Rs Aggarwal 2018 for Class 8 Math Chapter 1 - Rational Numbers

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Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 33:

Question 1:

Evaluate:
(i) 4⁻³
(ii) ${(\frac{1}{2})}^{- 5}$
(iii) ${(\frac{4}{3})}^{- 3}$
(iv) ${(- 3)}^{- 4}$
(v) ${(\frac{- 2}{3})}^{- 5}$

Answer:

(i) $4^{- 3} = \frac{1}{4^{3}} = \frac{1}{64}$

(ii) ${(\frac{1}{2})}^{- 5} = 2^{5} = 32$

(iii) ${(\frac{4}{3})}^{- 3} = {(\frac{3}{4})}^{3} = \frac{3^{3}}{4^{3}} = \frac{27}{64}$

(iv) ${(- 3)}^{- 4} = {(\frac{- 1}{3})}^{4} = \frac{(- 1)^{4}}{3^{4}} = \frac{1}{81}$

(v) ${(\frac{- 2}{3})}^{- 5} = {(\frac{- 3}{2})}^{5} = \frac{{(- 3)}^{5}}{2^{5}} = \frac{- 243}{32}$

Page No 33:

Question 2:

Evaluate:
(i) ${(\frac{5}{3})}^{2} \times {(\frac{5}{3})}^{2}$
(ii) ${(\frac{5}{6})}^{6} \times {(\frac{5}{6})}^{- 4}$
(iii) ${(\frac{2}{3})}^{- 3} \times {(\frac{2}{3})}^{- 2}$
(iv) ${(\frac{9}{8})}^{- 3} \times {(\frac{9}{8})}^{2}$

Answer:

(i) ${(\frac{5}{3})}^{2} \times {(\frac{5}{3})}^{2} = {(\frac{5}{3})}^{4} = \frac{5^{4}}{3^{4}} = \frac{625}{81}$

(ii) ${(\frac{5}{6})}^{6} \times {(\frac{5}{6})}^{- 4} = {(\frac{5}{6})}^{(6 + (- 4))} = {(\frac{5}{6})}^{(6 - 4)} = {(\frac{5}{6})}^{2} = \frac{5^{2}}{6^{2}} = \frac{25}{36}$

(iii) ${(\frac{2}{3})}^{- 3} \times {(\frac{2}{3})}^{- 2} = {(\frac{2}{3})}^{(- 3 - 2)} = {(\frac{2}{3})}^{- 5} = {(\frac{3}{2})}^{5} = \frac{3^{5}}{2^{5}} = \frac{243}{32}$

(iv) ${(\frac{9}{8})}^{- 3} \times {(\frac{9}{8})}^{2} = {(\frac{9}{8})}^{(- 3 + 2)} = {(\frac{9}{8})}^{- 1} = \frac{8}{9}$

Page No 33:

Question 3:

Evaluate:
(i) ${(\frac{5}{9})}^{- 2} \times {(\frac{3}{5})}^{- 3} \times {(\frac{3}{5})}^{0}$
(ii) ${(\frac{- 3}{5})}^{- 4} \times {(\frac{- 2}{5})}^{2}$
(iii) ${(\frac{- 2}{3})}^{- 3} \times {(\frac{- 2}{3})}^{- 2}$

Answer:

(i)

${(\frac{5}{9})}^{- 2} \times {(\frac{3}{5})}^{- 3} \times {(\frac{3}{5})}^{0} = {(\frac{5}{9})}^{- 2} \times {(\frac{3}{5})}^{- 3 + 0} = {(\frac{5}{9})}^{- 2} \times {(\frac{3}{5})}^{- 3} = {(\frac{9}{5})}^{2} \times {(\frac{5}{3})}^{3} = \frac{9^{2}}{5^{2}} \times \frac{5^{3}}{3^{3}} = \frac{{(3^{2})}^{2}}{5^{2}} \times \frac{5^{3}}{3^{3}} = \frac{3^{4}}{5^{2}} \times \frac{5^{3}}{3^{3}} = (3^{(4 - 3)}) \times (5^{(3 - 2)}) = 3 \times 5 = 15$

(ii)

${(\frac{- 3}{5})}^{- 4} \times {(\frac{- 2}{5})}^{2} = {(\frac{5}{- 3})}^{4} \times {(\frac{- 2}{5})}^{2} = \frac{5^{4}}{- 3^{4}} \times \frac{- 2^{2}}{5^{2}} = 5^{(4 - 2)} \times \frac{- 2^{2}}{- 3^{4}} = 5^{2} \times \frac{- 2^{2}}{- 3^{4}} = 25 \times \frac{4}{81} = \frac{100}{81}$

(iii)

${(\frac{- 2}{3})}^{- 3} \times {(\frac{- 2}{3})}^{- 2} = {(\frac{3}{- 2})}^{3} \times {(\frac{3}{- 2})}^{2} = \frac{3^{3}}{- 2^{3}} \times \frac{3^{2}}{- 2^{2}} = \frac{3^{(3 + 2)}}{- 2^{(3 + 2)}} = \frac{3^{5}}{- 2^{5}} = \frac{- 243}{32}$

