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

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

$$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$

$$\eqalign{ & \Leftrightarrow \sqrt y = 4x \cr & \Leftrightarrow y = {\left( {4x} \right)^2} \cr & \Leftrightarrow y = 16{x^2} \cr & \Leftrightarrow \frac{{{x^2}}}{y} = \frac{1}{{16}} \cr} $$

$$\eqalign{ & {\text{We have to find}} \cr & = \sqrt {9{x^2} + 25 - 30x} \cr} $$
 $$ = \sqrt {{{\left( {3x} \right)}^2} - 2.3x.5 + {{\left( { - 5} \right)}^2}} $$   $$\,\,\,\,\,\,\,\left\{ {\because {a^2} - 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right\}$$
$$\eqalign{ & = \sqrt {{{\left( {3x - 5} \right)}^2}} \cr & = 3x - 5 \cr} $$

$$\eqalign{ & = \sqrt {\frac{3}{{11}}} \cr & = \sqrt {\frac{{3 \times 11}}{{11 \times 11}}} \cr & = \frac{{\sqrt {33} }}{{11}} \cr & = \frac{{5.745}}{{11}} \cr & = 0.5223 \cr} $$

$$\eqalign{ & a = \frac{{\sqrt 3 }}{2}{\text{ (given)}} \cr & \therefore \sqrt {1 + a} + \sqrt {1 - a} \cr & = \sqrt {1 + \frac{{\sqrt 3 }}{2}} + \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2 + \sqrt 3 }}{2}} + \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & = \sqrt {\frac{{2\left( {2 + \sqrt 3 } \right)}}{4}} + \sqrt {\frac{{2\left( {2 - \sqrt 3 } \right)}}{4}} \cr & = \sqrt {\frac{{4 + 2\sqrt 3 }}{4}} + \sqrt {\frac{{4 - 2\sqrt 3 }}{4}} \cr} $$
$$ = \sqrt {\frac{{3 + 1 + 2 \times \sqrt 3 \times 1}}{2}} + $$     $$\sqrt {\frac{{3 + 1 - 2 \times \sqrt 3 \times 1}}{2}} $$     \[\because \left\{ \begin{gathered} {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} - 2.\sqrt 3 .1 = {\left( {\sqrt 3 - 1} \right)^2} \hfill \\ {\left( {\sqrt 3 } \right)^2} + {\left( 1 \right)^2} + 2.\sqrt 3 .1 = {\left( {\sqrt 3 + 1} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2} \hfill \\ {a^2} + {b^2} - 2ab = {\left( {a + b} \right)^2} \hfill \\ \end{gathered} \right\}\]
$$\eqalign{ & = \sqrt {\frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{2}} + \sqrt {\frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} \cr & = \frac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{2} \cr & = \frac{{2\sqrt 3 }}{2} \cr & = \sqrt 3 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2\sqrt {20} - \sqrt {32} + \sqrt {50} }}{\text{ }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 2 \times 2\sqrt 5 - 2 \times 2\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{5\sqrt 5 - 4\sqrt 5 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{5 + \sqrt {10} }}{{\sqrt 5 + \sqrt 2 }} \cr & = \frac{{\sqrt 5 \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\sqrt 5 + \sqrt 2 }} \cr & = \sqrt 5 \cr} $$

$$\frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}}$$ $${\text{ + }}$$$$\left[ {0.75 + {{\left( {0.75} \right)}^2} + 1} \right]$$
$$ = \frac{{{{\left( {0.75} \right)}^2} \times 0.75}}{{0.25}}$$   $${\text{ + }}$$ $$\left[ {0.75 + 0.5625 + 1} \right]$$
$$ = 0.5625 \times 3 \,\, + $$   $$\left[ {0.75 + 0.5625 + 1} \right]$$
$$\eqalign{ & = 1.6875 + 2.3125 \cr & = 4 \cr} $$
Square root of 4 = 2