Practice MCQ Questions and Answer on Surds and Indices

Solution:

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

Solution:

 $$\eqalign{ & 2 + \frac{6}{{\sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 - 2}} \cr & \Rightarrow 2 + \frac{{2 \times 3\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }} - \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{\left( {2 - \sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 + \left( {\frac{{2 - \sqrt 3 - 2 - \sqrt 3 }}{{4 - 3}}} \right) \cr & \Rightarrow 2 + 2\sqrt 3 - 2\sqrt 3 \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr & {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr & \Leftrightarrow {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr & \Leftrightarrow {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr & \Leftrightarrow {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr & \Leftrightarrow {2^x} = {2^{5.7}} \cr & \Leftrightarrow x = 5.7 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\text{I}}.\sqrt {64} + \sqrt {0.0064} + \sqrt {0.81} + \sqrt {0.0081} = 9.07 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = 8 + 0.08 + 0.9 + 0.09 \cr & = 9.07 = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & {\text{Hence, statement I is true}} \cr & {\text{II}}.\sqrt {0.010201} + \sqrt {98.01} + \sqrt {0.25} = 11.51 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sqrt {{{\left( {0.101} \right)}^2}} + \sqrt {{{\left( {9.9} \right)}^2}} + \sqrt {{{\left( {0.5} \right)}^2}} \cr & = 0.101 + 9.9 + 0.5 \cr & = 10.501 \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} \cr & \Rightarrow {\text{Statement II is not true}} \cr} $$

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:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & \Leftrightarrow x = 2.1 \cr} $$

Solution:

 $$1 + \left( {3 + 1} \right)$$  $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$ $$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$ $$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$ $$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$ $$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \left( {\sqrt 6 + \sqrt {10} - \sqrt {21} - \sqrt {35} } \right)\left( {\sqrt 6 - \sqrt {10} + \sqrt {21} - \sqrt {35} } \right) \cr & = \left\{ {\left( {\sqrt 6 - \sqrt {35} } \right) + \left( {\sqrt {10} - \sqrt {21} } \right)} \right\}\left\{ {\left( {\sqrt 6 - \sqrt {35} } \right) - \left( {\sqrt {10} - \sqrt {21} } \right)} \right\} \cr & = {\left( {\sqrt 6 - \sqrt {35} } \right)^2} - {\left( {\sqrt {10} - \sqrt {21} } \right)^2} \cr & = \left( {6 - 35 - 2\sqrt {210} } \right) - \left( {10 + 21 - 2\sqrt {210} } \right) \cr & = 41 - 2\sqrt {210} - 31 + 2\sqrt {210} \cr & = 41 - 31 \cr & = 10 \cr} $$