Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & a = \frac{1}{{2 - \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} \times \frac{{3 + \sqrt 8 }}{{3 + \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} + \frac{{3 + \sqrt 8 }}{{9 - 8}} + \frac{{4 + \sqrt {15} }}{{16 - 15}} \cr & \Rightarrow 2 + \sqrt 3 + 3 + \sqrt 8 + 4 + \sqrt {15} \cr & \Rightarrow 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} \cr & a = 9 < 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} < 18 \cr & \sqrt 3 = 1.73,\,\sqrt 2 = 1.41,\,\sqrt {15} = 3.9 \cr & \Rightarrow 9 < 9 + 1.73 + \left( {2 \times 1.41} \right) + 3.9 = 17.4 < 18 \cr} $$

$$\eqalign{ & {\text{Expression}} \cr & = \frac{{3\sqrt 8 - 2\sqrt {12} + \sqrt {20} }}{{3\sqrt {18} - 2\sqrt {27} + \sqrt {45} }} \cr & = \frac{{3\sqrt {2 \times 2 \times 2} - 2\sqrt {2 \times 2 \times 3} + \sqrt {2 \times 2 \times 5} }}{{3\sqrt {3 \times 3 \times 2} - 2\sqrt {3 \times 3 \times 3} + \sqrt {3 \times 3 \times 5} }} \cr & = \frac{{6\sqrt 2 - 4\sqrt 3 + 2\sqrt 5 }}{{9\sqrt 2 - 6\sqrt 3 + 3\sqrt 5 }} \cr & = \frac{{2\left( {3\sqrt 2 - 2\sqrt 3 + \sqrt 5 } \right)}}{{3\left( {3\sqrt 2 - 2\sqrt 3 + \sqrt 5 } \right)}} \cr & = \frac{2}{3} \cr} $$

$$\eqalign{ & \frac{1}{{{{\left( {0.1} \right)}^2}}} + \frac{1}{{{{\left( {0.01} \right)}^2}}} + \frac{1}{{{{\left( {0.5} \right)}^2}}} + \frac{1}{{{{\left( {0.05} \right)}^2}}} \cr & = \frac{1}{{0.1 \times 0.1}} + \frac{1}{{0.01 \times 0.01}} + \frac{1}{{0.5 \times 0.5}} + \frac{1}{{0.05 \times 0.05}} \cr & = 100 + 10000 + 4 + 20 \times 20 \cr & = 100 + 10000 + 4 + 400 \cr & = 10504 \cr} $$

$$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & \Rightarrow {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & \Rightarrow 36 + 49 + 64 + 81 + 100 \cr & \Rightarrow 330 \cr} $$

$$\eqalign{ & \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3} \cr & \therefore \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{x^{\frac{2}{3}}} - 1 + {x^{ - \frac{2}{3}}}} \right) \cr & = \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{{\left( {{x^{\frac{1}{3}}}} \right)}^2} - {x^{ - \frac{1}{3}}}.{x^{\frac{1}{3}}} + {{\left( {{x^{ - \frac{1}{3}}}} \right)}^2}} \right) \cr & = {\left( {{x^{\frac{1}{3}}}} \right)^3} + {\left( {{x^{ - \frac{1}{3}}}} \right)^3} \cr & = x + {x^{ - 1}} \cr & = x + \frac{1}{x} \cr} $$

$$\eqalign{ & {\text{Given,}}\,4x = \sqrt 5 + 2 \cr & \Rightarrow 16x = 4\left( {\sqrt 5 + 2} \right) \cr & \Rightarrow 16x = 4\sqrt 5 + 8 \cr & \therefore \frac{1}{{16x}} = \frac{1}{{4\sqrt 5 + 8}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{\left( {4\sqrt 5 + 8} \right)\left( {4\sqrt 5 - 8} \right)}} \cr & \left[ {{\text{Rationalising the denominator}}} \right] \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{80 - 64}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\left( {\sqrt 5 - 2} \right)}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{\sqrt 5 - 2}}{4} \cr & \therefore \left( {x - \frac{1}{{16x}}} \right) \cr & = \frac{{\sqrt 5 + 2}}{4} - \frac{{\sqrt 5 - 2}}{4} \cr & = \frac{{\sqrt 5 + 2 - \sqrt 5 + 2}}{4} \cr & = \frac{4}{4} \cr & = 1 \cr} $$

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

$$\eqalign{ & {\left( {\sqrt {11} + \sqrt 5 } \right)^2} = 16 + 2\sqrt {55} \cr & {\left( {\sqrt {14} + \sqrt 2 } \right)^2} = 16 + 2\sqrt {28} \cr & {\left( {\sqrt 8 + \sqrt 8 } \right)^2} = 16 + 2\sqrt {64} \cr & {\text{It is clear that }}16 + 2\sqrt {64} {\text{ is largest}} \cr & {\text{So, }}\left( {\sqrt 8 + \sqrt 8 } \right){\text{ will be largest}}{\text{.}} \cr} $$

$$\eqalign{ & \frac{{5 + \sqrt {11} }}{{3 - 2\sqrt {11} }} = x + y\sqrt {11} \cr & \frac{{\left( {5 + \sqrt {11} } \right)\left( {3 + 2\sqrt {11} } \right)}}{{\left( {3 - 2\sqrt {11} } \right)\left( {3 + 2\sqrt {11} } \right)}} = x + y\sqrt {11} \cr & \frac{{15 + 10\sqrt {11} + 3\sqrt {11} + 22}}{{9 - 44}} = x + y\sqrt {11} \cr & \frac{{37 - 13\sqrt {11} }}{{ - 35}} = x + y\sqrt {11} \cr & \frac{{ - 37}}{{35}} + \frac{{\left( { - 13} \right)\sqrt {11} }}{{35}} = x + y\sqrt {11} \cr & {\text{Comparing both side}} \cr & x = \frac{{ - 37}}{{35}} \cr & y = \frac{{ - 13}}{{35}} \cr} $$