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
$${a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$   $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$ = \frac{{{{(0.9)}^3} + {{(0.2)}^3} + {{(0.3)}^3} - 3 \times 0.9 \times 0.2 \times 0.3}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$ = \frac{{\left( {0.9 + 0.2 + 0.3} \right)\left[ {{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9} \right]}}{{{{(0.9)}^2} + {{(0.2)}^2} + {{(0.3)}^2} - 0.9 \times 0.2 - 0.2 \times 0.3 - 0.3 \times 0.9}}$$
$$\eqalign{ & = 0.9 + 0.2 + 0.3 \cr & = 1.4 \cr} $$