Page No 33:

Question 4:

Evaluate:
(i) ${\{{(\frac{- 2}{3})}^{2}\}}^{- 2}$
(ii) ${[{\{{(\frac{- 1}{3})}^{2}\}}^{- 2}]}^{- 1}$
(iii) ${\{{(\frac{3}{2})}^{- 2}\}}^{2}$

Answer:

(i) ${\{{(\frac{- 2}{3})}^{2}\}}^{- 2} = {(\frac{- 2}{3})}^{2 \times (- 2)} = {(\frac{- 2}{3})}^{- 4} = {(\frac{3}{- 2})}^{4} = \frac{3^{4}}{(- 2)^{4}} = \frac{3^{4}}{2^{4}} = \frac{81}{16}$
(ii) ${[{\{{(\frac{- 1}{3})}^{2}\}}^{- 2}]}^{- 1} = {[{(\frac{- 1}{3})}^{2 \times (- 2)}]}^{- 1} = {[{(\frac{- 1}{3})}^{- 4}]}^{- 1} = {(\frac{- 1}{3})}^{- 4 \times - 1} = {(\frac{- 1}{3})}^{4} = \frac{- 1^{4}}{3^{4}} = \frac{1^{4}}{3^{4}} = \frac{1}{81}$
(iii) ${\{{(\frac{3}{2})}^{- 2}\}}^{2} = {(\frac{3}{2})}^{- 2 \times 2} = {(\frac{3}{2})}^{- 4} = {(\frac{2}{3})}^{4} = \frac{2^{4}}{3^{4}} = \frac{16}{81}$

Page No 33:

Question 5:

Evaluate $\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} .$

Answer:

$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = \{3^{3} - 2^{3}\} \div 4^{3} = \{27 - 8\} \div 64 = \frac{19}{64}$

Page No 33:

Question 6:

Evaluate ${\{{(\frac{4}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}\}}^{- 1} .$

Answer:

${\{{(\frac{4}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}\}}^{- 1} = {\{{(\frac{3}{4})}^{1} - {(\frac{4}{1})}^{1}\}}^{- 1} = {\{(\frac{3}{4}) - (\frac{4}{1})\}}^{- 1} The L . C . M . of 4 and 1 is 4 . ∴ {\{(\frac{3 \times 1}{4 \times 1}) - (\frac{4 \times 4}{1 \times 4})\}}^{- 1} = {\{\frac{3}{4} - \frac{16}{4}\}}^{- 1} = {\{\frac{3 - 16}{4}\}}^{- 1} = {\{\frac{- 13}{4}\}}^{- 1} = {\{\frac{4}{- 13}\}}^{1} = \frac{4}{- 13} = \frac{4 \times - 1}{- 13 \times - 1} = \frac{- 4}{13}$

Page No 33:

Question 7:

Evaluate $[{(5^{- 1} \times 3^{- 1})}^{- 1} \div 6^{- 1}] .$

Answer:

$[{(5^{- 1} \times 3^{- 1})}^{- 1} \div 6^{- 1}] = [{(\frac{1}{5} \times \frac{1}{3})}^{- 1} \div \frac{1}{6}] = [{(\frac{1}{15})}^{- 1} \div \frac{1}{6}] = [15 \times 6] = 90$

Page No 33:

Question 8:

Find the value of:
(i) (2⁰ + 3⁻¹) × 3²
(ii) (2⁻¹ × 3⁻¹) ÷ 2⁻³
(iii) ${(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}$

Answer:

(i)
$(2^{0} + 3^{- 1}) \times 3^{2} = (1 + \frac{1}{3}) \times 3^{2} (because 2^{0} = 1 and 3^{- 1} = \frac{1}{3}) = (\frac{1 \times 3}{1 \times 3} + \frac{1 \times 1}{3 \times 1}) \times 3^{2} = (\frac{3}{3} + \frac{1}{3}) \times 3^{2} = (\frac{4}{3}) \times 3^{2} = 4 \times 3^{(2 - 1)} = 4 \times 3 = 12$

(ii)

$(2^{- 1} \times 3^{- 1}) \div 2^{- 3} = (\frac{1}{2} \times \frac{1}{3}) \div {(\frac{1}{2})}^{3} (\frac{1}{6}) \div \frac{1^{3}}{2^{3}} = (\frac{1}{6}) \div (\frac{1}{8}) = \frac{1}{6} \times 8 = \frac{8}{6} = \frac{4}{3}$

(iii)

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

Page No 34:

Question 9:

Find the value of x for which ${(\frac{5}{3})}^{- 4} \times {(\frac{5}{3})}^{- 5} = {(\frac{5}{3})}^{3 x} .$

Answer:

Consider the left side:
${(\frac{5}{3})}^{- 4} \times {(\frac{5}{3})}^{- 5} = {(\frac{5}{3})}^{(- 4 + (- 5))} = {(\frac{5}{3})}^{- 9}$

Given:
${(\frac{5}{3})}^{- 9} = {(\frac{5}{3})}^{3 x}$
Comparing the powers:
$- 9 = 3 x \Rightarrow x = - 3$

Page No 34:

Question 10:

Find the value of x for which ${(\frac{4}{9})}^{4} \times {(\frac{4}{9})}^{- 7} = {(\frac{4}{9})}^{2 x - 1} .$

Answer:

Given:
${(\frac{4}{9})}^{4} \times {(\frac{4}{9})}^{- 7} = {(\frac{4}{9})}^{2 x - 1}$

$∴$ ${(\frac{4}{9})}^{(4 - 7)} = {(\frac{4}{9})}^{- 3} = {(\frac{4}{9})}^{2 x - 1}$

$\Rightarrow 2 x - 1 = - 3 2 x = - 3 + 1 = - 2 \Rightarrow x = - 1$

Page No 34:

Question 11:

By what number should (−6)⁻¹ be multiplied so that the product becomes 9⁻¹?

Answer:

Let the required number be $x$ .

$∴$ $x \times (- 6)^{- 1} = 9^{- 1}$

$x \times \frac{1}{- 6} = \frac{1}{9} \Rightarrow \frac{x}{- 6} = \frac{1}{9} or x = \frac{- 6}{9}$
The greatest common divisor for the numerator and the denominator is 3.

$∴ x = \frac{- 6}{9} = \frac{(- 6) \div 3}{9 \div 3} = \frac{- 2}{3}$

Page No 34:

Question 12:

By what number should ${(\frac{- 2}{3})}^{- 3}$ be divided so that the quotient may be ${(\frac{4}{27})}^{- 2}$ ?

Answer:

Let the number be $x$ .

$∴$ ${(\frac{- 2}{3})}^{- 3} \div x = {(\frac{4}{27})}^{- 2}$
$\Rightarrow {(\frac{3}{- 2})}^{3} \div x = {(\frac{27}{4})}^{2} \Rightarrow {(\frac{- 3}{2})}^{3} \div x = {(\frac{27}{4})}^{2} \Rightarrow {(\frac{- 3}{2})}^{3} \times \frac{1}{x} = {(\frac{27}{4})}^{2} \Rightarrow \frac{- 3^{3}}{2^{3}} \times \frac{1}{x} = \frac{27^{2}}{4^{2}} \Rightarrow \frac{- 27}{8} \times \frac{1}{x} = \frac{27^{2}}{4^{2}} = \frac{27 \times 27}{4 \times 4} = \frac{27 \times 27}{4 \times 2 \times 2} = \frac{27 \times 27}{8 \times 2} ∴ \frac{1}{x} = \frac{(\frac{27 \times 27}{8 \times 2})}{(\frac{- 27}{8})} \Rightarrow x = \frac{(\frac{- 27}{8})}{(\frac{27 \times 27}{8 \times 2})} = (\frac{- 27}{8}) \times (\frac{8 \times 2}{27 \times 27}) = \frac{- 2}{27}$

Page No 34:

Question 13:

If 5^2x^{+ 1} ÷ 25 = 125, find the value of x.

Answer:

Given:
$5^{2 x + 1} \div 25 = 125$
$We know : 25 = 5 \times 5 = 5^{2} 125 = 5 \times 5 \times 5 = 5^{3} ∴ \frac{5^{2 x + 1}}{5^{2}} = 5^{3} \Rightarrow 5^{[(2 x + 1) - 2]} = 5^{3} or 5^{[(2 x + 1) - 2]} = 5^{[2 x - 1]} = 5^{3} \Rightarrow 2 x - 1 = 3 2 x = 3 + 1 = 4 x = \frac{4}{2} = 2$

$∴$ $x = 2$

Page No 36:

Question 1:

Write each of the following numbers in standard form:
(i) 57.36
(ii) 3500000
(iii) 273000
(iv) 168000000
(v) 4630000000000
(vi) 345 × 10⁵

Answer:

(i) $57.36 = 5.736 \times 10$
(ii) $3500000 = 3.5 \times 10^{6}$
(iii) $273000 = 2.73 \times 10^{5}$
(iv) $168000000 = 1.68 \times 10^{8}$
(v) $4630000000000 = 4.63 \times 10^{12}$
(vi) $345 \times 10^{5} = 3.45 \times 10^{7}$

Page No 36:

Question 2:

Write each of the following numbers in usual form:
(i) 3.74 × 10⁵
(ii) 6.912 × 10⁸
(iii) 4.1253 × 10⁷
(iv) 2.5 × 10⁴
(v) 5.17 × 10⁶
(vi) 1.679 × 10⁹

Answer:

