Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }} + \frac{1}{{\sqrt 5 - 2}} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{3 - \sqrt 8 }} \cr & \Rightarrow {\text{Rationalising}} \cr & \Rightarrow \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 - \sqrt 6 } \right)\left( {\sqrt 7 + \sqrt 6 } \right)}} - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 - \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 - \sqrt 4 } \right)\left( {\sqrt 5 + \sqrt 4 } \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 - \sqrt 7 } \right)\left( {\sqrt 8 + \sqrt 7 } \right)}} + \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 - \sqrt 8 } \right)\left( {\sqrt 9 + \sqrt 8 } \right)}} \cr & \Rightarrow \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{1} - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{1} + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{1} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{1} + \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{1} \cr & \Rightarrow \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 + \sqrt 5 + \sqrt 4 - \sqrt 8 - \sqrt 7 + \sqrt 9 + \sqrt 8 \cr & \Rightarrow \sqrt 4 + \sqrt 9 \cr & \Rightarrow 2 + 3 \cr & \Rightarrow 5 \cr} $$

$$\eqalign{ & \left( {\sqrt 2 + \sqrt 5 - \sqrt 3 } \right) \times {\text{k}} = - 12 \cr & {\text{In these type of questions, we can go through option}} \cr & {\text{From option D}} \cr & \Rightarrow \left( {\sqrt 2 + \sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 5 + \sqrt 3 } \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr & \Rightarrow \left[ {{{\left( {\sqrt 2 + \sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right]\left( {2 - \sqrt {10} } \right) = - 12 \cr & \Rightarrow \left( {2 + 5 + 2\sqrt {10} - 3} \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr & \Rightarrow 2\left( {2 + \sqrt {10} } \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr & \Rightarrow 2\left[ {{{\left( 2 \right)}^2} - {{\left( {\sqrt {10} } \right)}^2}} \right] = - 12 \cr & \Rightarrow 2 \times \left[ { - 6} \right] = - 12 \cr & \Rightarrow - 12 = - 12\,\,\,\,\left( {{\text{Satisfy}}} \right) \cr} $$

$$\eqalign{ & {\text{I}}.\sqrt {12} > \root 3 \of {16} > \root 4 \of {24} \cr & {12^{\frac{1}{2}}},\,{16^{\frac{1}{3}}},\,{24^{\frac{1}{4}}} \cr & 2{\left( 3 \right)^{\frac{1}{2}}} > 2{\left( 2 \right)^{\frac{1}{3}}} > 2{\left( {\frac{3}{2}} \right)^{\frac{1}{4}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{I}} \right){\text{ True}} \cr & {\text{II}}.\root 3 \of {25} > \root 4 \of {32} > \root 6 \of {48} \cr & {25^{\frac{1}{3}}},\,{32^{\frac{1}{4}}},\,{48^{\frac{1}{6}}} \cr & 5{\left( {\frac{1}{5}} \right)^{\frac{1}{3}}} > 2{\left( 2 \right)^{\frac{1}{4}}} > 2{\left( {\frac{3}{4}} \right)^6}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{II}}} \right){\text{ True}} \cr & {\text{III}}.\root 4 \of 9 > \root 3 \of {15} > \root 6 \of {24} \cr & {9^{\frac{1}{4}}},\,{15^{\frac{1}{3}}},\,{24^{\frac{1}{6}}} \cr & {\text{3}}{\left( {\frac{1}{9}} \right)^{\frac{1}{4}}} > {\left( {15} \right)^{\frac{1}{3}}} > {\left( {24} \right)^{\frac{1}{6}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{III}}} \right){\text{ False}} \cr} $$

