Quantitative Aptitude
SURDS AND INDICES MCQs
Surds & Indices, Indices And Surds, Power
Total Questions : 753
| Page 70 of 76 pages
Answer: Option A. -> $$\frac{3}{7}$$
$$\eqalign{
& \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr
& = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr
& = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr
& = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr
& = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr
& = \frac{{{3^1}}}{{{7^1}}} \cr
& = \frac{3}{7} \cr} $$
$$\eqalign{
& \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr
& = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr
& = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr
& = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr
& = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr
& = \frac{{{3^1}}}{{{7^1}}} \cr
& = \frac{3}{7} \cr} $$
Answer: Option D. -> 1
$$\eqalign{
& \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr
& = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr
& = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr
& = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr
& = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr
& = \frac{8}{8} \cr
& = 1 \cr} $$
$$\eqalign{
& \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr
& = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr
& = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr
& = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr
& = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr
& = \frac{8}{8} \cr
& = 1 \cr} $$
Answer: Option B. -> 400
$$\eqalign{
& \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr
& = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr
& = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr
& = 400 \cr} $$
$$\eqalign{
& \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr
& = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr
& = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr
& = 400 \cr} $$
Answer: Option C. -> 7
$$\eqalign{
& \sqrt {40 + \sqrt {9\sqrt {81} } } \cr
& \Rightarrow \sqrt {40 + \sqrt {9 \times 9} } \cr
& \Rightarrow \sqrt {40 + 9} \cr
& \Rightarrow \sqrt {49} \cr
& \Rightarrow 7 \cr} $$
$$\eqalign{
& \sqrt {40 + \sqrt {9\sqrt {81} } } \cr
& \Rightarrow \sqrt {40 + \sqrt {9 \times 9} } \cr
& \Rightarrow \sqrt {40 + 9} \cr
& \Rightarrow \sqrt {49} \cr
& \Rightarrow 7 \cr} $$
Answer: Option B. -> 5
$$\eqalign{
& \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr
& \Rightarrow x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr
& \Rightarrow x = 5{\text{ }} \cr} $$
$$\eqalign{
& \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr
& \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr
& \Rightarrow x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr
& \Rightarrow x = 5{\text{ }} \cr} $$
Answer: Option B. -> $$\sqrt 3 - 1$$
$$\eqalign{
& \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr
& = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr
& = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr
& = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr
& = \sqrt 3 - 1 \cr} $$
$$\eqalign{
& \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr
& = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr
& = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr
& = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr
& = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr
& = \sqrt 3 - 1 \cr} $$
Answer: Option C. -> 1
$$\eqalign{
& {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr
& {\text{Then,}} \cr
& {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
$${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{
& {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr
& {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr
& {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr
& {{\text{X}}^2} = 2 \cr
& {\text{X = }}\sqrt 2 \cr
& \therefore {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr
& {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr
& {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr
& {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr
& {\text{N}} = 1 \cr} $$
$$\eqalign{
& {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr
& {\text{Then,}} \cr
& {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
$${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{
& {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr
& {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr
& {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr
& {{\text{X}}^2} = 2 \cr
& {\text{X = }}\sqrt 2 \cr
& \therefore {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr
& {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr
& {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr
& {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr
& {\text{N}} = 1 \cr} $$
Answer: Option B. -> $$\sqrt 5 - \sqrt 3 $$
$$\eqalign{
& \sqrt {8 - 2\sqrt {15} } \cr
& = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr
& = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr
& = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr
& = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$
$$\eqalign{
& \sqrt {8 - 2\sqrt {15} } \cr
& = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr
& = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr
& = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr
& = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$
Answer: Option A. -> 1
$$\eqalign{
& {\text{let}} \cr
& a = 0.73 \cr
& b = 0.27 \cr
& = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr
& = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr
& = \left( {a + b} \right) \cr
& = \left( {0.73 + 0.27} \right) \cr
& = 1 \cr} $$
$$\eqalign{
& {\text{let}} \cr
& a = 0.73 \cr
& b = 0.27 \cr
& = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr
& = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr
& = \left( {a + b} \right) \cr
& = \left( {0.73 + 0.27} \right) \cr
& = 1 \cr} $$
Answer: Option B. -> x
$$\eqalign{
& {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr
& = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr
& = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr
& = x \cr} $$
$$\eqalign{
& {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr
& = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr
& = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr
& = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr
& = x \cr} $$
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