Practice MCQ Questions and Answer on Surds and Indices

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:

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

Solution:

 $$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {169} } } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {121} } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {36} } }}{2} \cr & \Rightarrow \frac{{\sqrt {16} }}{2} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr & = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr & = \frac{{{3^1}}}{{{7^1}}} \cr & = \frac{3}{7} \cr} $$

Solution:

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

Solution:

 We know that 112 = 121. Putting m = 11 and n = 2, we get: (m - 1)n + 1 = (11 - 1)(2 + 1) = 103 = 1000

Solution:

 $$\frac{2}{{\sqrt 7 + \sqrt 5 }} + \frac{7}{{\sqrt {12} - \sqrt 5 }} - \frac{5}{{\sqrt {12} - \sqrt 7 }}$$ $$ = \frac{2}{{\sqrt 7 + \sqrt 5 }} \times $$  $$\frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} + $$  $$\frac{7}{{\sqrt {12} - \sqrt 5 }} \times $$  $$\frac{{\sqrt {12} + \sqrt 5 }}{{\sqrt {12} + \sqrt 5 }} - $$  $$\left( {\frac{5}{{\sqrt {12} - \sqrt 7 }} \times \frac{{\sqrt {12} + \sqrt 7 }}{{\sqrt {12} + \sqrt 7 }}} \right)$$ $$ = \frac{{2\left( {\sqrt 7 - \sqrt 5 } \right)}}{2} + $$   $$\frac{{7\left( {\sqrt {12} + \sqrt 5 } \right)}}{7} - $$   $$\frac{{5\left( {\sqrt {12} + \sqrt 7 } \right)}}{5}$$ $$\eqalign{ &= \sqrt 7 - \sqrt 5 + \sqrt {12} + \sqrt 5 - \sqrt {12} - \sqrt 7 \cr &= 0 \cr} $$

Solution:

 a = bp and b = cq ∴ c = ar = (bp)r = (b)pr = (cq)pr = cpqr ⇒ pqr = 1

Solution:

 $$\frac{{{{1.5}^3} + {{4.7}^3} + {{3.8}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}$$ $$ = \frac{{\left( {1.5 + 4.7 + 3.8} \right)\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}$$ $$\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right]$$ $$\eqalign{ & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr} $$