Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & N = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} \cr & = \frac{{7 + 3 - 2\sqrt {21} }}{4} \cr & = \frac{{10 - 2\sqrt {21} }}{4} \cr & = \frac{5}{2} - \frac{{\sqrt {21} }}{2} \cr & {\text{Similarly }}\frac{1}{N} = \frac{5}{2} + \frac{{\sqrt {21} }}{2} \cr & \therefore N + \frac{1}{N} \cr & = \frac{5}{2} - \frac{{\sqrt {21} }}{2} + \frac{5}{2} + \frac{{\sqrt {21} }}{2} \cr & = \frac{5}{2} + \frac{5}{2} \cr & = 5 \cr} $$

Solution:

 $$\eqalign{ & {\text{We know that}} \cr & {\text{1}}{{\text{1}}^2} = 121 \cr & {\text{Putting}} \cr & m = 11\& n = 2 \cr & {\text{we get}} \cr & {\left( {m - 1} \right)^{n + 1}} \cr & = {\left( {11 - 1} \right)^{\left( {2 + 1} \right)}} \cr & = {10^3} \cr & = 1000 \cr} $$

Solution:

Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr & {\text{Then,}} \cr & {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr & \Rightarrow {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr & \Rightarrow x = 210 \cr} $$

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:

 $$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right) \times 2\left( {5 + \sqrt 3 } \right) \times 2\left( {\sqrt 3 - 1} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 - \sqrt 3 } \right)\left( {5 + \sqrt 3 } \right) \cr & \Rightarrow 4\left( {3 - 1} \right)\left( {25 - 3} \right)\,\,\,\,\,\left[ {\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr & \therefore {\text{The required answer is }}176 \cr} $$

Solution:

 $$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$   $$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$      $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$ $$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$

Solution:

 $$\eqalign{ & {\left( {\sqrt {11} + \sqrt 5 } \right)^2} = 16 + 2\sqrt {55} \cr & {\left( {\sqrt {14} + \sqrt 2 } \right)^2} = 16 + 2\sqrt {28} \cr & {\left( {\sqrt 8 + \sqrt 8 } \right)^2} = 16 + 2\sqrt {64} \cr & {\text{It is clear that }}16 + 2\sqrt {64} {\text{ is largest}} \cr & {\text{So, }}\left( {\sqrt 8 + \sqrt 8 } \right){\text{ will be largest}}{\text{.}} \cr} $$

Solution:

 (333 + 333 + 333)(233 + 233) = 6x (3.333)(2.233) = 6x (334)(234) = 6x 634 = 6x x = 34

Solution:

 $$2\sqrt 3 = {\left( {4 \times 3} \right)^{\frac{1}{2}}} \to {12^{\frac{1}{2}}} \to {12^{\frac{2}{4}}} \to \root 4 \of {144} $$ $$2\root 4 \of 5 = \root 4 \of {\left( {5 \times 16} \right)} \to {80^{\frac{1}{4}}} \to \root 4 \of {80} $$ $$\sqrt 8 = {8^{\frac{1}{2}}} \to {8^{\frac{1}{2}}} \to \boxed{\root 4 \of {64} }\,{\text{smallest}}$$ $$3\sqrt 2 = \sqrt {18} \to {18^{\frac{1}{2}}} \to {18^{\frac{2}{4}}} \to \root 4 \of {324} $$ $$\sqrt 8 \,{\text{ is answer}}$$

Solution:

 $$\eqalign{ & \frac{1}{{\root 3 \of {12} }},\,\frac{1}{{\root 4 \of {29} }},\,\frac{1}{{\sqrt 5 }} \cr & {\text{Take LCM of roots}} = 12 \cr & \frac{1}{{{{12}^{\frac{1}{3}}}}},\,\frac{1}{{{{29}^{\frac{1}{4}}}}},\,\frac{1}{{{5^{\frac{1}{2}}}}} \cr & {\text{By multiplying by 12}} \cr & \frac{1}{{{{12}^4}}},\,\frac{1}{{{{29}^3}}},\,\frac{1}{{{5^6}}} \cr & \frac{1}{{20736}},\,\frac{1}{{24389}},\,\frac{1}{{15625}} \cr & {\text{Divisible by largest number gives lowest quotient}}{\text{.}} \cr} $$