(i) $3.74 \times 10^{5} = \frac{374}{100} \times 10^{5} = \frac{374 \times 10^{5}}{10^{2}} = 374 \times 10^{(5 - 2)} = 374 \times 10^{3} = 374000$
(ii) $6.912 \times 10^{8} = \frac{6912}{1000} \times 10^{8} = \frac{6912 \times 10^{8}}{10^{3}} = 6912 \times 10^{(8 - 3)} = 6912 \times 10^{5} = 691200000$
(iii) $4.1253 \times 10^{7} = \frac{41253}{10000} \times 10^{7} = \frac{41253 \times 10^{7}}{10^{4}} = 41253 \times 10^{(7 - 4)} = 41253 \times 10^{3} = 41253000$
(iv) $2.5 \times 10^{4} = \frac{25}{10} \times 10^{4} = \frac{25 \times 10^{4}}{10} = 25 \times 10^{(4 - 1)} = 25 \times 10^{3} = 25000$
(v) $5.17 \times 10^{6} = \frac{517}{100} \times 10^{6} = \frac{517 \times 10^{6}}{10^{2}} = 517 \times 10^{(6 - 2)} = 517 \times 10^{4} = 5170000$
(vi) $1.679 \times 10^{9} = \frac{1679}{1000} \times 10^{9} = \frac{1679 \times 10^{9}}{10^{3}} = 1679 \times 10^{(9 - 3)} = 1679 \times 10^{6} = 1679000000$

Page No 36:

Question 3:

(i) The height of Mount Everest is 8848 m. Write it in standard form.
(ii) The speed of light is 300000000 m/sec. Express it in standard form.
(iii) The distance from the earth to the sun is 149600000000 m. Write it in standard form.

Answer:

(i) The height of the Mount Everest is 8848 m.
In standard form, we have:
$8848 = 8.848 \times 1000 m = 8.848 \times 10^{3} m$ .
(ii) The speed of light is 300000000 m/s.
In standard form, we have:
$300000000 = 3 \times 100000000 m / s = 3 \times 10^{8} m / s$
(iii) The Sun $-$ Earth distance is 149600000000 m.
In standard form, we have: $149600000000 = 1496 \times 100000000 = 1.496 \times 1000 \times 100000000 = 1.496 \times 10^{3} \times 10^{8} = 1.496 \times 10^{11} m$

Page No 36:

Question 4:

Mass of earth is (5.97 × 10²⁴) kg and mass of moon is (7.35 × 10²²) kg. What is the total mass of the two?

Answer:

Mass of the Earth = $5.97 \times 10^{24} kg$
Now, $5.97 \times 10^{24} = 5.97 \times 10^{(2 + 22)} = 5.97 \times 10^{2} \times 10^{22} = 597 \times 10^{22}$
So, the mass of the Earth can also be written as $597 \times 10^{22} kg$ .

Mass of the Moon = $7.35 \times 10^{22} kg$
Sum of the masses of the Earth and the Moon: $= (597 \times 10^{22}) + (7.35 \times 10^{22}) = (597 + 7.35) \times 10^{22} = 604.35 \times 10^{22} kg = 6.0435 \times 100 \times 10^{22} = 6.0435 \times 10^{2} \times 10^{22} = 6.0435 \times 10^{(2 + 22)} = 6.0435 \times 10^{24} kg$

Page No 36:

Question 5:

Write each of the following numbers in standard form:
(i) 0.0006
(ii) 0.00000083
(iii) 0.0000000534
(iv) 0.0027
(v) 0.00000165
(vi) 0.00000000689

Answer:

(i) $0.0006 = \frac{6}{10^{4}} = 6 \times 10^{- 4}$

(ii) $0.00000083 = \frac{83}{10^{8}} = \frac{8.3 \times 10}{10^{8}} = 8.3 \times 10^{(1 - 8)} = 8.3 \times 10^{- 7}$

(iii) $0.0000000534 = \frac{534}{10^{10}} = \frac{5.34 \times 10^{2}}{10^{10}} = 5.34 \times 10^{(2 - 10)} = 5.34 \times 10^{- 8}$

(iv) $0.0027 = \frac{27}{10^{4}} = \frac{2.7 \times 10}{10^{4}} = 2.7 \times 10^{(1 - 4)} = 2.7 \times 10^{- 3}$

(v) 0.00000165 = $\frac{165}{10^{8}} = \frac{1.65 \times 10^{2}}{10^{8}} = 1.65 \times 10$ ^(2-8) = 1.65×10^-6

(vi) 0.00000000689 = $\frac{689}{10^{11}} = \frac{6.89 \times 10^{2}}{10^{11}} = 6.89 \times 10$ ^(2-11)= 6.89×10^-9

Page No 36:

Question 6:

(i) 1 micron = $\frac{1}{1000000}$ m. Express it in standard form.
(ii) Size of a bacteria = 0.0000004 m. Express it in standard form.
(iii) Thickness of a paper = 0.03 mm. Express it in standard form.

Answer:

(i) $1 micron = \frac{1}{1000000} m = 1 \times 10^{- 6} m$

(ii) $0.0000004 m = \frac{4}{10^{7}} m = (4 \times 10^{- 7}) m$
(iii) $Thickness of paper = 0.03 mm = \frac{3}{10^{2}} mm = (3 \times 10^{- 2}) mm$

Page No 36:

Question 7:

Write each of the following numbers in usual form:
(i) 2.06 × 10⁻⁵
(ii) 5 × 10⁻⁷
(iii) 6.82 × 10⁻⁶
(iv) 5.673 × 10⁻⁴
(v) 1.8 × 10⁻²
(vi) 4.129 × 10⁻³

Answer:

