Practice MCQ Questions and Answer on Square Root and Cube Root

81.

Which of the following is closest to $$\sqrt 3 = \,?$$

• (A) 1.69
• (B) $$\frac{{173}}{{100}}$$
• (C) 1.75
• (D) $$\frac{9}{5}$$

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 $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 3 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.732 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,27|\,\,2\,\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,1\,\,\,89 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,343\,|\,\,\,\,\,\,\,\,\,11\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,10\,\,29 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 3492\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,71\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,69\,\,84 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,16 \cr} $$ $$\eqalign{ & \therefore \sqrt 3 \cr & = 1.73 \cr & = \frac{{173}}{{100}} \cr} $$

82.

Solved $$\root 4 \of {{{\left( {625} \right)}^3}} = ?$$

• (A) $$\root 3 \of {1875} $$
• (B) 25
• (C) 125
• (D) None of these

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 $$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$

83.

If $$\sqrt x \div \sqrt {441} = 0.02{\text{,}}$$    then the value of x is ?

• (A) 0.1764
• (B) 1.764
• (C) 1.64
• (D) 2.64

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 $$\eqalign{ & \Leftrightarrow \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02 \cr & \Leftrightarrow \frac{{\sqrt x }}{{21}} = 0.02 \cr & \Leftrightarrow \sqrt x = 0.02 \times 21 \cr & \Leftrightarrow \sqrt x = 0.42 \cr & \Leftrightarrow x = {\left( {0.42} \right)^2} \cr & \Leftrightarrow x = 0.1764{\text{ }} \cr} $$

84.

What should come in place of both the question marks in the equation ?
$$\frac{?}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{?}$$

• (A) 12
• (B) 14
• (C) 144
• (D) 196

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 $$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr & \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr & \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$

85.

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

86.

If $$3\sqrt 5 + \sqrt {125} $$   = 17.88, then what will be the value of $$\sqrt {80} + 6\sqrt 5 \,\,?$$

• (A) 13.41
• (B) 20.46
• (C) 21.66
• (D) 22.35

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 $$\eqalign{ & 3\sqrt 5 + \sqrt {125} = 17.88 \cr & \Rightarrow 3\sqrt 5 + \sqrt {25 \times 5} = 17.88 \cr & \Rightarrow 3\sqrt 5 + 5\sqrt 5 = 17.88 \cr & \Rightarrow 8\sqrt 5 = 17.88 \cr & \Rightarrow \sqrt 5 = 2.235 \cr & \therefore \sqrt {80} + 6\sqrt 5 = \sqrt {16 \times 5} + 6\sqrt 5 \cr & = 4\sqrt 5 + 6\sqrt 5 \cr & = 10\sqrt 5 = \left( {10 \times 2.235} \right) = 22.35 \cr} $$

87.

The value of $$\sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {0.003} \right)}^2} + {{\left( {0.021} \right)}^2} + {{\left( {0.0065} \right)}^2}}}} $$       is ?

• (A) 0.1
• (B) 10
• (C) $${10^2}$$
• (D) $${10^3}$$

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 $$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {\frac{{0.03}}{{10}}} \right)}^2} + {{\left( {\frac{{0.21}}{{10}}} \right)}^2} + {{\left( {\frac{{0.065}}{{10}}} \right)}^2}}}} \cr & = \sqrt {\frac{{100\left[ {{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}} \right]}}{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}} \cr & = \sqrt {100} \cr & = 10 \cr} $$

88.

$${\left[ {{{\left( {\sqrt {81} } \right)}^2}} \right]^2} = {\left( ? \right)^2}$$

• (A) 8
• (B) 9
• (C) 4096
• (D) 6561

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let,}} \cr & {\left[ {{{\left( {\sqrt {81} } \right)}^2}} \right]^2} = {\left( x \right)^2} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = {\left( {81} \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$

89.

The square root of 41209 is equal to = ?

• (A) 103
• (B) 203
• (C) 303
• (D) 403

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 $$\eqalign{ & \,\,\,\,\,2|4\,\,\overline {12} \,\,\overline {09} \,(203 \cr & \,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,40|\,\,\,\,\,\,\,\,12 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,12\,09 \cr & 403|\,\,\,\,\,\,\,\,\,12\,09 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \, \cr & \therefore \sqrt {41209} = 203 \cr} $$

90.

What should come in place of both x in the equation $$\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}$$

• (A) 12
• (B) 14
• (C) 144
• (D) 196

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 $$\eqalign{ & {\text{Let}}\,\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then}}\,{x^2} = \sqrt {128 \times 162} \cr & = \sqrt {64 \times 2 \times 18 \times 9} \cr & = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & = 8 \times 6 \times 3 \cr & = 144 \cr & \therefore x = \sqrt {144} = 12 \cr} $$