Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & \sqrt {40 + \sqrt {9\sqrt {81} } } \cr & \Rightarrow \sqrt {40 + \sqrt {9 \times 9} } \cr & \Rightarrow \sqrt {40 + 9} \cr & \Rightarrow \sqrt {49} \cr & \Rightarrow 7 \cr} $$

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:

 $$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$

Solution:

 $$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ $$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$  $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$ $$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$

Solution:

 $$\eqalign{ & {\left( {0.03125} \right)^{ - \frac{2}{5}}} \cr & = {\left[ {{{\left( {0.5} \right)}^5}} \right]^{ - \frac{2}{5}}} \cr & = {0.5^{\left[ {5 \times \left( { - \frac{2}{5}} \right)} \right]}} \cr & = {\left( {0.5} \right)^{ - 2}} \cr & = \frac{1}{{{{\left( {0.5} \right)}^2}}} \cr & = \frac{1}{{0.25}} \cr & = 4 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$

Solution:

 $$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$

Solution:

 $$\eqalign{ & \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3} \cr & \therefore \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{x^{\frac{2}{3}}} - 1 + {x^{ - \frac{2}{3}}}} \right) \cr & = \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{{\left( {{x^{\frac{1}{3}}}} \right)}^2} - {x^{ - \frac{1}{3}}}.{x^{\frac{1}{3}}} + {{\left( {{x^{ - \frac{1}{3}}}} \right)}^2}} \right) \cr & = {\left( {{x^{\frac{1}{3}}}} \right)^3} + {\left( {{x^{ - \frac{1}{3}}}} \right)^3} \cr & = x + {x^{ - 1}} \cr & = x + \frac{1}{x} \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }} + \frac{1}{{\sqrt 5 - 2}} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{3 - \sqrt 8 }} \cr & \Rightarrow {\text{Rationalising}} \cr & \Rightarrow \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 - \sqrt 6 } \right)\left( {\sqrt 7 + \sqrt 6 } \right)}} - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 - \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 - \sqrt 4 } \right)\left( {\sqrt 5 + \sqrt 4 } \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 - \sqrt 7 } \right)\left( {\sqrt 8 + \sqrt 7 } \right)}} + \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 - \sqrt 8 } \right)\left( {\sqrt 9 + \sqrt 8 } \right)}} \cr & \Rightarrow \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{1} - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{1} + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{1} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{1} + \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{1} \cr & \Rightarrow \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 + \sqrt 5 + \sqrt 4 - \sqrt 8 - \sqrt 7 + \sqrt 9 + \sqrt 8 \cr & \Rightarrow \sqrt 4 + \sqrt 9 \cr & \Rightarrow 2 + 3 \cr & \Rightarrow 5 \cr} $$