(i) $2.06 \times 10^{- 5} = \frac{206}{100} \times \frac{1}{10^{5}} = \frac{206}{10^{2} \times 10^{5}} = \frac{206}{10^{(5 + 2)}} = \frac{206}{10^{7}} = \frac{206}{10000000} = 0.0000206$

(ii) $5 \times 10^{- 7} = \frac{5}{10^{7}} = \frac{5}{10000000} = 0.0000005$

(iii) $6.82 \times 10^{- 6} = \frac{682}{100} \times \frac{1}{10^{6}} = \frac{682}{10^{2} \times 10^{6}} = \frac{682}{10^{(2 + 6)}} = \frac{682}{10^{8}} = \frac{682}{100000000} = 0.00000682$

(iv) $5.673 \times 10^{- 4} = \frac{5673}{1000} \times \frac{1}{10^{4}} = \frac{5673}{10^{3} \times 10^{4}} = \frac{5673}{10^{(3 + 4)}} = \frac{5673}{10^{7}} = \frac{5673}{10000000} = 0.0005673$

(v) $1.8 \times 10^{- 2} = \frac{18}{10} \times \frac{1}{10^{2}} = \frac{18}{10 \times 10^{2}} = \frac{18}{10^{(1 + 2)}} = \frac{18}{10^{3}} = \frac{18}{1000} = 0.018$

(vi) $4.129 \times 10^{- 3} = \frac{4129}{1000} \times \frac{1}{10^{3}} = \frac{4129}{10^{3} \times 10^{3}} = \frac{4129}{10^{(3 + 3)}} = \frac{4129}{10^{6}} = \frac{4129}{1000000} = 0.004129$

Page No 37:

Question 1:

Tick (✓) the correct answer
The value of ${(\frac{2}{5})}^{- 3}$ is
(a) $- \frac{8}{125}$
(b) $\frac{25}{4}$
(c) $\frac{125}{8}$
(d) $- \frac{2}{5}$

Answer:

Page No 37:

Question 2:

Tick (✓) the correct answer
The value of (−3)⁻⁴ is
(a) 12
(b) 81
(c) $- \frac{1}{12}$
(d) $\frac{1}{81}$

Answer:

(d) $\frac{1}{81}$

${(- 3)}^{- 4} = \frac{1}{{(- 3)}^{4}} = \frac{1}{{(- 1)}^{4} \times {(3)}^{4}} = \frac{1}{{(3)}^{4}} = \frac{1}{81}$

Page No 37:

Question 3:

Tick (✓) the correct answer
The value of (−2)⁻⁵ is
(a) −32
(b) $\frac{- 1}{32}$
(c) 32
(d) $\frac{1}{32}$

Answer:

(b) $\frac{- 1}{32}$

${(- 2)}^{- 5} = \frac{1}{{(- 2)}^{5}} = \frac{1}{- 32} = \frac{1 \times (- 1)}{- 32 \times (- 1)} = \frac{- 1}{32}$

Page No 37:

Question 4:

Tick (✓) the correct answer
(2⁻⁵ ÷ 2⁻²) = ?
(a) $\frac{1}{128}$
(b) $\frac{- 1}{128}$
(c) $- \frac{1}{8}$
(d) $\frac{1}{8}$

Answer:

(d) $\frac{1}{8}$

$(2^{- 5} \div 2^{- 2}) = (\frac{1}{2^{5}} \div \frac{1}{2^{2}}) = (\frac{1}{32} \div \frac{1}{4}) = (\frac{1}{32} \times 4) = \frac{4}{32} = \frac{1}{8}$

Page No 37:

Question 5:

Tick (✓) the correct answer
The value of (3⁻¹ + 4⁻¹)⁻¹ ÷ 5⁻¹ is
(a) $\frac{7}{10}$
(b) $\frac{60}{7}$
(c) $\frac{7}{5}$
(d) $\frac{7}{15}$

Answer:

(b) $\frac{60}{7}$

${(3^{- 1} + 4^{- 1})}^{- 1} \div 5^{- 1} = {(\frac{1}{3} + \frac{1}{4})}^{- 1} \div \frac{1}{5} = {(\frac{4 + 3}{12})}^{- 1} \div \frac{1}{5} = {(\frac{7}{12})}^{- 1} \div \frac{1}{5} = (\frac{12}{7}) \div \frac{1}{5} = \frac{12}{7} \times 5 = \frac{60}{7}$

Page No 37:

Question 6:

Tick (✓) the correct answer
${(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2} = ?$
(a) $\frac{61}{144}$
(b) $\frac{144}{61}$
(c) 29
(d) $\frac{1}{29}$

Answer:

(c) 29

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

Page No 37:

Question 7:

Tick (✓) the correct answer
$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = ?$
(a) $\frac{19}{64}$
(b) $\frac{27}{16}$
(c) $\frac{64}{19}$
(d) $\frac{16}{25}$

Answer:

(a) $\frac{19}{64}$

$\{{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}\} \div {(\frac{1}{4})}^{- 3} = \{3^{3} - 2^{3}\} \div 4^{3} = \{27 - 8\} \div 64 = 19 \div 64 = \frac{19}{64}$

Page No 37:

Question 8:

