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

$$\eqalign{ & \Rightarrow \frac{1}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)}} \cr & = \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} \cr & = \frac{{\left( {2.2361 + 1.7321} \right)}}{2} \cr & = \frac{{3.9682}}{2} \cr & = 1.9841{\text{ }} \cr} $$

Given expression,
$$ = \frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \times \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)}}$$     $$ - \frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 + \sqrt 7 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 + \sqrt 6 } \right)}}$$   $$ - \frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 + \sqrt 4 } \right)}}$$
$$ = \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {9 - 8} \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {8 - 7} \right)}}$$     $$ + \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {7 - 6} \right)}}$$   $$ - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {6 - 5} \right)}}$$   $$ + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {5 - 4} \right)}}$$
$$ = \left( {\sqrt 9 + \sqrt 8 } \right) - \left( {\sqrt 8 + \sqrt 7 } \right)$$     $$ + \left( {\sqrt 7 + \sqrt 6 } \right)$$   $$ - \left( {\sqrt 6 + \sqrt 5 } \right)$$   $$ + \left( {\sqrt 5 + \sqrt 4 } \right)$$
$$ = \left( {\sqrt 9 + \sqrt 4 } \right)$$
$$ = 3 + 2$$
$$ = 5$$

Given expressing,
$$ = \frac{1}{{\sqrt 1 + \sqrt 2 }}{\text{ + }}\frac{1}{{\sqrt 2 + \sqrt 3 }}$$     $$ + \frac{1}{{\sqrt 3 + \sqrt 4 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {120} + \sqrt {121} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }}{\text{ + }}\frac{1}{{\sqrt 3 + \sqrt 2 }}$$     $$ + \frac{1}{{\sqrt 4 + \sqrt 3 }}$$   $$ + ...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }}$$
$$ = \frac{1}{{\sqrt 2 + \sqrt 1 }} \times $$   $$\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}{\text{ + }}$$   $$\frac{1}{{\sqrt 3 + \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} + $$   $$\frac{1}{{\sqrt 4 + \sqrt 3 }} \times $$   $$\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} + $$   $$...... + $$   $$\frac{1}{{\sqrt {121} + \sqrt {120} }} \times $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{\sqrt {121} - \sqrt {120} }}$$
$$ = \frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}$$     $$ + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}}$$   $$ + ...... + $$   $$\frac{{\sqrt {121} - \sqrt {120} }}{{121 - 120}}$$
$$ = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 $$     $$ + \sqrt 4 - \sqrt 3 $$   $$ + ...... + $$   $$\sqrt {121} - \sqrt {120} $$
$$ = - 1 + \sqrt {121} $$
$$ = - 1 + 11$$
$$ = 10$$

$$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \times \frac{{\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)}} \cr & = \frac{{3\sqrt 3 - 3\sqrt 2 + 3\sqrt 2 - 2\sqrt 3 }}{{\left( {3 - 2} \right)}} \cr & = \sqrt 3 \cr & = 1.732 \cr} $$

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

$$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$
$$\therefore {x^2} + \frac{1}{{{x^2}}}$$
$$ = \left( {17 + 12\sqrt 2 } \right)$$   $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$   $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$
$$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$

$$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$

$$\eqalign{ & {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr & {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr & \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr & \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr & \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = \frac{{28}}{4} \cr & \,\,\,\,\,\,\, = 7 \cr & {\text{Also, }} \cr & ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr & \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = 1 \cr & \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr & \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr & \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr & \,\,\,\,\,\,\, = \frac{8}{6} \cr & \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$

Let the total number of camels be x
Then,
$$\eqalign{ & \Leftrightarrow x - \left( {\frac{x}{4} + 2\sqrt x } \right) = 15 \cr & \Leftrightarrow \frac{{3x}}{4} - 2\sqrt x = 15 \cr & \Leftrightarrow 3x - 8\sqrt x = 60 \cr & \Leftrightarrow 8\sqrt x = 3x - 60 \cr & \Leftrightarrow 64x = {\left( {3x - 60} \right)^2} \cr & \Leftrightarrow 64x = 9{x^2} + 3600 - 360x \cr & \Leftrightarrow 9{x^2} - 424x + 3600 = 0 \cr & \Leftrightarrow 9{x^2} - 324x - 100x + 3600 = 0 \cr & \Leftrightarrow 9x\left( {x - 36} \right) - 100\left( {x - 36} \right) = 0 \cr & \Leftrightarrow \left( {x - 36} \right)\left( {9x - 100} \right) = 0 \cr & \Leftrightarrow x = 36\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because x \ne \frac{{100}}{9}} \right] \cr} $$