Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let }}{\left( {0.04} \right)^2} \div \left( {0.008} \right) \times {\left( {0.2} \right)^6} = {\left( {0.2} \right)^x} \cr & {\text{Then,}}{\left( {0.2} \right)^x} = {\left[ {{{\left( {0.2} \right)}^2}} \right]^2} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {2 \times 2} \right)}} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^4} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {4 - 3 + 6} \right)}} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^7} \cr & \Leftrightarrow x = 7 \cr} $$

$$\eqalign{ & \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr & = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr & = \frac{{{3^1}}}{{{7^1}}} \cr & = \frac{3}{7} \cr} $$

$$\eqalign{ & \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr & = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr & = \frac{8}{8} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr & = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr & = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr & = 400 \cr} $$

Shortcut method:
When the question is in form
$$\eqalign{ & \Rightarrow \sqrt {n\sqrt {n\sqrt n } } ........\infty \cr & \Rightarrow {\text{So }}n{\text{ is answer }} \cr & \Rightarrow {\text{3}} \cr} $$

$$\eqalign{ & \frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}} \cr & = \frac{{{2^{n - 1}}\left( {1 - 2} \right)}}{{{2^{n + 1}}\left( {{2^3} + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{\left( {n - 1} \right) - \left( {n + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{ - 2}} \cr & = \left( { - \frac{1}{9}} \right).\frac{1}{{{2^2}}} \cr & = \left( { - \frac{1}{9}} \right) \times \frac{1}{4} \cr & = - \frac{1}{{36}} \cr} $$

$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$

$$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$

$$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ \Rightarrow \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$
$$ \Rightarrow \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$
$$\eqalign{ & \Rightarrow \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr & \Rightarrow \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr & \Rightarrow \sqrt 5 - \sqrt 6 \cr} $$