Square Root And Cube Root - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \Leftrightarrow \sqrt {x + \frac{x}{y}} = x\sqrt {\frac{x}{y}} \cr & \Leftrightarrow x + \frac{x}{y} = {x^2}.\frac{x}{y} \cr & \Leftrightarrow \frac{{xy + x}}{y} = \frac{{{x^3}}}{y} \cr & \Leftrightarrow xy + x = {x^3} \cr & \Leftrightarrow y + 1 = {x^2} \cr & \Leftrightarrow y = {x^2} - 1 \cr} $$

$$\eqalign{ & \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr & \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$
∴ Sum of digits of n
= 2 + 0 + 4 + 8
= 14

$$\eqalign{ & \Leftrightarrow \frac{{0.13}}{{{p^2}}} = 13 \cr & \Leftrightarrow {p^2} = \frac{{0.13}}{{13}} \cr & \Leftrightarrow {p^2} = \frac{1}{{100}} \cr & \Leftrightarrow p = \sqrt {\frac{1}{{100}}} \cr & \Leftrightarrow p = \frac{1}{{10}} \cr & \Leftrightarrow p = 0.1 \cr} $$

$$\eqalign{ & \Leftrightarrow \sqrt {{3^n}} = 729 \cr & \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr & \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr & \Leftrightarrow {3^n} = {3^{12}} \cr & \Leftrightarrow n = 12 \cr} $$

$$\eqalign{ & \Leftrightarrow \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02 \cr & \Leftrightarrow \frac{{\sqrt x }}{{21}} = 0.02 \cr & \Leftrightarrow \sqrt x = 0.02 \times 21 \cr & \Leftrightarrow \sqrt x = 0.42 \cr & \Leftrightarrow x = {\left( {0.42} \right)^2} \cr & \Leftrightarrow x = 0.1764{\text{ }} \cr} $$

$$\eqalign{ & {\text{Let ,}} \cr & \sqrt {\frac{{0.0196}}{x}} = 0.2 \cr & {\text{Then,}} \cr & \Leftrightarrow \frac{{0.0196}}{x} = 0.04 \cr & \Leftrightarrow x = \frac{{0.0196}}{{0.04}} \cr & \Leftrightarrow x = \frac{{1.96}}{4} \cr & \Leftrightarrow x = 0.49 \cr} $$

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

$$\eqalign{ & \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {y + 2} \right)} = 7 \cr & \Leftrightarrow \left( {x - 1} \right)\left( {y + 2} \right) = {\left( 7 \right)^2} \cr & \Leftrightarrow \left( {x - 1} \right) = 7{\text{ and}}\left( {y + 2} \right) = 7 \cr & \Leftrightarrow x = 8{\text{ and }}y = 5 \cr} $$

$$\eqalign{ & {\text{Let the number be }}x \cr & {\text{Then,}} \cr & \Leftrightarrow \frac{3}{5}{x^2} = 126.15 \cr & \Leftrightarrow {x^2} = \left( {126.15 \times \frac{5}{3}} \right) \cr & \Leftrightarrow {x^2} = 210.25 \cr & \Leftrightarrow x = \sqrt {210.25} \cr & \Leftrightarrow x = 14.5 \cr} $$

$$\eqalign{ & \Rightarrow \sqrt {1 + \frac{x}{{169}}} = \frac{{14}}{{13}} \cr & \Rightarrow 1 + \frac{x}{{169}} = \frac{{196}}{{169}} \cr & \Rightarrow \frac{x}{{169}} = \left( {\frac{{196}}{{169}} - 1} \right) \cr & \Rightarrow \frac{x}{{169}} = \frac{{27}}{{169}} \cr & \Rightarrow x = 27 \cr} $$