Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Given}}\,{\text{Expression}} \cr & = \frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{{\left( {{3^2}} \right)}^n} \times {3^{n - 1}}}} \cr & = \frac{{\left( {{3^{5 \times \left( {n/5} \right)}} \times {3^{2n + 1}}} \right)}}{{\left( {{3^{2n}} \times {3^{n - 1}}} \right)}} \cr & = \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & = \frac{{{3^{\left( {n + 2n + 1} \right)}}}}{{{3^{\left( {2n + n - 1} \right)}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{\left( {3n + 1 - 3n + 1} \right)}} \cr & = {3^2} \cr & = 9 \cr} $$

$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

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

$$\eqalign{ & {\text{Given Exp}}{\text{.}} \cr & = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}} \cr} $$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

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

$$\eqalign{ & \frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{\frac{{3\left( {2 - \sqrt 3 } \right) - 2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 \,} \right)\left( {2 - \sqrt 3 } \right)}}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{6 - 3\sqrt 3 - 4 - 2\sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\left( {2 - 5\sqrt 3 } \right)}} \cr & = \frac{{2 - 5\sqrt 3 }}{{2 - 5\sqrt 3 }} \cr & = 1 \cr} $$

$$\eqalign{ & \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr & = 2\sqrt 2 - 2 - \sqrt 2 \cr & = 2\sqrt 2 - \sqrt 2 - 2 \cr & = \sqrt 2(2 - 1) - 2 \cr & = \sqrt 2 - 2 \cr} $$

$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$

$$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$

$$\eqalign{ & \frac{{{4^3} \times {5^4}}}{{{4^5}}} \cr & = \frac{{{5^4}}}{{{4^{\left( {5 - 3} \right)}}}} \cr & = \frac{{{5^4}}}{{{4^2}}} \cr & = \frac{{625}}{{16}} \cr & = 39.0625 \cr} $$