Practice MCQ Questions and Answer on Square Root and Cube Root

1.

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 $$\eqalign{ & {\text{Thus, he was born in }} \cr & = \left( {1849 - 43} \right) \cr & = 1806 \cr} $$

2.

The least number of 4 digits which is a perfect square is = ?

• (A) 1000
• (B) 1016
• (C) 1024
• (D) 1036

Show Answer

 Least number of 4 digits is 1000 $$\eqalign{ & \,\,\,\,3|\overline {10} \,\,\overline {00} \,\,(31 \cr & \,\,\,\,\,\,\,|\,\,\,\,9 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & 61|\,\,\,\,\,1\,00 \cr & \,\,\,\,\,\,\,|\,\,\,\,\,\,61 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,39 \cr & \therefore {\left( {31} \right)^2} 1000 {\left( {32} \right)^2} \cr & {\text{Hence, required number}} \cr & = {\left( {32} \right)^2} \cr & = 1024 \cr} $$

3.

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   $${5^2} \times {3^2} \times 2,i.e.,450$$

4.

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

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

Show Answer

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

5.

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 $$\eqalign{ & = \frac{2}{{{{10}^6}}} \cr & = 0.000002 \cr} $$

6.

$$\sqrt[3]{\sqrt{0.000064}}=?$$

• (A) 0.02
• (B) 0.2
• (C) 2
• (D) None of these

Show Answer

 $$\eqalign{ & = \sqrt {0.000064} \cr & = \sqrt {\frac{{64}}{{{{10}^6}}}} \cr & = \frac{8}{{{{10}^3}}} \cr & = \frac{8}{{1000}} \cr & = 0.008 \cr & \therefore \root 3 \of {\sqrt {0.000064} } \cr & = \root 3 \of {0.008} \cr & = \root 3 \of {\frac{8}{{1000}}} \cr & = \frac{2}{{10}} \cr & = 0.2 \cr} $$

7.

One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels ?

• (A) 32
• (B) 34
• (C) 35
• (D) 36

Show Answer

 Let the total number of camels be x Then, $$\eqalign{ & \Leftrightarrow x - \left( {\frac{x}{4} + 2\sqrt x } \right) = 15 \cr & \Leftrightarrow \frac{{3x}}{4} - 2\sqrt x = 15 \cr & \Leftrightarrow 3x - 8\sqrt x = 60 \cr & \Leftrightarrow 8\sqrt x = 3x - 60 \cr & \Leftrightarrow 64x = {\left( {3x - 60} \right)^2} \cr & \Leftrightarrow 64x = 9{x^2} + 3600 - 360x \cr & \Leftrightarrow 9{x^2} - 424x + 3600 = 0 \cr & \Leftrightarrow 9{x^2} - 324x - 100x + 3600 = 0 \cr & \Leftrightarrow 9x\left( {x - 36} \right) - 100\left( {x - 36} \right) = 0 \cr & \Leftrightarrow \left( {x - 36} \right)\left( {9x - 100} \right) = 0 \cr & \Leftrightarrow x = 36\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because x \ne \frac{{100}}{9}} \right] \cr} $$

8.

The least number by which 294 must be multiplied to make it a perfect square, is = ?

• (A) 2
• (B) 3
• (C) 6
• (D) 24

Show Answer

 294 = 7 × 7 × 2 × 3 To make it a perfect square, it must be multiplied by 2 × 3 i.e.,6 ∴ Required number = 6

9.

If $$2\ast 3=\sqrt{13}$$   and 3 * 4 = 5, then the value of 5 * 12 is ?

• (A) $$\sqrt{17}$$
• (B) $$\sqrt{29}$$
• (C) 12
• (D) 13

Show Answer

 $$\eqalign{ & {\text{Clearly, }}a*b = \sqrt {{a^2} + {b^2}} \cr & \therefore 5*12 \cr & = \sqrt {{5^2} + {{12}^2}} \cr & = \sqrt {25 + 144} \cr & = \sqrt {169} \cr & = 13 \cr} $$

10.

Determined the value of $$\frac{1}{\sqrt{1}+\sqrt{2}}\text{ + }$$  $$\frac{1}{\sqrt{2}+\sqrt{3}}+$$   $$\frac{1}{\sqrt{3}+\sqrt{4}}+$$   $$......+$$   $$\frac{1}{\sqrt{120}+\sqrt{121}}\text{ = ?}$$

• (A) 8
• (B) 10
• (C) $$\sqrt{120}$$
• (D) $$12\sqrt{2}$$

Show Answer

 Given expressing, $$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$ $$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$ $$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$ $$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$ $$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$ $$ = - 1 + \sqrt {121} $$ $$ = - 1 + 11$$ $$ = 10$$