Practice MCQ Questions and Answer on Square Root and Cube Root

51.

The approximate value of $$\frac{{3\sqrt {12} }}{{2\sqrt {28} }}$$  $$ \div $$ $$\frac{{2\sqrt {21} }}{{\sqrt {98} }}$$   is ?

• (A) 1.0605
• (B) 1.0727
• (C) 1.6007
• (D) 1.6026

Show Answer

 $$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3\sqrt {12} }}{{2\sqrt {28} }} \times \frac{{\sqrt {98} }}{{2\sqrt {21} }} \cr & = \frac{{3\sqrt {4 \times 3} }}{{2\sqrt {4 \times 7} }} \times \frac{{\sqrt {49 \times 2} }}{{2\sqrt {21} }} \cr & = \frac{{6\sqrt 3 }}{{4\sqrt 7 }} \times \frac{{7\sqrt 2 }}{{2\sqrt {21} }} \cr & = \frac{{21\sqrt 6 }}{{4\sqrt {7 \times 21} }} \cr & = \frac{{21\sqrt 6 }}{{28\sqrt 3 }} \cr & = \frac{3}{4}\sqrt 2 \cr & = \frac{3}{4} \times 1.414 \cr & = 3 \times 0.3535 \cr & = 1.0605 \cr} $$

52.

The least perfect square number divisible by 3, 4, 5, 6 and 8 is = ?

• (A) 900
• (B) 1200
• (C) 2500
• (D) 3600

Show Answer

 L.C.M. of 3, 4, 5, 6, 8 is 120 Now 120 = 2 × 2 × 2 × 3 × 5 To make it a perfect square, it must be multiplied by 2 × 3 × 5 So, required number $$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$

53.

$${\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right)^2}\,{\text{simplifies}}\,{\text{to:}}$$

• (A) $$\frac{3}{4}$$
• (B) $$\frac{4}{{\sqrt 3 }}$$
• (C) $$\frac{4}{3}$$
• (D) None of these

Show Answer

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

54.

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

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

55.

If $$a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$$   and $$b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}, $$   the value of $$\left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right)$$   is ?

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

Show Answer

 $$\eqalign{ & {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr & {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr & \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr & \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr & \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = \frac{{28}}{4} \cr & \,\,\,\,\,\,\, = 7 \cr & {\text{Also, }} \cr & ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr & \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = 1 \cr & \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr & \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr & \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr & \,\,\,\,\,\,\, = \frac{8}{6} \cr & \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$

56.

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) $$3\sqrt 5 $$

Show Answer

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

57.

The square root of 123454321 is = ?

• (A) 111111
• (B) 12341
• (C) 11111
• (D) 11211

Show Answer

 $$\therefore \sqrt {123454321} = 11111$$

58.

If the product of four consecutive natural numbers increased by a natural number p, is a perfect square, then the value of p is = ?

• (A) 1
• (B) 2
• (C) 4
• (D) 8

Show Answer

 $$\eqalign{ & {\text{We have,}} \cr & 1 \times 2 \times 3 \times 4 = 24 \cr & {\text{And }}24 + 1 = 25\left[ {25 = {5^2}} \right] \cr & 2 \times 3 \times 4 \times 5 = 120{\text{ }} \cr & {\text{And 1}}20 + 1 = 121\left[ {121 = {{11}^2}} \right] \cr & 3 \times 4 \times 5 \times 6 = 360{\text{ }} \cr & {\text{And }}360 + 1 = 361\left[ {361 = {{19}^2}} \right] \cr & 4 \times 5 \times 6 \times 7 = 840{\text{ }} \cr & {\text{And }}840 + 1 = 841\left[ {841 = {{29}^2}} \right] \cr & \therefore p = 1 \cr} $$

59.

A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

• (A) 57
• (B) 67
• (C) 77
• (D) 87

Show Answer

 Money collected = (59.29 x 100) paise = 5929 paise. ∴ Number of members = $$\sqrt {5929} $$   = 77

60.

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

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

Show Answer

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