Simplification - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{6}{{5 - \frac{5}{3}}} \div \frac{{4 - \frac{2}{{4 - \frac{1}{2}}}}}{{5 - \frac{3}{2}}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{2}{3} \times \frac{1}{2}} \right\}{\text{ }} \cr & = \frac{6}{{\left( {\frac{{10}}{3}} \right)}} \div \frac{{4 - \frac{2}{{\left( {\frac{7}{2}} \right)}}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{1}{3}} \right\} \cr & = \frac{{6 \times 3}}{{10}} \div \frac{{4 - \frac{{2 \times 2}}{7}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \times 1 \cr & = \frac{9}{5} \div \frac{{\left( {4 - \frac{4}{7}} \right)}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \cr & = \frac{9}{5} \div \left( {\frac{{24}}{7} \times \frac{2}{7}} \right) - \frac{2}{5} \cr & = \frac{9}{5} \times \frac{{49}}{{48}} - \frac{2}{5} \cr & = \frac{{147}}{{80}} - \frac{2}{5} \cr & = \frac{{147 - 32}}{{80}} \cr & = \frac{{115}}{{80}} \cr & = \frac{{23}}{{16}} \cr & = 1\frac{7}{{16}} \cr} $$

Answer: Option E

$$\eqalign{ & {\text{According to question,}} \cr & {\text{I = }}\frac{3}{5} \div \frac{5}{6} \cr & {\text{I = }}\frac{3}{5} \times \frac{6}{5} = \frac{18}{{25}} \cr & \cr & {\text{II = 3}} \div \left[ {\frac{4}{5} \times \frac{1}{6}} \right] \cr & {\text{II = 3}} \times \frac{{30}}{4} \cr & {\text{II = }}\frac{{45}}{2} \cr & \cr & {\text{III = }}\left[ {3 \div \left( {4 \div 5} \right)} \right] \div 6\, \cr & {\text{III = }}\left[ {3 \times \frac{5}{4}} \right] \times \frac{1}{6}\, \cr & {\text{III = }}\frac{5}{8} \cr & \cr & {\text{IV = 3}} \div {\text{4}}\left( {5 \div 6} \right) \cr & {\text{IV = 3}} \div {\text{4}} \times \frac{5}{6} \cr & {\text{IV = 3}} \times \frac{6}{{20}} \cr & {\text{IV = }}\frac{9}{{10}} \cr } $$

Answer: Option B

$$\eqalign{ & {\text{ = }}\, - 7m - \left[ {3n - \left\{ {8m - \left( {4n - 10m} \right)} \right\}} \right]\,\,\, \cr & = \, - 7m - \left[ {3n - \left\{ {8m - 4n + 10m} \right\}} \right]\, \cr & = - 7m - \left[ {3n - 18m + 4n} \right] \cr & = - 7m - \left[ {7n - 18m} \right] \cr & = - 7m - 7n + 18m\,\,\, \cr & = \,11m - 7n\,\, \cr} $$

Answer: Option C

$$\eqalign{ & \frac{2}{a} + \frac{1}{b} \cr & = \frac{{2b + a}}{{ab}} \cr & = \frac{6}{4} \cr & = \frac{3}{2} \cr} $$

Answer: Option A

According to question,
$$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}}$$
As we know that
$${a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$   $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$ = \frac{{{{(0.9)}^3} + {{(0.2)}^3} + {{(0.3)}^3} - 3 \times 0.9 \times 0.2 \times 0.3}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$ = \frac{{\left( {0.9 + 0.2 + 0.3} \right)\left[ {{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9} \right]}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$\eqalign{ & = 0.9 + 0.2 + 0.3 \cr & = 1.4 \cr} $$

Answer: Option C

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}} \div \frac{4}{7}\left( {\frac{2}{5} + \frac{3}{{10}}} \right){\text{of }}\frac{{\frac{1}{2} + \frac{1}{3}}}{{\frac{1}{2} - \frac{1}{3}}} \cr & = \frac{{\frac{3}{2}}}{{\frac{1}{2}}} \div \frac{4}{7}\left( {\frac{{20 + 15}}{{50}}} \right) \times \frac{{\frac{5}{6}}}{{\frac{1}{6}}} \cr & = \frac{3}{2} \times \frac{2}{1} \div \frac{4}{7} \times \frac{{35}}{{50}} \times \frac{5}{6} \times \frac{6}{1} \cr & = \frac{3}{2} \times \frac{2}{1} \div 2 \cr & = \frac{3}{2} \times 2 \div \frac{1}{2} \cr & = \frac{3}{2} \cr} $$

Answer: Option A

Given expression,
$$ = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ {2{a^3} + \left( { - 7{a^3}} \right)} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right]$$
$$\eqalign{ & = 2{a^3} - \left[ {3{a^3} + 4{b^3} - \left\{ { - 5{a^3}} \right\}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {3{a^3} + 4{b^3}{\text{ + 5}}{a^3}{\text{ + 5}}{a^3} - {\text{7}}{{\text{b}}^3}{\text{ }}} \right] \cr & = 2{a^3} - \left[ {13{a^3} - 3{b^3}} \right] \cr & = 2{a^3} - 13{a^3} + 3{b^3} \cr & = - 11{a^3} + 3{b^3} \cr} $$

Answer: Option A

$$\left[ {1 + \frac{1}{{x + 1}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 2}}} \right]$$ $$\,\left[ {1 + \frac{1}{{x + 3}}} \right]$$ $$\left[ {1 + \frac{1}{{x + 4}}} \right]$$
$$ = \left[ {\frac{{\left( {x + 1} \right) + 1}}{{x + 1}}} \right]$$  $$\left[ {\frac{{\left( {x + 2} \right) + 1}}{{x + 2}}} \right]$$ $$\left[ {\frac{{\left( {x + 3} \right) + 1}}{{x + 3}}} \right]$$ $$\left[ {\frac{{\left( {x + 4} \right) + 1}}{{x + 4}}} \right]$$
$$\eqalign{ & = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left( {\frac{{x + 3}}{{x + 2}}} \right)\left( {\frac{{x + 4}}{{x + 3}}} \right)\left( {\frac{{x + 5}}{{x + 4}}} \right) \cr & = \frac{{x + 5}}{{x + 1}} \cr} $$

Answer: Option B

$$\eqalign{ & \frac{a}{b} = \frac{4}{5}{\text{ and }}\frac{b}{c} = \frac{{15}}{{16}} \cr & \Rightarrow \left( {\frac{a}{b} \times \frac{b}{c}} \right) = \left( {\frac{4}{5} \times \frac{{15}}{{16}}} \right) \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \,\frac{{{c^2} - {a^2}}}{{{c^2} + {a^2}}} \cr & = \frac{{1 - \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}}{{1 + \left( {\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{1 - {{\left( {\frac{a}{c}} \right)}^2}}}{{1 + {{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & = \frac{{\left( {\frac{7}{{16}}} \right)}}{{\left( {\frac{{25}}{{16}}} \right)}} \cr & = \frac{7}{{25}} \cr} $$

Answer: Option A

$$\eqalign{ & {\text{According to question,}} \cr & \frac{1}{8} + \frac{1}{{{8^2}}} + \frac{1}{{{8^3}}} + \frac{1}{{{8^4}}} + \frac{1}{{{8^5}}} \cr & \Rightarrow \frac{1}{8} + \frac{1}{{64}} + \frac{1}{{512}} + \frac{1}{{4096}} + \frac{1}{{32768}} \cr} $$
  ⇒ 0.125 + 0.015625 + 0.00195313 + 0.00024414 + 0.0000305175
  ⇒ 0.143