Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & \root 3 \of 6 ,\,\root 2 \of 5 {\text{ and }}\root 6 \of {12} \cr & {\text{Take LCM of }}\left( {3,\,2{\text{ and }}6} \right) = 6 \cr & {6^{\frac{{1 \times 6}}{3}}},\,{5^{\frac{{1 \times 6}}{2}}},\,{12^{\frac{{1 \times 6}}{6}}} \cr & 36,\,125{\text{ and }}12 \cr & {\text{Clearly, }}\root 2 \of 5 {\text{ is the largest}}{\text{.}} \cr} $$

$$\eqalign{ & \frac{{{{\left( {243} \right)}^{\frac{n}{5}}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} \cr & \Rightarrow \frac{{{3^{5 \times \frac{n}{5}}} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & \Rightarrow \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & \Rightarrow \frac{{{3^{n + 2n + 1}}}}{{{3^{2n + n - 1}}}} \cr & \Rightarrow \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & \Rightarrow {3^{\left( {3n + 1} \right) - \left( {3n - 1} \right)}} \cr & \Rightarrow {3^{3n + 1 - 3n + 1}} \cr & \Rightarrow {3^2} \cr & \Rightarrow 9 \cr} $$

$$\eqalign{ & \frac{{\sqrt {72} \times \sqrt {363} \times \sqrt {175} }}{{\sqrt {32} \times \sqrt {147} \times \sqrt {252} }} \cr & \Rightarrow \frac{{\sqrt {2 \times 2 \times 2 \times 3 \times 3} \times \sqrt {11 \times 11 \times 3} \times }}{{\sqrt {2 \times 2 \times 2 \times 2 \times 2} \times \sqrt {3 \times 7 \times 7} \times }}\,.\,.\,.\,.\,\frac{{\sqrt {5 \times 5 \times 7} }}{{\sqrt {2 \times 2 \times 3 \times 3 \times 7} }} \cr & \Rightarrow \frac{{6\sqrt 2 \times 11\sqrt 3 \times 5\sqrt 7 }}{{4\sqrt 2 \times 7\sqrt 3 \times 6\sqrt 7 }} \cr & \Rightarrow \frac{{6 \times 11 \times 5}}{{4 \times 7 \times 6}} \cr & \Rightarrow \frac{{55}}{{28}} \cr} $$

$$\eqalign{ & \frac{{3x - 2y}}{{2x + 3y}} = \frac{5}{6} \cr & 18x - 12y = 10x + 15y \cr & 8x = 27y \cr & \frac{x}{y} = \frac{{27}}{8} \cr & {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & = {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & = {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & = {\left( 5 \right)^2} \cr & = 25 \cr} $$

$$\eqalign{ & a\sqrt 2 + b\sqrt 3 = \sqrt {98} + \sqrt {108} - \sqrt {48} - \sqrt {72} \cr & a\sqrt 2 + b\sqrt 3 = \sqrt {7 \times 7 \times 2} + \sqrt {3 \times 3 \times 3 \times 2 \times 2} - \sqrt {2 \times 2 \times 2 \times 2 \times 3} - \sqrt {3 \times 3 \times 2 \times 2 \times 2} \cr & a\sqrt 2 + b\sqrt 3 = 7\sqrt 2 + 6\sqrt 3 - 4\sqrt 3 - 6\sqrt 2 \cr & a\sqrt 2 + b\sqrt 3 = 1\sqrt 2 + 2\sqrt 3 \cr & a = 1 \cr & b = 2 \cr} $$

$$\eqalign{ & \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} \div \frac{{\sqrt {14} + \sqrt {10} }}{{\sqrt {14} - \sqrt {10} }} + \frac{{\sqrt {10} }}{{\sqrt 5 }} \cr & = \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} \div \frac{{\frac{{\sqrt {14} + \sqrt {10} }}{{\sqrt 2 }}}}{{\frac{{\sqrt {14} - \sqrt {10} }}{{\sqrt 2 }}}} + \frac{{\sqrt {10} }}{{\sqrt 5 }} \cr & = \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} \times \frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} + \sqrt 2 \cr & = 1 + \sqrt 2 \cr} $$

3^x - 3^{x - 1} = 486
⇒ 3^{x - 1}(3 - 1) = 486
⇒ 3^{x - 1} = 243
⇒ 3^{x - 1} = 3⁵
⇒ x = 6

$$\eqalign{ & 5\sqrt 3 + \sqrt {75} = 17.32 \cr & 5\sqrt 3 + 5\sqrt 3 = 17.32 \cr & 10\sqrt 3 = 17.32\,.\,.\,.\,.\,.\,\left( 1 \right) \cr & \therefore 14\sqrt 3 + \sqrt {108} \cr & = 14\sqrt 3 + \sqrt {12 \times 9} \cr & = 14\sqrt 3 + 6\sqrt 3 \cr & = 20\sqrt 3 \cr & = 2 \times 10\sqrt 3 \cr & = 2 \times 17.32 \cr & = 34.64 \cr} $$

$$\eqalign{ & \root 3 \of {11} ,\,\sqrt 7 ,\,\root 4 \of {45} \cr & {11^{\frac{1}{3}}},\,{7^{\frac{1}{2}}},\,{45^{\frac{1}{4}}} \cr & {\text{Take LCM of 3, 2, 4 is 12}} \cr & {11^4},\,{7^6},\,{45^3} \cr & {\text{14641,}}\,{\text{117649,}}\,{\text{91125}} \cr & 117649 > 91125 > 14641 \cr & {\text{Option C is true}} \cr & \sqrt 7 > \root 4 \of {45} > \root 3 \of {11} \cr} $$