{3}{7}\right)^{-1}\right]^{-7}$$
$$\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left[\left(\frac{3}{7}\right)^{-1}\right]^{-7}$$
⇒ $$\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left(\frac{3}{7}\right)^7$$
⇒ $$\left(\frac{3}{7}\right)^{-2 x+1}=\left(\frac{3}{7}\right)^7 \times\left(\frac{3}{7}\right)^{-1}$$
⇒ $$\left(\frac{3}{7}\right)^{-2 x+1}=\left(\frac{3}{7}\right)^6$$
⇒ -2x + 1 = 6
⇒ -2x = 6 – 1
⇒ -2x = 5
∴ x = $$\frac{-5}{2}$$

Question 16.
If $$\frac{p}{q}=\left(\frac{5}{6}\right)^{-2} \times\left(\frac{4}{3}\right)^0$$, find the value of $$\left(\frac{p}{q}\right)^{-2}$$

## DAV Class 7 Maths Chapter 4 HOTS

Question 1.
Express in exponential form.
$$\frac{256 \times 81}{64 \times 729}$$
$$\frac{256 \times 81}{64 \times 729}=\frac{(2)^8 \times(3)^4}{(2)^6 \times(3)^6}$$
= $$\frac{(2)^{8-6}}{(3)^{6-4}}=\frac{(2)^2}{(3)^2}$$
= $$\left(\frac{2}{3}\right)^2$$

Question 2.
If 22x-3= (64)x, find the value of x.
22x-3 = (64)x
⇒ 22x-3 = (26)x
⇒ 22x-3 = 26x
⇒ 2x – 3 = 6x
⇒ 2x – 6x = 3
⇒ -4x = 3
⇒ x = $$\frac{-3}{4}$$

## DAV Class 7 Maths Chapter 7 Enrichment Questions

Question 1.
Find the value of:
(i) $$\left(\frac{x}{y}\right)^a \times\left(\frac{y}{z}\right)^a \times\left(\frac{z}{x}\right)^a$$
$$\left(\frac{x}{y}\right)^a \times\left(\frac{y}{z}\right)^a \times\left(\frac{z}{x}\right)^a$$
= $$\left(\frac{x}{y} \times \frac{y}{z} \times \frac{z}{x}\right)^a$$
= (1)a
= 1

(ii) $$\left(\frac{x^a}{x^b}\right) \times\left(\frac{x^b}{x^c}\right) \times\left(\frac{x^c}{x^a}\right)$$
$$\left(\frac{x^a}{x^b}\right) \times\left(\frac{x^b}{x^c}\right) \times\left(\frac{x^c}{x^a}\right)$$
= (x)a-b × (x)b-c × (x)c-a
= (x)a-b+b-c+c-a
= (x)0
= x0
= 1

Question 1.
Evaluate the following:
(i) (125)$$\frac{2}{3}$$
(125)$$\frac{2}{3}$$
= (53)$$\frac{2}{3}$$
= 53×$$\frac{2}{3}$$
= 52
= 25

(ii) (512)$$\frac{-2}{9}$$
(512)$$\frac{-2}{9}$$
= (29)$$\frac{-2}{9}$$
= 29×$$\frac{-2}{9}$$
= 2-2
= $$\frac{1}{2^2}=\frac{1}{4}$$

(iii) (0.027)$$\frac{-2}{3}$$
(0.027)$$\frac{-2}{3}$$
= [(0.3)3]$$\frac{-2}{3}$$
= (0.3)3×$$\frac{-2}{3}$$
= $$\frac{1}{(0.3)^2}=\frac{1}{0.09}$$

Question 2.
Evaluate the following:
(i) $$\frac{\left(\frac{3}{2}\right)^{-5} \times\left(\frac{3}{2}\right)^4}{\left(\frac{3}{2}\right)^{-4} \div\left(\frac{3}{2}\right)^{-4}}$$

(ii) $$\left[6^{-2}+\left(\frac{3}{2}\right)^{-2}\right]^{-2}$$

(iii) $$\left(\frac{81}{16}\right)^{\frac{3}{4}} \times\left[\left(\frac{25}{9}\right)^{\frac{3}{2}} \div\left(\frac{5}{2}\right)^{-3}\right]$$

