Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ..... \propto \cr & {\text{Square both side}} \cr & {x^2} = a\,\root 3 \of {ab\,x} \,\,\,\,\,\left( {\because \sqrt {a\root 3 \of {ab} } ..... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}ab\,x \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr} $$

$$\eqalign{ & {\text{Expression}} \cr & \frac{{3 \times {9^{n + 1}} + 9 \times {3^{2n - 1}}}}{{9 \times {3^{2n}} - 6 \times {9^{n - 1}}}} \cr & = \frac{{3 \times {{\left( {{3^2}} \right)}^{n + 1}} + {3^2} \times {3^{2n - 1}}}}{{{3^2} \times {3^{2n}} - 6 \times {{\left( {{3^2}} \right)}^{n - 1}}}} \cr & = \frac{{{3^{2n + 2 + 1}} + {3^{2n - 1 + 2}}}}{{{3^{2n + 2}} - 6 \times {3^{2n - 2}}}} \cr & = \frac{{{3^{2n + 3}} + {3^{2n + 1}}}}{{{3^{2n + 2}} - 6 \times {3^{2n - 2}}}} \cr & = \frac{{{3^{2n + 1}}\left( {{3^2} + 1} \right)}}{{{3^{2n - 2}}\left( {{3^4} - 6} \right)}} \cr & = {3^{2n + 1 - 2n + 2}}\left( {\frac{{10}}{{75}}} \right) \cr & = \frac{{{3^3} \times 10}}{{75}} \cr & = \frac{{27 \times 10}}{{75}} \cr & = \frac{{18}}{5} \cr & = 3\frac{3}{5} \cr} $$

M = 0.1 + (0.1)² + (0.01)²
N = 0.3 + (0.03)² + (0.003)²
M + N = 0.1 + (0.1)² + (0.01)² + (0.3) + (3 × 0.01)² + (3 × 0.001)²
= 0.1 + (0.1)² + (0.01)² + 0.3 + 9 × (0.01)² + 9 × (0.001)²
= 0.4 + (0.1)² + 10 × (0.01)² + 9 × (0.001)²
= 0.4 + 0.01 + 0.001 + 0.000009
= 0.411009

a = b^p and b = c^q
∴ c = a^r = (b^p)^r = (b)^pr = (c^q)^pr = c^pqr
⇒ pqr = 1

$$\eqalign{ & x = \sqrt {1 + \frac{{\sqrt 3 }}{2}} - \sqrt {1 - \frac{{\sqrt 3 }}{2}} \cr & x = \sqrt {\frac{{2 + \sqrt 3 }}{2}} - \sqrt {\frac{{2 - \sqrt 3 }}{2}} \cr & {x^2} = \frac{{2 + \sqrt 3 }}{2} + \frac{{2 - \sqrt 3 }}{2} - \frac{2}{2}\sqrt {4 - 3} \cr & {x^2} = 1 + 1 - 1 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & {\text{At }}x = 1 \cr & \frac{{\sqrt 2 - x}}{{\sqrt 2 + x}} = \frac{{1.414 - 1}}{{1.414 + 1}} = 0.17 \cr} $$

$$\eqalign{ & {\text{I}}.\,\frac{3}{{71}} < \frac{5}{{91}} < \frac{7}{{99}} \cr & {\text{For }}{{\text{1}}^{{\text{st}}}}{\text{ 2 terms, }}\frac{3}{{71}}{\text{ and }}\frac{5}{{91}} \cr & {\text{3}} \times 91 < 5 \times 71\,\,\left( {{\text{Using cross product}}} \right) \cr & \frac{3}{{71}} < \frac{5}{{91}} \cr & {\text{Now, for }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}{{\text{3}}^{{\text{rd}}}}{\text{ term, }}\frac{5}{{91}}{\text{ and }}\frac{7}{{99}} \cr & 5 \times 99 < 7 \times 91\,\,\left( {{\text{Using cross product}}} \right) \cr & \frac{3}{{71}} < \frac{5}{{91}} < \frac{7}{{99}} \cr & {\text{Hence, statement I is true}} \cr & {\text{II}}.\,\frac{{11}}{{135}} > \frac{{12}}{{157}} > \frac{{13}}{{181}} \cr & {\text{For }}{{\text{1}}^{{\text{st}}}}{\text{ 2 terms, }}\frac{{11}}{{135}}{\text{ and }}\frac{{12}}{{157}} \cr & 157 \times 11 > 12 \times 135\,\,\left( {{\text{Using cross product}}} \right) \cr & \frac{{11}}{{135}} > \frac{{12}}{{157}} \cr & {\text{And for }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}{{\text{3}}^{{\text{rd}}}}{\text{ term}} \cr & \frac{{12}}{{157}} > \frac{{13}}{{181}} \cr & 12 \times 181 > 157 \times 13\,\,\left( {{\text{Using cross product}}} \right) \cr & \therefore \frac{{11}}{{135}} > \frac{{12}}{{157}} > \frac{{13}}{{181}} \cr & {\text{Hence, statement II is true}} \cr} $$

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

$$\eqalign{ & {\text{Expression}} \cr & = \left( {1 - \sqrt 2 } \right) + \left( {\sqrt 2 - \sqrt 3 } \right) + \left( {\sqrt 3 - \sqrt 4 } \right) + \,.\,.\,.\,.\, + \left( {\sqrt {15} - \sqrt {16} } \right) \cr & = 1 - \sqrt 2 + \sqrt 2 - \sqrt 3 + \sqrt 3 - \sqrt 4 + \,.\,.\,.\,.\, + \sqrt {15} - \sqrt {16} \cr & = 1 - \sqrt {16} \cr & = 1 - 4 \cr & = - 3 \cr} $$

$$\eqalign{ & \sqrt {20} + \sqrt {12} + \root 3 \of {729} - \frac{4}{{\sqrt 5 - \sqrt 3 }} - \sqrt {81} \cr & \Rightarrow 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{4}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right) - 9 \cr & \Rightarrow 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{{4\left( {\sqrt 5 + \sqrt 3 } \right)}}{2}} \right) - 9 \cr & \Rightarrow 2\sqrt 5 + 2\sqrt 3 + 9 - 2\sqrt 5 - 2\sqrt 3 - 9 \cr & \Rightarrow 0 \cr} $$

I. 3³³ > 33³
(3³)¹¹ > 33³
(27)¹¹ > 33³
It is clear that 3³³ > 33³, this statement is true
II. 33³ > 333
It is clear that this statement is true
III. 3³³ > 333
This statement is true
So, all I, II and III statement are true, option D is correct.