Practice MCQ Questions and Answer on Square Root and Cube Root

51.

In the equation $$\frac{4050}{\sqrt{x}}=450\text{,}$$   the value of x is = ?

• (A) 9
• (B) 49
• (C) 81
• (D) 100

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 $$\eqalign{ & \frac{{4050}}{{\sqrt x }} = 450 \cr & \Leftrightarrow \sqrt x = \frac{{4050}}{{450}} \cr & \Leftrightarrow \sqrt x = 9 \cr & \Leftrightarrow x = {\left( 9 \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$

52.

If $$\sqrt{1369}+\sqrt{0.0615+x}$$      = 37.25, the x is equal to ?

• (A) $${10}^{-1}$$
• (B) $${10}^{-2}$$
• (C) $${10}^{-3}$$
• (D) None of these

Show Answer

 $$\eqalign{ & \Leftrightarrow 37 + \sqrt {0.0615 + x} = 37.25 \cr & \Leftrightarrow \sqrt {0.0615 + x} = 0.25 \cr & \Leftrightarrow 0.0615 + x = {\left( {0.25} \right)^2} = 0.0625 \cr & \Leftrightarrow x = 0.001 \cr & \Leftrightarrow x = \frac{1}{{{{10}^3}}} \cr & \Leftrightarrow x = {10^{ - 3}} \cr} $$

53.

If $$y=5\text{,}$$   then what is the value of $$10y\sqrt{y^3-y^2}$$   = ?

• (A) $$50\sqrt{2}$$
• (B) 100
• (C) $$200\sqrt{5}$$
• (D) 500

Show Answer

 $$\eqalign{ & = 10y\sqrt {{y^3} - {y^2}} \cr & = 10 \times 5\sqrt {{5^3} - {5^2}} \cr & = 50 \times \sqrt {125 - 25} \cr & = 50 \times \sqrt {100} \cr & = 50 \times 10 \cr & = 500 \cr} $$

54.

Which number should replace both the question marks in the following equation ?
$$\frac{?}{1776}=\frac{111}{?}$$

• (A) 343
• (B) 414
• (C) 644
• (D) 543

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}\frac{x}{{1776}} = \frac{{111}}{x} \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 111 \times 1776 \cr & \Leftrightarrow {x^2} = 111 \times 111 \times 16 \cr & \Leftrightarrow x = \sqrt {{{\left( {111} \right)}^2} \times {{\left( 4 \right)}^2}} \cr & \Leftrightarrow x = 111 \times 4 \cr & \Leftrightarrow x = 444 \cr} $$

55.

$$\left(\frac{\sqrt{625}}{11}\times \frac{14}{\sqrt{25}}\times \frac{11}{\sqrt{196}}\right)$$     is equal to :

• (A) 5
• (B) 6
• (C) 8
• (D) 11

Show Answer

 $$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$

56.

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

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

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 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 $$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$

57.

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

• (A) 77
• (B) 98
• (C) 131
• (D) 221

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

58.

The value of $$\sqrt{\frac{{\left(0.03\right)}^2+{\left(0.21\right)}^2+{\left(0.065\right)}^2}{{\left(0.003\right)}^2+{\left(0.021\right)}^2+{\left(0.0065\right)}^2}}$$       is ?

• (A) 0.1
• (B) 10
• (C) $${10}^2$$
• (D) $${10}^3$$

Show Answer

 $$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {\frac{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}{{{{\left( {\frac{{0.03}}{{10}}} \right)}^2} + {{\left( {\frac{{0.21}}{{10}}} \right)}^2} + {{\left( {\frac{{0.065}}{{10}}} \right)}^2}}}} \cr & = \sqrt {\frac{{100\left[ {{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}} \right]}}{{{{\left( {0.03} \right)}^2} + {{\left( {0.21} \right)}^2} + {{\left( {0.065} \right)}^2}}}} \cr & = \sqrt {100} \cr & = 10 \cr} $$

59.

For what value of * the statement $$\left(\frac{\ast }{15}\right)$$ $$\left(\frac{\ast }{135}\right)$$  = 1 is true ?

• (A) 15
• (B) 25
• (C) 35
• (D) 45

Show Answer

 $$\eqalign{ & {\text{Method 1:}} \cr & {\text{Let the missing number be }}x \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 15 \times 135 \cr & \Leftrightarrow x = \sqrt {15 \times \left(15 \times 9 \right)} \cr & \Leftrightarrow x = \sqrt {{{15}^2} \times {3^2}} \cr & \Leftrightarrow x = 15 \times 3 \cr & \Leftrightarrow x = 45 \cr & \cr & {\text{Method 2:}} \cr & {\text{Let the missing number be }}x \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 15 \times 135 \cr & \Leftrightarrow {x^2} = 2025 \cr & \Leftrightarrow x = \sqrt {{2025} } \cr & \Leftrightarrow x = 45 \cr} $$

60.

If $$\sqrt{24}=4.889,$$   the value of $$\sqrt{\frac{8}{3}}$$   is = ?

• (A) 0.544
• (B) 1.333
• (C) 1.633
• (D) 2.666

Show Answer

 $$\eqalign{ & = \sqrt {\frac{8}{3}} \cr & = \sqrt {\frac{{8 \times 3}}{{3 \times 3}}} \cr & = \frac{{\sqrt {24} }}{3} \cr & = \frac{{4.899}}{3} \cr & = 1.633 \cr} $$