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

$$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {176 + 49} \cr & = \sqrt {225} \cr & = 15 \cr} $$

$$\eqalign{ & {\text{Let, }} \cr & {\left( {15} \right)^2} + {\left( {18} \right)^2} - 20 = \sqrt x \cr & {\text{Then, }} \cr & \sqrt x = 225 + 324 - 20 \cr & \Leftrightarrow x = {\left( {529} \right)^2} \cr & \Leftrightarrow x = 279841 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & \sqrt x \times \sqrt {484} = 1034 \cr & {\text{Then}}, \cr & \sqrt x \times 22 = 1034 \cr & \Leftrightarrow \sqrt x = \frac{{1034}}{{22}} = 47 \cr & \Leftrightarrow x = {\left( {47} \right)^2} \cr & \Leftrightarrow x = 2209 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & \sqrt {11881} \times \sqrt x = 10137 \cr & {\text{Then}}, \cr & 109 \times \sqrt x = 10137 \cr & \Leftrightarrow \sqrt x = \frac{{10137}}{{109}} = 93 \cr & \Leftrightarrow x = {\left( {93} \right)^2} \cr & \Leftrightarrow x = 8649 \cr} $$

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

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

$$\eqalign{ & = \sqrt {\left( {{{272}^2} - {{128}^2}} \right)} \cr & = \sqrt {\left( {272 + 128} \right)\left( {272 - 128} \right)} \cr & = \sqrt {400 \times 144} \cr & = \sqrt {57600} \cr & = 240 \cr} $$

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

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

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