$$\eqalign{ & {\text{Take LCM of }}\left( {2,\,3\,\& \,6} \right) = 6 \cr & {\text{I}}.{\left( {27} \right)^{\frac{1}{3}}} > {\left( {13} \right)^{\frac{1}{2}}} < {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {27} \right)^{\frac{2}{{3 \times 2}}}} > {\left( {13} \right)^{\frac{3}{{2 \times 3}}}} < {\left( {47} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {27} \right)^{\frac{2}{6}}} > {\left( {13} \right)^{\frac{3}{6}}} < {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {729} \right)^{\frac{1}{6}}} > {\left( {2197} \right)^{\frac{1}{6}}} < {\left( {47} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{II}}.{\left( {23} \right)^{\frac{1}{3}}} < {\left( {49} \right)^{\frac{1}{2}}} < {\left( {52} \right)^{\frac{1}{6}}} \cr & = {\left( {23} \right)^{\frac{2}{{3 \times 2}}}} < {\left( {49} \right)^{\frac{3}{{2 \times 3}}}} < {\left( {52} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {529} \right)^{\frac{1}{6}}} < {\left( {117649} \right)^{\frac{1}{6}}} < {\left( {52} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{III}}.{\left( {53} \right)^{\frac{1}{6}}} < {\left( {41} \right)^{\frac{1}{3}}} < {\left( {37} \right)^{\frac{1}{2}}} \cr & = {\left( {53} \right)^{\frac{1}{{6 \times 1}}}} < {\left( {41} \right)^{\frac{2}{{3 \times 2}}}} < {\left( {37} \right)^{\frac{3}{{2 \times 3}}}} \cr & = {\left( {53} \right)^{\frac{1}{6}}} < {\left( {1681} \right)^{\frac{1}{6}}} < {\left( {50653} \right)^{\frac{1}{6}}} \cr & {\text{This statement is true}} \cr} $$

$$\eqalign{ & x = \sqrt {2\root 3 \of {4\sqrt {2\root 3 \of 4 } } } ...... \cr & {\text{Squaring both side}} \cr & \Rightarrow {x^2} = 2\root 3 \of {4\sqrt {2\root 3 \of 4 } } ...... \cr & {\text{Now cubing both sides}} \cr & \Rightarrow {x^6} = 8 \times 4x \cr & \Rightarrow {x^5} = 32 \cr & \Rightarrow {x^5} = {2^5} \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & f\left( 1 \right) = 1 - \frac{1}{2} \cr & f\left( 2 \right) = \frac{1}{2} - \frac{1}{3} \cr & f\left( 3 \right) = \frac{1}{3} - \frac{1}{4} \cr & f\left( 4 \right) = \frac{1}{4} - \frac{1}{5} \cr & 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}\,.\,.\,.\,.\,.\,\frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{11}} \cr & = 1 - \frac{1}{{11}} \cr & = \frac{{10}}{{11}} \cr} $$

At the start t = 0
L = 2 - 2⁰
⇒ 2 - 1 = 1 cm

$$\eqalign{ & 2\sqrt {50} + \sqrt {18} - \sqrt {72} \cr & \Rightarrow 2 \times 5\sqrt 2 + 3\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 13\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 7\sqrt 2 \cr & \Rightarrow 7 \times 1.414 \cr & \Rightarrow 9.898 \cr} $$

$$\eqalign{ & \root 3 \of a = \root 3 \of {26} + \root 3 \of 7 + \root 3 \of {63} \cr & {\text{Take round figure}} \cr & \Rightarrow \root 3 \of a < \root 3 \of {27} + \root 3 \of 8 + \root 3 \of {64} \cr & \Rightarrow \root 3 \of a < 3 + 2 + 4 \cr & \Rightarrow \root 3 \of a < 9 \cr & \Rightarrow a < {9^3} \cr & \Rightarrow a < 729 \cr & {\text{Option A is answer}} \cr} $$

$$\eqalign{ & {{\text{x}}^{\frac{1}{6}}} = {{\text{y}}^{\frac{2}{3}}} \cr & \therefore {\text{LCM of }}\left( {3\,\& \,6} \right) = 6 \cr & \Rightarrow {{\text{x}}^{\frac{1}{6} \times 6}} = {{\text{y}}^{\frac{2}{3} \times 6}} \cr & \Rightarrow \boxed{{\text{x}} = {{\text{y}}^4}} \cr} $$