Practice MCQ Questions and Answer on Square Root and Cube Root

21.

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

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

Show Answer

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

22.

The cube root of .000216 is:

• (A) .6
• (B) .06
• (C) 77
• (D) 87

Show Answer

 $$\eqalign{ & {\left( {.000216} \right)^{\frac{1}{3}}} = {\left( {\frac{{216}}{{{{10}^6}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{6 \times 6 \times 6}}{{{{10}^2} \times {{10}^2} \times {{10}^2}}}} \right)^{\frac{1}{3}}} \cr & = \frac{6}{{{{10}^2}}} \cr & = \frac{6}{{100}} \cr & = 0.06 \cr} $$

23.

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

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

Show Answer

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

24.

1250 oranges were distributed among a group of girls of a class. Each girl got twice as many oranges as the number of girls in that group. The number of girls in the group was = ?

• (A) 25
• (B) 45
• (C) 50
• (D) 100

Show Answer

 Let the number of girls in the group be x Then, number of oranges given to each girl = 2x $$\eqalign{ & \therefore x \times 2x = 1250 \cr & \Leftrightarrow 2{x^2} = 1250 \cr & \Leftrightarrow {x^2} = 625 \cr & \Leftrightarrow x = 25 \cr} $$

25.

Which number can replace both the question marks in the equation ?
$$\frac{4\frac{1}{2}}{?}=\frac{?}{32}$$

• (A) 1
• (B) 7
• (C) $$7\frac{1}{2}$$
• (D) None of these

Show Answer

 $$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{{4\frac{1}{2}}}{x} = \frac{x}{{32}} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = 32 \times \frac{9}{2} \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$

26.

$$\sqrt{\frac{16}{25}}\times \sqrt{\frac{?}{25}}\times \frac{16}{25}=\frac{256}{625}$$

• (A) 5
• (B) 8
• (C) 16
• (D) None of these

Show Answer

 $$\eqalign{ & {\text{Let}}, \cr & \sqrt {\frac{{16}}{{25}}} \times \sqrt {\frac{x}{{25}}} \times \frac{{16}}{{25}} = \frac{{256}}{{625}} \cr & {\text{Then}}, \cr & \frac{4}{5} \times \frac{{\sqrt x }}{5} \times \frac{{16}}{{25}} = \frac{{256}}{{625}} \cr & \Leftrightarrow \frac{{64\sqrt x }}{{625}} = \frac{{256}}{{625}} \cr & \Leftrightarrow \sqrt x = \frac{{256}}{{625}} \times \frac{{625}}{{64}} \cr & \Leftrightarrow x = {4^2} \cr & \Leftrightarrow x = 16 \cr} $$

27.

$$\sqrt{\frac{0.081\times 0.484}{0.0064\times 6.25}}$$   is equal to ?

• (A) 0.9
• (B) 0.99
• (C) 9
• (D) 99

Show Answer

 Sum of decimal places in the numerator and denominator under the radical sign being the same, we remove the decimal. ∴ Given expression, $$\eqalign{ & = \sqrt {\frac{{0.081 \times 0.484}}{{0.0064 \times 6.25}}} \cr & = \sqrt {\frac{{81 \times 484}}{{64 \times 625}}} \cr & = \frac{{9 \times 22}}{{8 \times 25}} \cr & = 0.99 \cr} $$

28.

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.

29.

The square root of $$\text{0}\text{.}\bar{\text{4}}$$  is ?

• (A) $$0.\bar{6}$$
• (B) $$0.\bar{7}$$
• (C) $$0.\bar{8}$$
• (D) $$0.\bar{9}$$

Show Answer

 $$\eqalign{ & = \sqrt {{\text{0}}{\text{.}}\overline {\text{4}} {\text{ }}} \cr & = \sqrt {\frac{4}{9}} \cr & = \frac{2}{3} \cr & = 0.666...... \cr & = 0.\overline 6 \cr} $$

30.

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

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