Practice MCQ Questions and Answer on Square Root and Cube Root

1.

$$1728 \div \root 3 \of {262144} \, \times \,?\, - 288$$      = 4491

• (A) 148
• (B) 156
• (C) 173
• (D) 177

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 $$\eqalign{ & = 262144 \cr & = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \cr & = {8^6} \cr & \therefore \root 3 \of {262144} \cr & = {8^2} \cr & = 64 \cr & {\text{Let, }} \cr & {\text{1728}} \div \root 3 \of {262144} \times x - 288 = 4491 \cr & {\text{Then,}} \cr & \Leftrightarrow 1728 \div 64 \times x - 288 = 4491 \cr & \Leftrightarrow 27x = 4779 \cr & \Leftrightarrow x = \frac{{4779}}{{27}} \cr & \Leftrightarrow x = 177 \cr} $$ $$\eqalign{ & 8|262144 \cr & - - - - - - - - \cr & 8|\,\,\,32768 \cr & - - - - - - - - \cr & 8|\,\,\,\,4096 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,512 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,64 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,\,8 \cr & - - - - - - - - \cr} $$

2.

If a = 0.1039, then the value of $$\sqrt {4{a^2} - 4a + 1} + 3a$$     is:

• (A) 0.1039
• (B) 0.2078
• (C) 1.1039
• (D) 2.1039

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 $$\eqalign{ & \sqrt {4{a^2} - 4a + 1} + 3a \cr & = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2a} \right)}^2} - 2 \times 1 \times 2a} + 3a \cr & = \sqrt {{{\left( {1 - 2a} \right)}^2}} + 3a \cr & = \left( {1 - 2a} \right) + 3a \cr & = \left( {1 + a} \right) \cr & = \left( {1 + 0.1039} \right) \cr & = 1.1039 \cr} $$

3.

Given $$\sqrt 5 = 2.2361,$$   $$\sqrt 3 = 1.7321{\text{,}}$$   then $$\frac{1}{{\sqrt 5 - \sqrt 3 }}$$   is equal to ?

• (A) 1.98
• (B) 1.984
• (C) 1.9841
• (D) 2

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 $$\eqalign{ & \Rightarrow \frac{1}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)}} \cr & = \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} \cr & = \frac{{\left( {2.2361 + 1.7321} \right)}}{2} \cr & = \frac{{3.9682}}{2} \cr & = 1.9841{\text{ }} \cr} $$

4.

$$\sqrt {176 + \sqrt {2401} } $$    is equal to = ?

• (A) 14
• (B) 15
• (C) 18
• (D) 24

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 $$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {176 + 49} \cr & = \sqrt {225} \cr & = 15 \cr} $$

5.

The value of $$\frac{{1 + \sqrt {0.01} }}{{1 - \sqrt {0.1} }}$$   is close to = ?

• (A) 0.6
• (B) 1.1
• (C) 1.6
• (D) 1.7

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 $$\eqalign{ & = \frac{{1 + \sqrt {0.01} }}{{1 - \sqrt {0.1} }} \cr & = \frac{{1 + 0.1}}{{1 - 0.316}} \cr & = \frac{{1.1}}{{0.684}} \cr & = \frac{{1100}}{{684}} \cr & = 1.6 \cr & \,\,\,\,\,\,\,3|\overline 0 \,.\,\overline {10} \,\,\overline {00} \,\,\overline {00} \,(0.316 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,61|\,\,\,\,\,\,\,\,\,\,100 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,61 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 626\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,3900 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,3756 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr} $$

6.

The value of $${\text{ }}\root 3 \of {\frac{{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}}{{0.4 \times 0.4 \times 0.4 + 0.08 \times 0.08 \times 0.08}}} $$        is ?

• (A) 0.125
• (B) 0.25
• (C) 0.5
• (D) 0.75

Show Answer

 $$\eqalign{ & = \root 3 \of {\frac{{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}}{{0.4 \times 0.4 \times 0.4 + 0.08 \times 0.08 \times 0.08}}} \cr & = \root 3 \of {\frac{{0.008 + 0.000064}}{{0.064 + 0.000512}}} \cr & = \root 3 \of {\frac{{0.008064}}{{0.064512}}} \cr & = \root 3 \of {\frac{{8064}}{{64512}}} \cr & = \root 3 \of {\frac{1}{8}} \cr & = \frac{1}{2} \cr & = 0.5 \cr} $$

7.

A mobile company offered to pay the Indian Cricket Team as much money per run scored by the side as the total number it gets in a one-dayer against Australia. Which one of the following cannot be the total amount to be spent by the company in this deal ?

• (A) 21904
• (B) 56169
• (C) 101761
• (D) 121108

Show Answer

 Clearly, the required number must be a perfect square. Since a number having 8 as the unit's digit cannot be a perfect square, so 121108 is not a perfect square.

8.

The square root of $$\left( {{{272}^2} - {{128}^2}} \right)$$   is = ?

• (A) 144
• (B) 200
• (C) 240
• (D) 256

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 $$\eqalign{ & = \sqrt {\left( {{{272}^2} - {{128}^2}} \right)} \cr & = \sqrt {\left( {272 + 128} \right)\left( {272 - 128} \right)} \cr & = \sqrt {400 \times 144} \cr & = \sqrt {57600} \cr & = 240 \cr} $$

9.

Given $$\sqrt 2 = 1.414.$$   Then the value of $$\sqrt 8 $$  $$ + $$ $$2\sqrt {32} $$  $$ - $$ $$3\sqrt {128} $$  $$ + $$ $$4\sqrt {50} $$   is = ?

• (A) 8.426
• (B) 8.484
• (C) 8.526
• (D) 8.876

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 Given expression, $$\sqrt {4 \times 2} + 2\sqrt {16 \times 2} - 3\sqrt {64 \times 2} $$       $$ + $$ $$4\sqrt {25 \times 2} $$ $$\eqalign{ & = 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & = 6\sqrt 2 \cr & = 6 \times 1.414 \cr & = 8.484 \cr} $$

10.

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

Show Answer

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