(iv) $$\left(\frac{64}{125}\right)^{\frac{-2}{3}} \div\left(\frac{256}{625}\right)^{\frac{-1}{4}}+\left(\frac{\sqrt{36}}{3 \sqrt{64}}\right)^0$$

Question 3.
If $$\frac{x}{y}=\left(\frac{3}{4}\right)^{-3} \div\left(\frac{4}{5}\right)^0$$, find $$\left(\frac{x}{y}\right)^{-2}$$

Question 4.
Evaluate: $$\frac{(32)^{\frac{2}{5}} \times(4)^{\frac{-1}{2}} \times(8)^{\frac{1}{3}}}{(2)^{-2} \div(64)^{\frac{-1}{3}}}$$

Question 5.
By what number should (-10)-2 be multiplied so that the quotient may be equal to (-2)-1?
Let x be the required number.
x × (-10)-2 = (-2)-1
⇒ x = (-2)-1 ÷ (-10)-2
⇒ x = $$\frac{-1}{2} \div\left(\frac{-1}{10}\right)^2$$
⇒ x = $$\frac{-1}{2}$$ × (-10)2
⇒ x = $$\frac{-1}{2}$$ × 100 = -50
Hence x = -50

Question 6.
Express $$\frac{1.2 \times 10^3}{2.4 \times 10^{-4}}$$ in the form K × 10n when 1 ≤ K < 10.
$$\frac{1.2 \times 10^3}{2.4 \times 10^{-4}}$$
= $$\frac{1}{2}$$ × 103+4
= 0.5 × 107

Question 7.
Express (5.2 × 10-3) × (3.1 × 106) in the form k × 10n where 1 ≤ k < 10 and n is an integer.
(5.2 × 10-3) × (3.1 × 106)
= 5.2 ×3.1 × 10-3+6
= 16.12 × 103
= 1.612 × 104

Question 8.
Prove that: $$\left(\frac{x^a}{x^b}\right)^{a+b} \cdot\left(\frac{x^b}{x^c}\right)^{b+c} \cdot\left(\frac{x^c}{x^a}\right)^{c+a}$$ = 1
$$\left(\frac{x^a}{x^b}\right)^{a+b} \cdot\left(\frac{x^b}{x^c}\right)^{b+c} \cdot\left(\frac{x^c}{x^a}\right)^{c+a}$$
= x(a – b)(a + b). x(b – c)(b + c). x(c – a)(c + a)
= xa2-b2. xb2-c2. xc2-a2
= xa2-b2+b2-c2+c2-a2
= x0
= 1
Hence proved

Question 9.
Find the value of x if 32. (72 + 42x) = 9$$\frac{9}{256}$$
32. (72 + 42x) = 9$$\frac{9}{256}$$
⇒ 9(1 + 42x) = $$\frac{2313}{256}$$
⇒ 1 + 42x = $$\frac{2313}{256 \times 9}$$
⇒ 42x = $$\frac{257}{256}$$ – 1
⇒ 42x = $$\frac{1}{256}$$
⇒ 42x = $$\frac{1}{(4)^4}$$
⇒ 42x = 4-4
⇒ 2x = -4
∴ x = -2
Hence x = -2

Question 10.
Solve for x: $$\sqrt[3]{\left(\frac{4}{5}\right)^{1-2 x}}=1 \frac{61}{64}$$

⇒ 1 – 2x = -9
⇒ -2x = -9 – 1
⇒ -2x = -10
∴ x = 5

Question 11.
If 1960 = 2a × 5b × 7c, find the value of a, b and c.
Find the product of the cube of $$\left(\frac{-2}{3}\right)$$ and the square of $$\left(\frac{-4}{5}\right)$$.
$$\left(\frac{-2}{3}\right)^3 \times\left(\frac{-4}{5}\right)^2$$
= $$\frac{-8}{27} \times \frac{16}{25}$$
= $$\frac{-128}{675}$$