Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & 8 = {7^d} \cr & \,\,\,\,\,\, = {\left( {{6^c}} \right)^d} \cr & \,\,\,\,\,\, = {\left( {{5^b}} \right)^{cd}} \cr & \,\,\,\,\,\, = {5^{bcd}} \cr & \,\,\,\,\,\, = {\left( {{4^a}} \right)^{bcd}} \cr & \,\,\,\,\,\, = {4^{abcd}} \cr & \Rightarrow {4^{abcd}} = 8 \cr & \Rightarrow {\left( {{2^2}} \right)^{abcd}} = {2^3} \cr & \Rightarrow 2abcd = 3 \cr & \Rightarrow abcd = \frac{3}{2} \cr} $$

$$\eqalign{ & {\text{5}}\sqrt 5 \times {{\text{5}}^3} \div {{\text{5}}^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow {{\text{5}}^1} \times {5^{\frac{1}{2}}} \times {{\text{5}}^3} \div {{\text{5}}^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow {5^{1 + \frac{1}{2} + 3 - \left( { - \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & \Rightarrow {5^{1 + \frac{1}{2} + 3 + \frac{3}{2}}} = {5^{a + 2}} \cr & \Rightarrow {5^6} = {5^{a + 2}} \cr & \Rightarrow a + 2 = 6 \cr & \Rightarrow a = 4 \cr} $$

$$\frac{1}{{\sqrt 7 - \sqrt 6 }} - $$  $$\frac{1}{{\sqrt 6 - \sqrt 5 }} + $$  $$\frac{1}{{\sqrt 5 - 2 }} - $$  $$\frac{1}{{\sqrt 8 - \sqrt 7 }} + $$  $$\frac{1}{{3 - \sqrt 8 }} = ?$$
⇒ Rationalising,
$$ \Rightarrow \frac{{\sqrt 7 + \sqrt 6 }}{{\left( {\sqrt 7 + \sqrt 6 } \right)\left( {\sqrt 7 - \sqrt 6 } \right)}} - $$     $$\frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} \times $$   $$\frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}} + $$   $$\frac{{\sqrt 5 + \sqrt 4 }}{{\left( {\sqrt 5 - \sqrt 4 } \right)\left( {\sqrt 5 + \sqrt 4 } \right)}} - $$     $$\frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 - \sqrt 7 } \right)\left( {\sqrt 8 + \sqrt 7 } \right)}} + $$     $$\frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)\left( {\sqrt 9 - \sqrt 8 } \right)}}$$
$$ \Rightarrow \frac{{\sqrt 7 + \sqrt 6 }}{1} - $$   $$\frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{1} + $$   $$\frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{1} - $$   $$\frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{1} + $$   $$\frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{1}$$
$$ \Rightarrow \sqrt 7 \,+ $$  $$\sqrt 6 \,- $$  $$\sqrt 6 \,- $$   $$\sqrt 5 \,+ $$  $$\sqrt 5 \,+ $$  $$\sqrt 4 \,- $$  $$\sqrt 8 \,- $$  $$\sqrt 7 \,+ $$  $$\sqrt 9 \,+ $$  $$\sqrt 8 $$
$$\eqalign{ & \Rightarrow \sqrt 4 + \sqrt 9 \cr & \Rightarrow 2 + 3 \cr & \Rightarrow 5 \cr} $$

Given expression,
$${\frac{1}{{1 + a + {b^{ - 1}}}} + }$$   $${\frac{1}{{1 + b + {c^{ - 1}}}} + }$$   $${\frac{1}{{1 + c + {a^{ - 1}}}}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{b^{ - 1}}{{{b^{ - 1}} + 1 + {b^{ - 1}}{c^{ - 1}}}} + $$    $$\frac{1}{{a + ac + 1}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{{{b^{ - 1}}}}{{1 + {b^{ - 1}} + a}} + $$   $$\frac{a}{{a + {b^{ - 1}} + 1}}$$
$$\eqalign{ & = \frac{{1 + a + {b^{ - 1}}}}{{1 + a + {b^{ - 1}}}} \cr & = 1 \cr} $$
$$\left[ {\because abc = 1 \Rightarrow {{\left( {bc} \right)}^{ - 1}} = a \Rightarrow {b^{ - 1}}{c^{ - 1}} = a,{\text{and }}ac = {b^{ - 1}}} \right]$$

$$\eqalign{ & {{\text{3}}^{x - y}} = 27 = {3^3} \cr & \Leftrightarrow x - y = 3........(i) \cr & {3^{x + y}} = 243 = {3^5} \cr & \Leftrightarrow x + y = 5........(ii) \cr & {\text{On solving (i) and (ii) ,}} \cr & {\text{we get }}x = 4 \cr} $$

$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$

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

$$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & \Rightarrow {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & \Rightarrow 36 + 49 + 64 + 81 + 100 \cr & \Rightarrow 330 \cr} $$

$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & \Leftrightarrow x = 3y + 3 \cr & \Leftrightarrow x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & \Leftrightarrow 2y = x - 9 \cr & \Leftrightarrow x - 2y = 9......({\text{ii}}) \cr & {\text{Subtracting (i) from (ii),}} \cr & {\text{we}}\,{\text{get}}\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & \therefore x + y = 21 + 6 = 27 \cr} $$

$$\eqalign{ & {{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}} \cr & \Leftrightarrow \frac{{{{\text{2}}^{0.7x}}}}{{{{\text{3}}^{ 1.25y}}}}{\text{ = }}\frac{{{2^3}{{.2}^{\frac{1}{2}}}{{.3}^{\frac{1}{2}}}}}{{{3^3}}} \cr & \Leftrightarrow \frac{{{2^{\left( {3 + \frac{1}{2}} \right)}}}}{{{3^{\left( {3 - \frac{1}{2}} \right)}}}} = \frac{{{2^{\frac{7}{2}}}}}{{{2^{\frac{5}{2}}}}} = \frac{{{2^{3.5}}}}{{{3^{2.5}}}} \cr & \therefore 0.7x = 3.5 \Rightarrow x = \frac{{3.5}}{{0.7}}{\text{ = 5}} \cr & {\text{and }}1.25y = 2.5 \cr & \Rightarrow y = \frac{{2.5}}{{1.25}} = 2 \cr} $$