Practice MCQ Questions and Answer on Square Root and Cube Root

21.

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

• (A) 0.9
• (B) 0.99
• (C) 9
• (D) 99

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

22.

Which smallest number must be added to 710 so that the sum is a perfect cube ?

• (A) 11
• (B) 19
• (C) 21
• (D) 29

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 $$\eqalign{ & {\text{Required number to be added}} \cr & = {9^3} - 710 \cr & = 729 - 710 \cr & = 19 \cr} $$

23.

$${\left[{\left(\sqrt{81}\right)}^2\right]}^2={(?)}^2$$

• (A) 8
• (B) 9
• (C) 4096
• (D) 6561

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let,}} \cr & {\left[ {{{\left( {\sqrt {81} } \right)}^2}} \right]^2} = {\left( x \right)^2} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = {\left( {81} \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$

24.

The value of $$\sqrt{0.01}\text{ + }$$ $$\sqrt{0.81}\text{ + }$$ $$\sqrt{1.21}\text{ + }$$ $$\sqrt{0.0009}$$   is = ?

• (A) 2.03
• (B) 2.1
• (C) 2.11
• (D) 2.13

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 Given expression, $$ = \sqrt {\frac{1}{{100}}} + \sqrt {\frac{{81}}{{100}}} + \sqrt {\frac{{121}}{{100}}} + $$      $$\sqrt {\frac{9}{{10000}}} $$ $$\eqalign{ & = \frac{1}{{10}} + \frac{9}{{10}} + \frac{{11}}{{10}} + \frac{3}{{100}} \cr & = 0.1 + 0.9 + 1.1 + 0.03 \cr & = 2.13 \cr} $$

25.

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

• (A) $$16-\sqrt{3}$$
• (B) $$4-\sqrt{3}$$
• (C) $$2-\sqrt{3}$$
• (D) $$2+\sqrt{3}$$

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 Given expression, $$ = \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$     $$ + \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$   $$ \times \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}}$$   $$ + \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$   $$ \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}$$ $$ = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}} + \frac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}}$$     $$ + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}}$$ $$ = \left[ {{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}} \right]$$     $$ + \frac{{4 - 2\sqrt 3 }}{2}$$ $$ = 2\left( {4 + 3} \right) + 2 - \sqrt 3 $$ $$ = 16 - \sqrt 3 $$

26.

What should come in place of both x in the equation $$\frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$$

• (A) 12
• (B) 14
• (C) 144
• (D) 196

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 $$\eqalign{ & {\text{Let}}\,\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then}}\,{x^2} = \sqrt {128 \times 162} \cr & = \sqrt {64 \times 2 \times 18 \times 9} \cr & = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & = 8 \times 6 \times 3 \cr & = 144 \cr & \therefore x = \sqrt {144} = 12 \cr} $$

27.

$$\sqrt{\frac{25}{81}-\frac{1}{9}}=?$$

• (A) $$\frac{2}{3}$$
• (B) $$\frac{4}{9}$$
• (C) $$\frac{16}{81}$$
• (D) $$\frac{25}{81}$$

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 $$\eqalign{ & = \sqrt {\frac{{25}}{{81}} - \frac{1}{9}} \cr & = \sqrt {\frac{{25 - 9}}{{81}}} \cr & = \sqrt {\frac{{16}}{{81}}} \cr & = \frac{{\sqrt {16} }}{{\sqrt {81} }} \cr & = \frac{4}{9} \cr} $$

28.

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

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

29.

The number of perfect square numbers between 50 and 1000 is = ?

• (A) 21
• (B) 22
• (C) 23
• (D) 24

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 The first perfect square number after 50 is 64 $$\left( {64 = {8^2}} \right)$$   and the last perfect square number before 1000 is 961 $$\left[ {961 = {{\left( {31} \right)}^2}} \right]$$ So, the perfect squares between 50 and 1000 are the squares of numbers from 8 to 31. (31 - 8) + 1 = 24 Clearly, these are 24 in number.

30.

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

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

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