Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$

$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$
Formula :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$     $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$

$$\eqalign{ & {\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow 2x - 1 = - 3 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow \frac{{5 \times {5^{\frac{1}{2}}} \times {5^3}}}{{{5^{ - \frac{3}{2}}}}} = {5^{a + 2}} \cr & \Rightarrow {5^{\left( {1 + \frac{1}{2} + 3 + \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & \Rightarrow {5^6} = {5^{a + 2}} \cr & \Rightarrow a + 2 = 6 \cr & \Rightarrow a = 4 \cr} $$

$$\eqalign{ & {{\text{2}}^x} = \root 3 \of {32} \cr & \Rightarrow {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}} \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$
(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

$$\eqalign{ & {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = 2}}{{\text{5}}^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{\left( {{5^2}} \right)^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^2}^{(3x - 4)} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^{(6x - 8)}} \cr & \Rightarrow x + 3 = 6x - 8 \cr & \Rightarrow 5x = 11 \cr & \Rightarrow x = \frac{{11}}{5} \cr} $$

$$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr} $$

$$\eqalign{ & a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2} + {{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left[ {{{\left( {\sqrt 5 } \right)}^2} + 1} \right]}}{{5 - 1}} \cr & = \frac{{2\left( {5 + 1} \right)}}{4} \cr & = 3 \cr & a.b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = 1 \cr & {\text{Put value in expression}} \cr & \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & = \frac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a + b} \right)}^2} - 3ab}} \cr & = \frac{{{3^2} - 1}}{{{3^2} - 3}} \cr & = \frac{{9 - 1}}{{9 - 3}} \cr & = \frac{4}{3} \cr} $$