Quantitative Aptitude
SURDS AND INDICES MCQs
Surds & Indices, Indices And Surds, Power
Total Questions : 753
| Page 71 of 76 pages
Answer: Option E. -> 210
$$\eqalign{
& {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr
& {\text{Then,}} \cr
& {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr
& \Rightarrow {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr
& \Rightarrow x = 210 \cr} $$
$$\eqalign{
& {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr
& {\text{Then,}} \cr
& {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr
& \Rightarrow {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr
& \Rightarrow x = 210 \cr} $$
Answer: Option A. -> 8
$$\eqalign{
& x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr
& \Rightarrow x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr
& {\text{Similarly}} \cr
& y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr
& \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr
& {\text{Now, }}x + y \cr
& = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr
& = \frac{{16}}{2} \cr
& = 8 \cr} $$
$$\eqalign{
& x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr
& \Rightarrow x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr
& {\text{Similarly}} \cr
& y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr
& \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr
& {\text{Now, }}x + y \cr
& = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr
& = \frac{{16}}{2} \cr
& = 8 \cr} $$
Answer: Option B. -> 5
$$\eqalign{
& {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}} \cr
& = {5^{0.25}} \times {\left( {{5^3}} \right)^{0.25}} \cr
& = {5^{0.25}} \times {5^{\left( {3 \times 0.25} \right)}} \cr
& = {5^{0.25}} \times {5^{0.75}} \cr
& = {5^{\left( {0.25 + 0.75} \right)}} \cr
& = {5^1} \cr
& = 5 \cr} $$
$$\eqalign{
& {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}} \cr
& = {5^{0.25}} \times {\left( {{5^3}} \right)^{0.25}} \cr
& = {5^{0.25}} \times {5^{\left( {3 \times 0.25} \right)}} \cr
& = {5^{0.25}} \times {5^{0.75}} \cr
& = {5^{\left( {0.25 + 0.75} \right)}} \cr
& = {5^1} \cr
& = 5 \cr} $$
Answer: Option D. -> 0
$$\eqalign{
& \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr
& = \left[ {8 - \left( {{2^3}} \right)} \right] \cr
& = \left[ {8 - 8} \right] \cr
& = 0 \cr} $$
$$\eqalign{
& \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr
& = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr
& = \left[ {8 - \left( {{2^3}} \right)} \right] \cr
& = \left[ {8 - 8} \right] \cr
& = 0 \cr} $$
Answer: Option B. -> $${\text{1}} + \sqrt 5 + \sqrt 2 - \sqrt {10} $$
$$\eqalign{
& \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr
& = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr
& = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr
& = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr
& = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr
& = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr
& = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr
& {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr} $$
$$\eqalign{
& \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr
& = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr
& = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr
& = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr
& = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr
& = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr
& = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr
& = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr
& {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr} $$
Answer: Option C. -> 0
$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$
$$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$ $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$ $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$ $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$ $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
$$\eqalign{
& = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr
& = \sqrt {12} - 2\sqrt 3 \cr
& = 2\sqrt 3 - 2\sqrt 3 \cr
& = 0 \cr} $$
$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$
$$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$ $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$ $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$ $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$ $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
$$\eqalign{
& = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr
& = \sqrt {12} - 2\sqrt 3 \cr
& = 2\sqrt 3 - 2\sqrt 3 \cr
& = 0 \cr} $$
Answer: Option B. -> $$\frac{1}{{48}}$$
$$\eqalign{
& {\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16 \times 3} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16} \right)^{ - \frac{2}{7}}} \times {\left( 3 \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16} \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \cr
& = {\left( {16} \right)^{\left( { - \frac{7}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{7}{7}} \right)}} \cr
& = {\left( {16} \right)^{ - 1}} \times {\left( 3 \right)^{ - 1}} \cr
& = \frac{1}{{16}} \times \frac{1}{3} \cr
& = \frac{1}{{48}} \cr} $$
$$\eqalign{
& {\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16 \times 3} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16} \right)^{ - \frac{2}{7}}} \times {\left( 3 \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr
& = {\left( {16} \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \cr
& = {\left( {16} \right)^{\left( { - \frac{7}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{7}{7}} \right)}} \cr
& = {\left( {16} \right)^{ - 1}} \times {\left( 3 \right)^{ - 1}} \cr
& = \frac{1}{{16}} \times \frac{1}{3} \cr
& = \frac{1}{{48}} \cr} $$
Answer: Option A. -> 102
$$\eqalign{
& \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr
& = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr
& = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr
& = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr
& = \left( {{6^2} + {4^3} + {2^1}} \right) \cr
& = \left( {36 + 64 + 2} \right) \cr
& = 102 \cr} $$
$$\eqalign{
& \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr
& = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr
& = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr
& = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr
& = \left( {{6^2} + {4^3} + {2^1}} \right) \cr
& = \left( {36 + 64 + 2} \right) \cr
& = 102 \cr} $$
Answer: Option D. -> $$\frac{5}{3}$$
$$\eqalign{
& {{\text{2}}^x} = \root 3 \of {32} \cr
& \Rightarrow {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}} \cr
& \Rightarrow x = \frac{5}{3} \cr} $$
$$\eqalign{
& {{\text{2}}^x} = \root 3 \of {32} \cr
& \Rightarrow {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}} \cr
& \Rightarrow x = \frac{5}{3} \cr} $$
Answer: Option B. -> 4
$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$
(3, 4) are the factor of 4
If there is '+' in $$\sqrt {\,\,\,} $$ Answer is Highest value
If there is '-' in $$\sqrt {\,\,\,} $$ Answer is lowest value.
So the answer is 4
$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$
(3, 4) are the factor of 4
If there is '+' in $$\sqrt {\,\,\,} $$ Answer is Highest value
If there is '-' in $$\sqrt {\,\,\,} $$ Answer is lowest value.
So the answer is 4
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