Practice MCQ Questions and Answer on Surds and Indices

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:

 $$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \sqrt 3 = 1.732 \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}} \cr & \Rightarrow 4 + 3 + 4\sqrt 3 \cr & \Rightarrow 7 + 4 \times 1.732 \cr & \Rightarrow 7 + 6.928 \cr & \Rightarrow 13.928 \cr} $$

Solution:

 $$\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr} $$

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:

 $$\eqalign{ & \frac{1}{{\sqrt {11 - 2\sqrt {30} } }} \cr & = \frac{1}{{\sqrt {6 + 5 - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}} }} \cr & = \frac{1}{{\sqrt 6 - \sqrt 5 }} \cr & = \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 - \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} \cr & = \sqrt 6 + \sqrt 5 \cr & \frac{3}{{\sqrt {7 - 2\sqrt {10} } }} \cr & = \frac{3}{{\sqrt {5 + 2 - 2 \times \sqrt 5 \times \sqrt 2 } }} \cr & = \frac{3}{{\sqrt 5 - \sqrt 2 }} \cr & = \frac{{3 \times \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}} \cr & = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} \cr & = \sqrt 5 + \sqrt 2 \cr & \frac{4}{{\sqrt {8 + 4\sqrt 3 } }} \cr & = \frac{4}{{\sqrt {8 + 2\sqrt {12} } }} \cr & = \frac{4}{{\sqrt {6 + 2 + 2 \times \sqrt 6 \times \sqrt 2 } }} \cr & = \frac{4}{{\sqrt {{{\left( {\sqrt 6 + \sqrt 2 } \right)}^2}} }} \cr & = \frac{{4 \times \left( {\sqrt 6 - \sqrt 2 } \right)}}{{\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 6 - \sqrt 2 } \right)}} \cr & = \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{{6 - 2}} \cr & = \sqrt 6 - \sqrt 2 \cr & \therefore {\text{Expression}} \cr & = \left( {\sqrt 6 + \sqrt 5 } \right) - \left( {\sqrt 5 + \sqrt 2 } \right) - \left( {\sqrt 6 - \sqrt 2 } \right) \cr & = \sqrt 6 + \sqrt 5 - \sqrt 5 - \sqrt 2 - \sqrt 6 + \sqrt 2 \cr & = 0 \cr} $$

Solution:

 $$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$ $$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$      $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$     $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$ $$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$    $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$   $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$ $$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$

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:

 $$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$