Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{{{9^3} \times {6^2}}}{{{3^3}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {3 \times 2} \right)}^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {3 \times 2} \right)}} \times {3^2} \times {2^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {6 + 2} \right)}} \times {2^2}}}{{{3^3}}} \cr & = {3^{\left( {8 - 3} \right)}} \times {2^2} \cr & = {3^5} \times {2^2} \cr & = 243 \times 4 \cr & = 972 \cr} $$

$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$
$$ = \left\{ {\frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}} \right\}$$
$$ = \left\{ {\frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 }}{{4 - 3}} + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}} \right\}$$
$$\eqalign{ & = \left\{ {14 + \frac{{3 + 1 - 2\sqrt 3 }}{2}} \right\} \cr & = 14 + \frac{{2\left( {2 - \sqrt 3 } \right)}}{2} \cr & = 14 + 2 - \sqrt 3 \cr & = 16 - \sqrt 3 \cr} $$

$$\eqalign{ & \sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{\sqrt {36} + \sqrt {24} - \sqrt {24} - \sqrt {16} }}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{6 - 4}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{2}{{5 + \sqrt {24} }} \times \frac{{5 - \sqrt {24} }}{{5 - \sqrt {24} }}} \cr & = \sqrt {\frac{{2\left( {5 - \sqrt {24} } \right)}}{{25 - 24 }}} \cr & = \sqrt {2\left( {5 - 2\sqrt 6 } \right)} \cr & = \sqrt {2\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 3 \times \sqrt 2 } \right\}} \cr & = \sqrt {2{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \cr & = \sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right) \cr & = \sqrt 6 - 2 \cr} $$

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

$$\eqalign{ & {\left( {64} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( {{8^2}} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( 8 \right)^{\left( {2 \times 4} \right)}} \div {8^5} \cr & = \frac{{{8^8}}}{{{8^5}}} \cr & = {8^{\left( {8 - 5} \right)}} \cr & = {8^3} \cr} $$

$$\frac{1}{{\sqrt {\left( {12 - \sqrt {140} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {60} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {84} } \right)} }}$$
$$ = \frac{1}{{\sqrt {\left( {12 - \sqrt {4 \times 35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {4 \times 15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {4 \times 21} } \right)} }}$$
$$ = \frac{1}{{\sqrt {\left( {12 - 2\sqrt {35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - 2\sqrt {15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + 2\sqrt {21} } \right)} }}$$
$$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 7 .\sqrt 5 } }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 5 .\sqrt 3 } }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 7 .\sqrt 3 } }}$$
$$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}} }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} }}$$
$$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }}$$
Rationalizing in above equation,
$$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} \times \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }}$$
$$ = \frac{{\sqrt 7 + \sqrt 5 }}{2} - \frac{{\sqrt 5 + \sqrt 3 }}{2} - \frac{{\sqrt 7 - \sqrt 3 }}{2}$$
$$ = \frac{{\sqrt 7 + \sqrt 5 - \sqrt 5 - \sqrt 3 - \sqrt 7 + \sqrt 3 }}{2}$$
$$ = 0$$

$$\eqalign{ & \frac{{{{\left( {1000} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {{{10}^3}} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{\left( {3 \times 12} \right)}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{36}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = {\left( {10} \right)^{\left( {36 - 30} \right)}} \cr & = {10^6} \cr & = {\left( {{{10}^3}} \right)^2} \cr & = {\left( {1000} \right)^2} \cr} $$

$$\frac{{{{1.5}^3} + {{4.7}^3} + {{3.8}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}$$
$$ = \frac{{\left( {1.5 + 4.7 + 3.8} \right)\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}$$
$$\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right]$$
$$\eqalign{ & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr} $$

$$\eqalign{ & \frac{{0.41 \times 0.41 \times 0.41 + 0.69 \times 0.69 \times 0.69}}{{0.41 \times 0.41 - 0.41 \times 0.69 + 0.69 \times 0.69}} \cr & = \frac{{{{\left( {0.41} \right)}^3} + {{\left( {0.69} \right)}^3}}}{{{{\left( {0.41} \right)}^2} - 0.41 \times 0.69 + {{\left( {0.69} \right)}^2}}} \cr & .....\left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.41 + 0.69 \cr & = 1.1 \cr} $$

$$\eqalign{ & {{\text{3}}^{10}} \times {\text{2}}{{\text{7}}^2}{\text{ = }}{{\text{9}}^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10}} \times {\left( {{3^3}} \right)^2}{\text{ = }}{\left( {{3^2}} \right)^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10 + 6}}{\text{ = }}{{\text{3}}^{4 + n}} \cr & \Rightarrow 10 + 6 = 4 + n \cr & \Rightarrow n = 12 \cr} $$