Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr} $$

$$\eqalign{ & {\text{The exponential form of }} \cr & \sqrt {\sqrt 2 \times \sqrt 3 } \cr & = \sqrt {{6^{\frac{1}{2}}}} \cr & = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr & = {6^{\frac{1}{4}}} \cr} $$

$$\eqalign{ & {\text{Expression,}} \cr & {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr} $$

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

$$\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr} $$

$$\eqalign{ & {\left( {32 \times {{10}^{ - 5}}} \right)^{ 2}} \times 64 \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) = {10^?} \cr & \Rightarrow {\left( {{2^5} \times {{10}^{ - 5}}} \right)^{ 2}} \times {2^6} \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right] \cr & \Rightarrow \frac{{{2^{10}} \times {{10}^{ - 10}} \times {2^6}}}{{{2^{16}} \times {{10}^{ - 4}}}} \cr & \,\,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow \frac{{{2^{16}} \times {{10}^4}}}{{{2^{16}} \times {{10}^{10}}}} \cr & \,\,\,\,\,\,\,\, = {10^?}.....\left[ {{a^{ - m}} = \frac{1}{{{a^m}}}} \right] \cr & \Rightarrow {10^{4 - 10}} = {10^?} \cr & \Rightarrow {10^{ - 6}} = {10^?} \cr & \Rightarrow ? = - 6 \cr} $$

$$\eqalign{ & 2\root 3 \of {32} - 3\root 3 \of 4 + \root 3 \of {500} \cr & = 2\root 3 \of {{2^3} \times 4} - 3\root 3 \of 4 + \root 3 \of {{5^3} \times 4} \cr & = 2 \times 2\root 3 \of 4 - 3\root 3 \of 4 + 5\root 3 \of 4 \cr & = 9\root 3 \of 4 - 3\root 3 \of 4 \cr & = 6\root 3 \of 4 \cr} $$

$$2\sqrt 3 = {\left( {4 \times 3} \right)^{\frac{1}{2}}} \to {12^{\frac{1}{2}}} \to {12^{\frac{2}{4}}} \to \root 4 \of {144} $$
$$2\root 4 \of 5 = \root 4 \of {\left( {5 \times 16} \right)} \to {80^{\frac{1}{4}}} \to \root 4 \of {80} $$
$$\sqrt 8 = {8^{\frac{1}{2}}} \to {8^{\frac{1}{2}}} \to \boxed{\root 4 \of {64} }\,{\text{smallest}}$$
$$3\sqrt 2 = \sqrt {18} \to {18^{\frac{1}{2}}} \to {18^{\frac{2}{4}}} \to \root 4 \of {324} $$
$$\sqrt 8 \,{\text{ is answer}}$$

$$\eqalign{ & \sqrt 2 \to {2^{\frac{1}{2}}} \to {2^{\frac{{15}}{{30}}}} = \root {30} \of {{2^{15}}} = \root {30} \of {32768} \cr & \root 3 \of 3 \to {3^{\frac{1}{3}}} \to {3^{\frac{{10}}{{30}}}} = \root {30} \of {{3^{10}}} = \root {30} \of {59049} \cr & \root 6 \of 6 \to {6^{\frac{1}{6}}} \to {6^{\frac{5}{{30}}}} = \root {30} \of {{6^5}} = \root {30} \of {7776} \cr & \root 5 \of 5 \to {5^{\frac{1}{5}}} \to {5^{\frac{6}{{30}}}} = \root {30} \of {{5^6}} = \root {30} \of {15625} \cr & {\text{So }}\root 3 \of 3 {\text{ is the greatest}}{\text{.}} \cr} $$

$$\eqalign{ & {\left( {2.89} \right)^{0.5}} = {\left( {2.89} \right)^{\frac{5}{{10}}}} \to \sqrt {2.89} \to 1.7 \cr & {\text{2}} - {\left( {0.5} \right)^2} = 2 - 0.25 \to 1.75 \cr & 1 + \frac{{0.5}}{{1 - \frac{1}{2}}} = 1 + \frac{{0.5}}{{0.5}} \to 1 + 1 \to 2 \leftarrow {\text{Greatest}} \cr & \sqrt 3 = 1.732 \cr} $$