Tick (✓) the correct answer
${[{\{{(- \frac{1}{2})}^{2}\}}^{- 2}]}^{- 1} = ?$
(a) $\frac{1}{16}$
(b) 16
(c) $- \frac{1}{16}$
(d) −16

Answer:

(a) $\frac{1}{16}$

${[{\{{(- \frac{1}{2})}^{2}\}}^{- 2}]}^{- 1} = {[{\{- \frac{1}{2}\}}^{- 4}]}^{- 1} = {(- \frac{1}{2})}^{(- 4 \times - 1)} = {(- \frac{1}{2})}^{4} = \frac{1}{16}$

Page No 37:

Question 9:

Tick (✓) the correct answer
The value of x for which ${(\frac{7}{12})}^{- 4} \times {(\frac{7}{12})}^{3 x} = {(\frac{7}{12})}^{5},$ is
(a) −1
(b) 1
(c) 2
(d) 3

Answer:

(d) 3

${(\frac{7}{12})}^{- 4} \times {(\frac{7}{12})}^{3 x} = {(\frac{7}{12})}^{5} \Rightarrow {(\frac{7}{12})}^{- 4 + 3 x} = {(\frac{7}{12})}^{5} \Rightarrow 3 x - 4 = 5 3 x = 9 or x = \frac{9}{3} = 3$

Page No 37:

Question 10:

Tick (✓) the correct answer
If (2^3x^{− 1} + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2

Answer:

(d) 2

$(2^{3 x - 1} + 10) \div 7 = 6 \Rightarrow \frac{(2^{3 x - 1} + 10)}{7} = \frac{6}{1} On cross multiplying : (2^{3 x - 1} + 10) \times 1 = 6 \times 7 = 42$

⇒ 2^3x-1 = 42 -10
⇒2^3x-1 = 32
⇒ 2^3x-1 = 2⁵
⇒ 3x-1 = 5
⇒ 3x = 6
Therefore, x = 2

Page No 37:

Question 11:

Tick (✓) the correct answer
${(\frac{2}{3})}^{0} = ?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 1
(d) 0

Answer:

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Question 12:

Tick (✓) the correct answer
${(\frac{- 5}{3})}^{- 1} = ?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{- 3}{5}$
(d) none of these

Answer:

(c) $\frac{- 3}{5}$

${(\frac{- 5}{3})}^{- 1} = {(\frac{3}{- 5})}^{1} = \frac{3}{- 5} = \frac{3 \times (- 1)}{- 5 \times (- 1)} = \frac{- 3}{5}$

Page No 38:

Question 13:

Tick (✓) the correct answer
${(\frac{- 1}{2})}^{3} = ?$
(a) $\frac{- 1}{6}$
(b) $\frac{1}{6}$
(c) $\frac{1}{8}$
(d) $\frac{- 1}{8}$

Answer:

(d) $\frac{- 1}{8}$

${(\frac{- 1}{2})}^{3} = \frac{- 1^{3}}{2^{3}} = \frac{- 1}{8}$

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Question 14:

Tick (✓) the correct answer
${(\frac{- 3}{4})}^{2} = ?$
(a) $\frac{- 9}{16}$
(b) $\frac{9}{16}$
(c) $\frac{16}{9}$
(d) $\frac{- 16}{9}$

Answer:

(b) $\frac{9}{16}$

${(\frac{- 3}{4})}^{2} = \frac{{(- 3)}^{2}}{{(4)}^{2}} = \frac{9}{16}$

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Question 15:

Tick (✓) the correct answer
3670000 in standard form is
(a) 367 × 10⁴
(b) 36.7 × 10⁵
(c) 3.67 × 10⁶
(d) none of these

Answer:

(c) $3.67 \times 10^{6}$

$3670000 = 367 \times 10^{4} = 3.67 \times 100 \times 10^{4} = 3.67 \times 10^{2} \times 10^{4} = 3.67 \times 10^{(2 + 4)} = 3.67 \times 10^{6}$

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Question 16:

Tick (✓) the correct answer
0.0000463 in standard form is
(a) 463 × 10⁻⁷
(b) 4.63 × 10⁻⁵
(c) 4.63 × 10⁻⁹
(d) 46.3 × 10⁻⁶

Answer:

(b) $4.63 \times 10^{- 5}$

$0.0000463 = \frac{463}{10^{7}} = \frac{4.63 \times 10^{2}}{10^{7}} = 4.63 \times 10^{(2 - 7)} = 4.63 \times 10^{- 5}$

Page No 38:

Question 17:

Tick (✓) the correct answer
0.000367 × 10⁴ in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367

Answer:

(a) 3.67

$0.000367 \times 10^{4} = \frac{367}{10^{6}} \times 10^{4} = 367 \times 10^{(4 - 6)} = 367 \times 10^{- 2} = \frac{367}{10^{2}} = \frac{367}{100} = 3.67$

Page No 39:

Question 1:

Evaluate:
(i) 3⁻⁴
(ii) (−4)³
(iii) ${(\frac{3}{4})}^{- 2}$
(iv) ${(\frac{- 2}{3})}^{- 5}$
(v) ${(\frac{5}{7})}^{0}$

Answer:

