Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$

Solution:

 $$\eqalign{ & \root 3 \of 6 ,\,\root 2 \of 5 {\text{ and }}\root 6 \of {12} \cr & {\text{Take LCM of }}\left( {3,\,2{\text{ and }}6} \right) = 6 \cr & {6^{\frac{{1 \times 6}}{3}}},\,{5^{\frac{{1 \times 6}}{2}}},\,{12^{\frac{{1 \times 6}}{6}}} \cr & 36,\,125{\text{ and }}12 \cr & {\text{Clearly, }}\root 2 \of 5 {\text{ is the largest}}{\text{.}} \cr} $$

Solution:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & \frac{{{{\left( {19} \right)}^{12}} \times {{\left( {19} \right)}^8}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{19}^{\left( {12 + 8} \right)}}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{\left( {19} \right)}^{20}}}}{{{{\left( {19} \right)}^4}}} \cr & = {\left( {19} \right)^{\left( {20 - 4} \right)}} \cr & = {\left( {19} \right)^{16}} \cr} $$ Hence the missing number = 16

Solution:

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

Solution:

 $$\eqalign{ & {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr & = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr & = 2 + 0.3 - 0.2 - 2.10 \cr & = 2.3 - 2.3 \cr & = 0 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr & = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr & = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr & = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr & {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr} $$

Solution:

 $$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given Exp}}{\text{.}} \cr & = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}} \cr} $$   $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$ $$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} = \frac{1}{{\sqrt 6 }} \cr & {\text{On squaring both sides,}} \cr & \Rightarrow {\left( {\sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} } \right)^2} = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} - 2 = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = 2 + \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{{{\left( {1 + x} \right)}^2} + {x^2}}}{{x\left( {1 + x} \right)}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{1 + {x^2} + 2x + {x^2}}}{{x + {x^2}}} = \frac{{13}}{6} \cr & \Rightarrow \left( {2{x^2} + 2x + 1} \right)6 = 13\left( {x + {x^2}} \right) \cr & \Rightarrow 12{x^2} + 12x + 6 = 13x + 13{x^2} \cr & \Rightarrow {x^2} + x - 6 = 0 \cr & \Rightarrow {x^2} + 3x - 2x - 6 = 0 \cr & \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2{\text{ as }}x \ne - 3 \cr & {\text{Or, of the given options, when }}x = 2 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} \cr & = \sqrt {\frac{{1 + 2}}{2}} - \sqrt {\frac{2}{{1 + 2}}} \cr & = \frac{{\sqrt 3 }}{{\sqrt 2 }} - \frac{{\sqrt 2 }}{{\sqrt 3 }} \cr & = \frac{{3 - 2}}{{\sqrt 6 }} \cr & = \frac{1}{{\sqrt 6 }} \cr} $$