Practice MCQ Questions and Answer on Square Root and Cube Root

81.

A man born in the first half of the nineteenth century was x years old in the year x². He was born in ?

• (A) 1806
• (B) 1812
• (C) 1825
• (D) 1836

Show Answer

 Clearly, the man was born between 1800 and 1850 is 1849. $${\text{And, }}1849 = {\left( {43} \right)^2}$$ So, the man was 43 years old in 1849 $$\eqalign{ & {\text{Thus, he was born in }} \cr & = \left( {1849 - 43} \right) \cr & = 1806 \cr} $$

82.

What is the least number which should be subtracted 0.000326 in order to make it a perfect square = ?

• (A) 0.000002
• (B) 0.000004
• (C) 0.02
• (D) 0.04

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 $$\eqalign{ & 0.000326 = \frac{{326}}{{{{10}^6}}} \cr & \,\,\,\,1|\overline 3 \,\,\overline {26} \,\,(18 \cr & \,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & 28|\,\,\,2\,26 \cr & \,\,\,\,\,\,\,|\,\,\,2\,24 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr} $$ ∴ Required number to be subtracted $$\eqalign{ & = \frac{2}{{{{10}^6}}} \cr & = 0.000002 \cr} $$

83.

Given that $$\sqrt{13}=3.605$$   and $$\sqrt{130}=11.40\text{,}$$   find the value of $$\sqrt{1.30}$$  $$+$$ $$\sqrt{1300}$$  $$+$$ $$\sqrt{0.0130}$$   = ?

• (A) 36.164
• (B) 36.304
• (C) 37.164
• (D) 37.304

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 $$\eqalign{ & {\text{Given expression,}} \cr & \sqrt {1.30} + \sqrt {1300} + \sqrt {0.0130} \cr & = \sqrt {\frac{{130}}{{100}}} + \sqrt {13 \times 100} + \sqrt {\frac{{130}}{{10000}}} \cr & = \frac{{\sqrt {130} }}{{10}} + \sqrt {13} \times 10 + \frac{{\sqrt {130} }}{{100}} \cr & = \frac{{11.40}}{{10}} + 3.605 \times 10 + \frac{{11.40}}{{100}} \cr & = 1.14 + 36.05 + 0.114 \cr & = 37.304 \cr} $$

84.

Value of $$\sqrt{0.01}\times \sqrt[3]{0.008}-0.02$$     is = ?

• (A) 0
• (B) 1
• (C) 2
• (D) 3

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 $$\eqalign{ & = \sqrt {0.01} \times \root 3 \of {0.008} - 0.02{\text{ }} \cr & = \sqrt {{{\left( {0.1} \right)}^2}} \times \root 3 \of {{{\left( {0.2} \right)}^3}} - 0.02 \cr & = 0.1 \times 0.2 - 0.02 \cr & = 0.02 - 0.02 \cr & = 0 \cr} $$

85.

The square root of $$\left(7+3\sqrt{5}\right)$$  $$\left(7-3\sqrt{5}\right)$$   is ?

• (A) $$\sqrt{5}$$
• (B) 2
• (C) 4
• (D) $$3\sqrt{5}$$

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 $$\eqalign{ & = \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr & = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr & = \sqrt {49 - 45} \cr & = \sqrt 4 \cr & = 2 \cr} $$

86.

The square root of $$\text{0}\text{.}\bar{\text{4}}$$  is ?

• (A) $$0.\bar{6}$$
• (B) $$0.\bar{7}$$
• (C) $$0.\bar{8}$$
• (D) $$0.\bar{9}$$

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 $$\eqalign{ & = \sqrt {{\text{0}}{\text{.}}\overline {\text{4}} {\text{ }}} \cr & = \sqrt {\frac{4}{9}} \cr & = \frac{2}{3} \cr & = 0.666...... \cr & = 0.\overline 6 \cr} $$

87.

$$\sqrt{110.25}\times \sqrt{0.01}÷$$    $$\sqrt{0.0025}$$   $$-$$ $$\sqrt{420.25}$$  equals ?

• (A) 0.50
• (B) 0.64
• (C) 0.73
• (D) 0.75

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 Given expression, $$ = \sqrt {\frac{{11025}}{{100}}} \times \sqrt {\frac{1}{{100}}} \, \div \,$$    $$\sqrt {\frac{{25}}{{10000}}} \, - \,$$ $$\,\sqrt {\frac{{42025}}{{100}}} $$ $$\eqalign{ & = \frac{{105}}{{10}} \times \frac{1}{{10}} \div \frac{5}{{100}} - \frac{{205}}{{10}} \cr & = \frac{{105}}{{100}} \times \frac{{100}}{5} - \frac{{205}}{{10}} \cr & = 21 - \frac{{205}}{{10}} \cr & = \frac{{210 - 205}}{{10}} \cr & = \frac{5}{{10}} \cr & = \frac{1}{2} \cr & = 0.50 \cr} $$

88.

Which number should replace both the question marks in the following equation ?
$$\frac{?}{1776}=\frac{111}{?}$$

• (A) 343
• (B) 414
• (C) 644
• (D) 543

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}\frac{x}{{1776}} = \frac{{111}}{x} \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 111 \times 1776 \cr & \Leftrightarrow {x^2} = 111 \times 111 \times 16 \cr & \Leftrightarrow x = \sqrt {{{\left( {111} \right)}^2} \times {{\left( 4 \right)}^2}} \cr & \Leftrightarrow x = 111 \times 4 \cr & \Leftrightarrow x = 444 \cr} $$

89.

The number of digits in the square root of 625685746009 is = ?

• (A) 4
• (B) 5
• (C) 6
• (D) 7

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 The number of digits of the square root of a perfect square number of n digits is $$\eqalign{ & {\text{(i)}}\frac{n}{2}{\text{, if n is even}} \cr & {\text{(ii)}}\frac{{n + 1}}{2}{\text{, if n is odd}} \cr & {\text{Here, }}n = 12 \cr & {\text{So, required number of digits}} \cr & = \frac{n}{2} \cr & = \frac{{12}}{2} \cr & = 6{\text{ }} \cr} $$

90.

What is the smallest number by which 3600 be divided to make it a perfect cube ?

• (A) 9
• (B) 50
• (C) 300
• (D) 450

Show Answer

 $$3600 = {2^3} \times {5^2} \times {3^2} \times 2$$ To make it a perfect cube, it must be divided by   $${5^2} \times {3^2} \times 2,i.e.,450$$