Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \left( {a + \frac{1}{a}} \right) = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = {6^2} \cr & \,\,\,\,\,\,\,\, = 36 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = {\left( {34} \right)^2} \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow \left( {{a^4} + \frac{1}{{{a^4}}}} \right) = 1154 \cr} $$

$$\eqalign{ & \left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = }}{\left( {\sqrt {21} } \right)^2} = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 23 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 25 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{5}}^2} \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right) \cr & = 23 \times 5 \cr & = 115 \cr} $$

  $${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{ }}$$
  $$ = 1 - \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - $$     $$\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}$$
$$\eqalign{ & = 1 - \sqrt 2 + 1 - \sqrt 2 - 1 \cr & = 1 - 2\sqrt 2 \cr} $$

$$\eqalign{ & \left( {{a^4} + \frac{1}{{{a^4}}}} \right){\text{ = 1154}} \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^4} + \frac{1}{{{a^4}}} + 2 = 1156 \cr & \Rightarrow {\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = 1156 \cr & \Rightarrow \left( {{a^2} + \frac{1}{{{a^2}}}} \right) = 34 \cr & \left( {{\text{Adding}}\,{\text{2}}\,{\text{in}}\,{\text{both}}\,{\text{sides}}} \right) \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 36 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^2} = 36 \cr & \Rightarrow a + \frac{1}{a} = 6 \cr & \Rightarrow {\left( {a + \frac{1}{a}} \right)^3} = {6^3} = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 6 = 216 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 216 - 18 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 198 \cr} $$

$$\eqalign{ & \left( {x + \frac{1}{x}} \right) = \sqrt {13} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 4 = {\left( {\sqrt {13} } \right)^2} - 4 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) = 3 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {3^3} = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times 3 = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 = 36 \cr} $$

$$\eqalign{ & \left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - {\text{4 = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow \left( {2b + \frac{1}{b}} \right) = \sqrt 6 \cr & \left( {{\text{Cubeing the both sides}}} \right) \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3\times2b\times\frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) = 0 \cr} $$

Let the missing number is x
$$\eqalign{ & {\text{Given,}} \cr & \frac{{20 + 8 \times 0.5}}{{20 - x}}{\text{ = 12}} \cr & \Rightarrow \frac{{20 + 4}}{{20 - x}} = 12 \cr & \Rightarrow \frac{{24}}{{20 - x}} = 12 \cr & \Rightarrow 20 - x = \frac{{24}}{{12}} = 2 \cr & \Rightarrow x = 20 - 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr & {\text{Hence, the number is 18}} \cr} $$

$$\eqalign{ & 0 < x < 1, \cr & {\text{Let }}x = \frac{4}{{10}} \cr & {\text{So}},\sqrt x = \frac{2}{{\sqrt {10} }}\,\& \, \cr & \,\,{x^2} = \frac{{16}}{{100}} = 0.16 \cr & {\text{Now}}, \cr & \because 0.16 < \frac{4}{{10}} < \frac{2}{{\sqrt {10} }} \cr & \therefore {x^2} < x < \sqrt x \cr} $$

$$\eqalign{ & {\text{Given,}}\frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 2}} \cr & \Rightarrow \frac{{{a^2} + {b^2}}}{{ab}} = 2 \cr & \Rightarrow {a^2} + {b^2} = 2ab \cr & \Rightarrow {a^2} + {b^2} - 2ab = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} = 0 \cr & \Rightarrow \left( {a - b} \right){\text{ = 0 }} \cr} $$

Given,
24.96² ÷ (34.11 ÷ 20.05) + 67.96 - 89.11
≈ (25)² ÷ (34 ÷ 20) + 68 - 89
≈ (25)² ÷ $$\frac{{34}}{{20}}$$ + 68 - 89
≈ 625 ÷ 1.7 + 68 - 89
≈ 367.6 + 68 - 89
≈ 367 + 68 - 89
≈ 346