Practice MCQ Questions and Answer on Square Root and Cube Root

101.

The square root of 64009 is:

• (A) 253
• (B) 347
• (C) 363
• (D) 803

Show Answer

 $$\eqalign{ & \,\,\,\,\,\,\,\,2|\overline 6 \overline {40} \overline {09} \,(\,\,253 \cr & \,\,\,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,45\,|240 \cr & \,\,\,\,\,\,\,\,\,\,\,|225 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & 503\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \therefore \sqrt {64009} = 253 \cr} $$

102.

What is the least number to be added to 7700 to make it a perfect square ?

• (A) 77
• (B) 98
• (C) 131
• (D) 221

Show Answer

Answer & Solution Answer: Option E Solution: $$\eqalign{ & \,\,\,\,\,8|\overline {77} \,\,\overline {00} \,\,(87 \cr & \,\,\,\,\,\,\,\,\,|\,\,64 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & 167|\,\,\,13\,00 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,11\,69 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,31 \cr & \therefore {\text{Number to be added}} \cr & = {\left( {88} \right)^2} - 7700 \cr & = 7744 - 7700 \cr & = 44 \cr} $$

103.

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

• (A) 14
• (B) 15
• (C) 18
• (D) 24

Show Answer

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

104.

What is $$\frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2\sqrt {20} - \sqrt {32} + \sqrt {50} }}$$      equal to ?

• (A) 5
• (B) $$5\sqrt 2 $$
• (C) $$5\sqrt 5 $$
• (D) $$\sqrt 5 $$

Show Answer

 $$\eqalign{ & {\text{Given,}} \cr & \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2\sqrt {20} - \sqrt {32} + \sqrt {50} }}{\text{ }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2 \times 2\sqrt 5 - 2 \times 2\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 4\sqrt 5 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{\sqrt 5 + \sqrt 2 }} \cr & = \frac{{\sqrt 5 \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\sqrt 5 + \sqrt 2 }} \cr & = \sqrt 5 \cr} $$

105.

How many perfect squares lie between 120 and 300 ?

• (A) 5
• (B) 6
• (C) 7
• (D) 8

Show Answer

 $$\eqalign{ & {\left( {11} \right)^2} = 121{\text{ }} \cr & {\text{And }} \cr & {\left( {17} \right)^2} = 289 \cr} $$ So, the perfect squares between 120 and 300 are the squares of numbers from 11 to 17. Clearly, these are 7 in number.

106.

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

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

Show Answer

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

107.

$$\sqrt {\sqrt {17956} + \sqrt {24025} } = ?$$

• (A) 19
• (B) 155
• (C) 256
• (D) 289

Show Answer

Answer & Solution Answer: Option E Solution: $$\eqalign{ & \sqrt {\sqrt {17956} + \sqrt {24025} } \cr & = \sqrt {134 + 155} \cr & = \sqrt {289} \cr & = 17 \cr} $$

108.

If $$a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }},$$   $$b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$   then the value of $${a^2} + {b^2}$$   would be = ?

• (A) 10
• (B) 98
• (C) 99
• (D) 100

Show Answer

 $$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$

109.

If 1537* is a perfect square, then the digit which replace * is = ?

• (A) 2
• (B) 4
• (C) 5
• (D) 6

Show Answer

 $$\eqalign{ & {\text{Let the missing digit be x}} \cr & \,\,\,\,\,\,\,1|\overline 1 \,\,\overline {53} \,\,\overline {7{\text{x}}} \,\,(124 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,22|\,\,\,\,\,\,\,53 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 244\,|\,\,\,\,\,\,\,\,\,\,\,\,\,97{\text{x}} \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,976 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & {\text{Then, x}} = 6 \cr} $$

110.

If $$\sqrt 2 = 1.414{\text{,}}$$   the square root of $$\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$$   is nearest to = ?

• (A) 0.172
• (B) 0.414
• (C) 0.586
• (D) 1.414

Show Answer

 $$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$