(i) $3^{- 4} = \frac{1}{3^{4}} = \frac{1}{81}$

(ii) ${(- 4)}^{3} = {(- 1)}^{3} \times {(4)}^{3} = - 1 \times 64 = - 64$

(iii) ${(\frac{3}{4})}^{- 2} = {(\frac{4}{3})}^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$

(iv) ${(\frac{- 2}{3})}^{- 5} = {(\frac{3}{- 2})}^{5} = \frac{3^{5}}{- 2^{5}} = \frac{243}{- 32} = \frac{243 \times - 1}{- 32 \times - 1} = \frac{- 243}{32}$

(v) Using the property ${(\frac{a}{b})}^{0} = 1$ , we have:
${(\frac{5}{7})}^{0} = 1$

Page No 39:

Question 2:

Evaluate: ${\{{(\frac{- 2}{3})}^{3}\}}^{- 2} .$

Answer:

${\{{(\frac{- 2}{3})}^{3}\}}^{- 2} = {(\frac{- 2}{3})}^{- 6} = {(\frac{3}{- 2})}^{6} = \frac{3^{6}}{- 2^{6}} = \frac{729}{64}$

Page No 39:

Question 3:

Simplify: $(3^{- 1} + 6^{- 1}) \div {(\frac{3}{4})}^{- 1} .$

Answer:

$(3^{- 1} + 6^{- 1}) \div {(\frac{3}{4})}^{- 1} = (\frac{1}{3} + \frac{1}{6}) \div {(\frac{4}{3})}^{1} = ([\frac{1 \times 2}{3 \times 2}] + [\frac{1 \times 1}{6 \times 1}]) \div (\frac{4}{3}) = (\frac{2 + 1}{6}) \div (\frac{4}{3}) = (\frac{3}{6}) \div (\frac{4}{3}) = (\frac{1}{2}) \div (\frac{4}{3}) = (\frac{1}{2}) \times (\frac{3}{4}) = \frac{3}{8}$

$∴ (3^{- 1} + 6^{- 1}) \div {(\frac{3}{4})}^{- 1} = \frac{3}{8}$

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Question 4:

By what number should ${(\frac{- 2}{3})}^{- 3}$ be divided so that the quotient is ${(\frac{4}{9})}^{- 2}$ ?

Answer:

Let the number be $x$ .

$∴$ ${(\frac{- 2}{3})}^{- 3} \div x = {(\frac{4}{9})}^{- 2}$

$\Rightarrow {(\frac{3}{- 2})}^{3} \div x = {(\frac{9}{4})}^{2} \Rightarrow \frac{{(\frac{3}{- 2})}^{3}}{x} = {(\frac{9}{4})}^{2} \Rightarrow \frac{\frac{3^{3}}{- 2^{3}}}{x} = \frac{9^{2}}{4^{2}} \Rightarrow x = \frac{(\frac{3^{3}}{- 2^{3}})}{(\frac{9^{2}}{4^{2}})} = \frac{(\frac{3^{3}}{- 2^{3}})}{(\frac{{(3^{2})}^{2}}{{(2^{2})}^{2}})} = (\frac{3^{3}}{- 2^{3}}) \times (\frac{{(2^{2})}^{2}}{{(3^{2})}^{2}}) = (\frac{3^{3}}{- 2^{3}}) \times (\frac{2^{4}}{3^{4}}) = (\frac{3^{3}}{- 2^{3}}) \times (\frac{2^{3}}{3^{3}}) \times (\frac{2^{1}}{3^{1}}) \Rightarrow (\frac{1}{- 1}) \times (\frac{2^{1}}{3^{1}}) = \frac{2}{- 3} = \frac{2 \times - 1}{- 3 \times - 1} = \frac{- 2}{3}$

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Question 5:

By what number should (−3)⁻¹ be multiplied so that the product becomes 6⁻¹?

Answer:

Let the number be $x$ .

$∴$ ${(- 3)}^{- 1} \times x = {(6)}^{- 1}$
$\Rightarrow \frac{1}{- 3} \times x = \frac{1}{6} \Rightarrow \frac{1 \times - 1}{- 3 \times - 1} \times x = \frac{1}{6} ∴ \frac{- x}{3} = \frac{1}{6} On cross multiplying : - x \times 6 = 1 \times 3 \Rightarrow - 6 x = 3 \Rightarrow 6 x = - 3 ∴ x = \frac{- 3}{6} = \frac{- 1}{2}$

Page No 39:

Question 6:

Express each of the following in standard form:
(i) 345
(ii) 180000
(iii) 0.000003
(iv) 0.000027

Answer:

(i) $345 = 3.45 \times 100 = 3.45 \times 10^{2}$

(ii) $180000 = 18 \times 10000 = 18 \times 10^{4} = 1.8 \times 10 \times 10^{4} = 1.8 \times 10^{(1 + 4)} = 1.8 \times 10^{5}$

(iii) $0.000003 = \frac{3}{1000000} = 3 \times 10^{- 6}$

(iv) $0.000027 = \frac{27}{100000} = \frac{27}{10^{6}} = \frac{2.7 \times 10}{10^{6}} = 2.7 \times 10^{(1 - 6)} = 2.7 \times 10^{- 5}$

