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
Answer:
(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
Answer:
(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
Answer:
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
Answer:
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) \(\frac{1}{2}\)
Answer:
\(\left(\frac{3}{5}\right)^{2 x}\) = 1
⇒ \(\left(\frac{3}{5}\right)^{2 x}=\left(\frac{3}{5}\right)^0\)
⇒ 2x =1
⇒ x = 0
Hence, (b) is the correct answer.

B. Answer the following questions:
(i) Simplify \(\left[\left\{\left(\frac{4}{9}\right)^2\right\}^0\right]^5\)
Answer:
\(\left[\left\{\left(\frac{4}{9}\right)^2\right\}^0\right]^5=\left[\left\{\frac{16}{81}\right\}^0\right]^5\)
= [1]
= 1

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

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

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

(v) Write in usual form 11.2 × (10)^-7.
Answer:
= 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) \(\left(\frac{1}{9}\right)^{-4}\)
Answer:
\(\left(\frac{1}{9}\right)^{-4}\)
Base = \(\frac{1}{9}\)
and exponent = -4

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

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

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

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

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

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

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

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

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

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

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

Question 4.
Express the following in the exponential form:
(i) (2.5)² ÷ \(\left(\frac{1}{2.5}\right)^2\)
Answer:
(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
Answer:
(5² × 5⁵) ÷ 5¹⁰
= 5²⁺⁵ ÷ 5¹⁰
= 5⁷ ÷ 5¹⁰
= 5^7-10
= 5^-3

Question 5.
Find the reciprocals of:
(i) \(\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3\)
Answer:
\(\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3\)
= \(\left(\frac{2}{3}\right)^{-2} \div\left(\frac{2}{3}\right)^3\)
∴ 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) \(\left(\frac{3}{4}\right)^{-2} \times\left(\frac{3}{4}\right)^3\)
Answer:
\(\left(\frac{3}{4}\right)^{-2} \times\left(\frac{3}{4}\right)^3\)
= \(\left(\frac{3}{4}\right)^{-2+3}=\frac{3}{4}\)
∴ Reciprocal of \(\frac{3}{4}\)
= \(\frac{1}{3 / 4}=\frac{4}{3}\)

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

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

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

Question 6.
Express (15³)^-16 as a single exponent of 15.
Solution:
(15³)¹⁶ = 15^{3 × (-16)}
= (15)^-48

Question 7.
By what number should (7)^-2 be multiplied so that the product may be equal to (545T1?
Answer:
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 \(\left(\frac{-64}{27}\right)^{-1}\)?
Answer:
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 \(\left(\frac{12}{15}\right)^{-4}\)?
Answer:
Let the required number be x

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

(ii) 2.9 × 10^-10
Answer:
2.9 × 10^-10 = 2.9 × \(\frac{1}{10000000000}\)
= 0.00000000029

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

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

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

(vi) 6.00005 × 10⁹
Answer:
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
Answer:
762850 = 7.62850 × 10⁵

(ii) 2500000
Answer:
2500000 = 2.5 × 10⁶

(iii) 0.09
Answer:
0.09 = 9.0 × 10^-2

(iv) 0.0000076
Answer:
0.0000076 = 7.6 × 10^-6

(v) 0.0000008
Answer:
0.00000008 = 8.0 × 10^-8

(vi) 4592000000000
Answer:
459200000000 = 4.592 × 10¹¹

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

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

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

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

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

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

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

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.
Answer:
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.
Answer:
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.
Answer:
5.5 millions = 5500000 = 5.5 × 10⁶

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

(ii) \(\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left[\left(\frac{3}{7}\right)^{-1}\right]^{-7}\)
Answer:
\(\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left[\left(\frac{3}{7}\right)^{-1}\right]^{-7}\)
⇒ \(\left(\frac{3}{7}\right)^{-2 x+1} \div\left(\frac{3}{7}\right)^{-1}=\left(\frac{3}{7}\right)^7\)
⇒ \(\left(\frac{3}{7}\right)^{-2 x+1}=\left(\frac{3}{7}\right)^7 \times\left(\frac{3}{7}\right)^{-1}\)
⇒ \(\left(\frac{3}{7}\right)^{-2 x+1}=\left(\frac{3}{7}\right)^6\)
⇒ -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 \(\left(\frac{p}{q}\right)^{-2}\)
Answer:

DAV Class 7 Maths Chapter 4 HOTS

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

Question 2.
If 22^x-3= (64)^x, find the value of x.
Answer:
2^2x-3 = (64)^x
⇒ 2^2x-3 = (2⁶)^x
⇒ 2^2x-3 = 2^6x
⇒ 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) \(\left(\frac{x}{y}\right)^a \times\left(\frac{y}{z}\right)^a \times\left(\frac{z}{x}\right)^a\)
Answer:
\(\left(\frac{x}{y}\right)^a \times\left(\frac{y}{z}\right)^a \times\left(\frac{z}{x}\right)^a\)
= \(\left(\frac{x}{y} \times \frac{y}{z} \times \frac{z}{x}\right)^a\)
= (1)^a
= 1

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

Additional Questions

Question 1.
Evaluate the following:
(i) (125)^{\(\frac{2}{3}\)}
Answer:
(125)^{\(\frac{2}{3}\)}
= (5³)^{\(\frac{2}{3}\)}
= 5^{3×\(\frac{2}{3}\)}
= 5²
= 25

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

(iii) (0.027)^{\(\frac{-2}{3}\)}
Answer:
(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) \(\frac{\left(\frac{3}{2}\right)^{-5} \times\left(\frac{3}{2}\right)^4}{\left(\frac{3}{2}\right)^{-4} \div\left(\frac{3}{2}\right)^{-4}}\)
Answer:

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

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

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

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

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

Question 5.
By what number should (-10)^-2 be multiplied so that the quotient may be equal to (-2)^-1?
Answer:
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.
Answer:
\(\frac{1.2 \times 10^3}{2.4 \times 10^{-4}}\)
= \(\frac{1}{2}\) × 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.
Answer:
(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
Answer:
\(\left(\frac{x^a}{x^b}\right)^{a+b} \cdot\left(\frac{x^b}{x^c}\right)^{b+c} \cdot\left(\frac{x^c}{x^a}\right)^{c+a}\)
= x^{(a – b)(a + b)}. x^{(b – c)(b + c)}. x^{(c – a)(c + a)}
= x^a2-b2. x^b2-c2. x^c2-a2
= x^{a2-b2+b2-c2+c2-a2}
= x⁰
= 1
Hence proved

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

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

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

Question 11.
If 1960 = 2^a × 5^b × 7^c, find the value of a, b and c.
Answer:
1960 = 2^a × 5^b × 7^c
⇒ 2 × 2 × 2 × 5 × 7 × 7 = 2^a × 5^b × 7^c
⇒ 2³ × 5¹ × 7² = 2^a × 5^b × 7C
⇒ Equating the powers of corresponding numbers,
a = 3, b = 1 and c = 2.

Question 12.
Find the product of the cube of \(\left(\frac{-2}{3}\right)\) and the square of \(\left(\frac{-4}{5}\right)\).
Answer:
\(\left(\frac{-2}{3}\right)^3 \times\left(\frac{-4}{5}\right)^2\)
= \(\frac{-8}{27} \times \frac{16}{25}\)
= \(\frac{-128}{675}\)