Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$

$$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$

$$\eqalign{ & {\text{ 2}}{{\text{7}}^{ - \frac{2}{3}}} \cr & = {\left( {{3^3}} \right)^{ - \frac{2}{3}}} \cr & = {3^{\left[ {3 \times \left( { - \frac{2}{3}} \right)} \right]}} \cr & = {3^{ - 2}} \cr & = \frac{1}{{{3^2}}} \cr & = \frac{1}{9} \cr & {\text{Clearly}},\,\,0 < \frac{1}{9} < 1 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt {3.25} + \sqrt {2.25} }} \times \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{\sqrt {3.25} - \sqrt {2.25} }} \cr & = \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{3.25 - 2.25}} \cr & = \sqrt {3.25} - \sqrt {2.25} \,......(i) \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }} \cr & = \sqrt {4.25} - \sqrt {3.25} \,.......(ii) \cr & \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }} \cr & = \sqrt {5.25} - \sqrt {4.25} \,.......(iii) \cr & \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }} \cr & = \sqrt {6.25} - \sqrt {5.25} \,.......(iv) \cr & {\text{Now}}\,{\text{add}}\,{\text{all}}\,{\text{them}} \cr & \sqrt {3.25} - \sqrt {2.25} + \sqrt {4.25} - \sqrt {3.25} + \sqrt {5.25} - \sqrt {4.25} + \sqrt {6.25} - \sqrt {5.25} \cr & = \sqrt {6.25} - \sqrt {2.25} \cr & = 2.5 - 1.5 \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}} \cr & = \frac{{1 + \frac{1}{3}}}{{\frac{1}{3} - 1}} \cr & = \frac{{\frac{4}{3}}}{{ - \frac{2}{3}}} \cr & = - 2 \cr} $$

$$\frac{2}{{\sqrt 7 + \sqrt 5 }} + \frac{7}{{\sqrt {12} - \sqrt 5 }} - \frac{5}{{\sqrt {12} - \sqrt 7 }}$$
$$ = \frac{2}{{\sqrt 7 + \sqrt 5 }} \times $$  $$\frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} + $$  $$\frac{7}{{\sqrt {12} - \sqrt 5 }} \times $$  $$\frac{{\sqrt {12} + \sqrt 5 }}{{\sqrt {12} + \sqrt 5 }} - $$  $$\left( {\frac{5}{{\sqrt {12} - \sqrt 7 }} \times \frac{{\sqrt {12} + \sqrt 7 }}{{\sqrt {12} + \sqrt 7 }}} \right)$$
$$ = \frac{{2\left( {\sqrt 7 - \sqrt 5 } \right)}}{2} + $$   $$\frac{{7\left( {\sqrt {12} + \sqrt 5 } \right)}}{7} - $$   $$\frac{{5\left( {\sqrt {12} + \sqrt 7 } \right)}}{5}$$
$$\eqalign{ &= \sqrt 7 - \sqrt 5 + \sqrt {12} + \sqrt 5 - \sqrt {12} - \sqrt 7 \cr &= 0 \cr} $$

$$\eqalign{ & \root 3 \of {{2^4}\sqrt {{2^{ - 5}}\sqrt {{2^6}} } } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( {{2^6}} \right)}^{\frac{1}{2}}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( 2 \right)}^{\left( {6 \times \frac{1}{2}} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{.2}^3}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{\left( { - 5 + 3} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 2}}} } {\text{ }} \cr & = \root 3 \of {{2^4}.{{\left( {{2^2}} \right)}^{\frac{1}{2}}}} \cr & {\text{ = }}\root 3 \of {{2^4}{{.2}^{\left( { - 2 \times \frac{1}{2}} \right)}}} \cr & = \root 3 \of {{2^4}{{.2}^{\left( { - 1} \right)}}} \cr & = \root 3 \of {{2^{\left( {4 - 1} \right)}}} \cr & {\text{ = }}\root 3 \of {{2^3}} \cr & {\text{ = }}{\left( {{2^3}} \right)^{\frac{1}{3}}} \cr & {\text{ = }}{{\text{2}}^{\left( {3 \times \frac{1}{3}} \right)}} \cr & {\text{ = 2 }} \cr} $$

$$\eqalign{ & \sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt 2 } } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{1}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{1}{2}} \right)}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\frac{3}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2.{{\left( {{2^{\frac{3}{2}}}} \right)}^{\frac{1}{2}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\left( {\frac{3}{2} \times \frac{1}{2}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{3}{4}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{3}{4}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\frac{7}{4}}}} } } \cr & = \sqrt {2\sqrt {2{{\left( {{2^{\frac{7}{4}}}} \right)}^{\frac{1}{2}}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\left( {\frac{7}{4} \times \frac{1}{2}} \right)}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\frac{7}{8}}}} } \cr & = \sqrt {2\sqrt {{2^{\left( {1 + \frac{7}{8}} \right)}}} } \cr & = \sqrt {2\sqrt {{2^{\frac{{15}}{8}}}} } \cr & = \sqrt {2{{\left( {{2^{\frac{{15}}{8}}}} \right)}^{\frac{1}{2}}}} \cr & = \sqrt {{{2.2}^{\frac{{15}}{{16}}}}} \cr & = \sqrt {{2^{\left( {1 + \frac{{15}}{{16}}} \right)}}} \cr & = \sqrt {{2^{\frac{{31}}{{16}}}}} \cr & = {\left( {{2^{\frac{{31}}{{16}}}}} \right)^{\frac{1}{2}}} \cr & = {2^{\left( {\frac{{31}}{{16}} \times \frac{1}{2}} \right)}} \cr & = {2^{\frac{{31}}{{32}}}} \cr} $$

$$\eqalign{ & {\left( {0.03125} \right)^{ - \frac{2}{5}}} \cr & = {\left[ {{{\left( {0.5} \right)}^5}} \right]^{ - \frac{2}{5}}} \cr & = {0.5^{\left[ {5 \times \left( { - \frac{2}{5}} \right)} \right]}} \cr & = {\left( {0.5} \right)^{ - 2}} \cr & = \frac{1}{{{{\left( {0.5} \right)}^2}}} \cr & = \frac{1}{{0.25}} \cr & = 4 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 3 + \sqrt 4 }} \cr & = \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} \cr & = \frac{{\sqrt 4 - \sqrt 3 }}{1} \cr & = \sqrt 4 - \sqrt 3 \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt 4 + \sqrt 5 }} = \sqrt 5 - \sqrt 4 \cr & \Rightarrow \frac{1}{{\sqrt 5 + \sqrt 6 }} = \sqrt 6 - \sqrt 5 \cr & \Rightarrow \frac{1}{{\sqrt 6 + \sqrt 7 }} = \sqrt 7 - \sqrt 6 \cr & \Rightarrow \frac{1}{{\sqrt 7 + \sqrt 8 }} = \sqrt 8 - \sqrt 7 \cr & \Rightarrow \frac{1}{{\sqrt 8 + \sqrt 9 }} = \sqrt 9 - \sqrt 8 \cr & {\text{Now put values}} \cr & \Rightarrow \sqrt 4 - \sqrt 3 + \sqrt 5 - \sqrt 4 + \sqrt 6 - \sqrt 5 + \sqrt 7 - \sqrt 6 + \sqrt 8 - \sqrt 7 + \sqrt 9 - \sqrt 8 \cr & \Rightarrow \sqrt 9 - \sqrt 3 \cr & \Rightarrow 3 - \sqrt 3 \cr} $$