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Question 7:

Mark (✓) against the correct answer
The value of (−3)⁻³ is
(a) −27
(b) 9
(c) $\frac{- 1}{27}$
(d) $\frac{1}{27}$

Answer:

(c) $\frac{- 1}{27}$

${(- 3)}^{- 3} = {(\frac{1}{- 3})}^{3} = \frac{1^{3}}{- 3^{3}} = \frac{1}{- 27} = \frac{1 \times - 1}{- 27 \times - 1} = \frac{- 1}{27}$

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Question 8:

Mark (✓) against the correct answer
The value of ${(\frac{3}{4})}^{- 3}$ is
(a) $\frac{- 27}{64}$
(b) $\frac{64}{27}$
(c) $\frac{- 9}{4}$
(d) $\frac{27}{64}$

Answer:

(b) $\frac{64}{27}$

${(\frac{3}{4})}^{- 3} = {(\frac{4}{3})}^{3} = \frac{4^{3}}{3^{3}} = \frac{64}{27}$

Page No 39:

Question 9:

Mark (✓) against the correct answer
(3⁻⁶ ÷ 3⁴) = ?
(a) 3⁻²
(b) 3²
(c) 3⁻¹⁰
(d) 3¹⁰

Answer:

(c) $3^{- 10}$

$(3^{- 6} \div 3^{4}) = (\frac{1}{3^{6}} \div 3^{4}) = \frac{1}{3^{6}} \times \frac{1}{3^{4}} = \frac{1}{3^{(6 + 4)}} = \frac{1}{3^{10}} = 3^{- 10}$

Page No 39:

Question 10:

Mark (✓) against the correct answer
If ${(\frac{5}{12})}^{- 4} \times {(\frac{5}{12})}^{3 x} = {(\frac{5}{12})}^{5}, then x = ?$
(a) −1
(b) 1
(c) 2
(d) 3

Answer:

(d) 3

${(\frac{5}{12})}^{- 4} \times {(\frac{5}{12})}^{3 x} = {(\frac{5}{12})}^{5} \Rightarrow {(\frac{5}{12})}^{- 4 + 3 x} = {(\frac{5}{12})}^{5} \Rightarrow - 4 + 3 x = 5 \Rightarrow 3 x = 5 + 4 = 9 \Rightarrow x = \frac{9}{3} = 3$

Page No 39:

Question 11:

Mark (✓) against the correct answer
${(\frac{3}{5})}^{0} = ?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) 1
(d) 0

Answer:

Page No 39:

Question 12:

Mark (✓) against the correct answer
${(\frac{- 6}{5})}^{- 1} = ?$
(a) $\frac{6}{5}$
(b) $\frac{- 6}{5}$
(c) $\frac{5}{6}$
(d) $\frac{- 5}{6}$

Answer:

(d) $(\frac{- 5}{6})$

${(\frac{- 6}{5})}^{- 1} = {(\frac{5}{- 6})}^{1} = \frac{5}{- 6} = \frac{5 \times - 1}{- 6 \times - 1} = \frac{- 5}{6}$

Page No 39:

Question 13:

Mark (✓) against the correct answer
${(\frac{- 1}{3})}^{3} = ?$
(a) $\frac{- 1}{9}$
(b) $\frac{1}{9}$
(c) $\frac{- 1}{27}$
(d) $\frac{1}{27}$

Answer:

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Question 14:

Fill in the blanks.
(i) 360000 written in standard form is .........
(ii) 0.0000123 written in standard form is .........
(iii) ${(\frac{- 2}{3})}^{- 2} = . . . . . . . . .$
(iv) 3 × 10⁻³ in usual form is .........
(v) 5.32 × 10⁻⁴ in usual form is .........

Answer:

(i) The standard form of 36000 is $3.6 \times 10^{5}$ .
$360000 = 36 \times 10^{4} = 3.6 \times 10 \times 10^{4} = 3.6 \times 10^{(1 + 4)} = 3.6 \times 10^{5}$

(ii) The standard form of 0.0000123 is $1.23 \times 10^{- 5}$ .
$0.0000123 = \frac{123}{10000000} = \frac{123}{10^{7}} = \frac{1.23 \times 100}{10^{7}} = \frac{1.23 \times 10^{2}}{10^{7}} = 1.23 \times 10^{(2 - 7)} = 1.23 \times 10^{- 5}$

(iii) ${(\frac{- 2}{3})}^{- 2} = \frac{9}{4}$
${(\frac{- 2}{3})}^{- 2} = {(\frac{3}{- 2})}^{2} = \frac{3^{2}}{- 2^{2}} = \frac{9}{4}$

(iv) $The usual form of 3 \times 10^{- 3} is 0.003 .$
$3 \times 10^{- 3} = \frac{3}{10^{3}} = \frac{3}{1000} = 0.003$

(v) $The usual form of 5.32 \times 10^{- 4} is 0.000532 .$
$5.32 \times 10^{- 4} = \frac{5.32}{10^{4}} = \frac{5.32}{10000} = 0.000532$

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