Simplification - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

$$\eqalign{ & {\text{I = }}\frac{3}{4} \div \frac{5}{6} \cr & \,\,\,\, = \frac{3}{4} \times \frac{6}{5} \cr & \,\,\,\, = \frac{9}{{10}} \cr & {\text{II = 3}} \div \left[ {\left( {4 \div 5} \right) \div 6} \right] \cr & \,\,\,\,\,\, = 3 \div \left( {\frac{4}{5} \times \frac{1}{6}} \right) \cr & \,\,\,\,\,\, = 3 \div \frac{4}{{30}} \cr & \,\,\,\,\,\, = 3 \times \frac{{30}}{4} \cr & \,\,\,\,\,\, = \frac{{45}}{2} \cr & {\text{III = }}\left[ {3 \div \left( {4 \div 5} \right)} \right] \div {\text{6}} \cr & \,\,\,\,\,\,\,\,{\text{ = }}\left[ {3 \div \frac{4}{5}} \right] \div 6 \cr & \,\,\,\,\,\,\,\, = \left[ {3 \times \frac{5}{4}} \right] \div 6 \cr & \,\,\,\,\,\,\,\, = \frac{{15}}{4} \times \frac{1}{6} \cr & \,\,\,\,\,\,\,\, = \frac{5}{8} \cr & {\text{IV = 3}} \div 4 \div \left( {5 \div 6} \right) \cr & \,\,\,\,\,\,\,\, = 3 \div 4 \div \frac{5}{6} \cr & \,\,\,\,\,\,\,\, = \frac{3}{4} \times \frac{6}{5} \cr & \,\,\,\,\,\,\,\, = \frac{9}{{10}} \cr & {\text{So, I and IV are equal}}{\text{.}} \cr} $$

Answer: Option A

According to the question,
As we know that the square of 252 is which that is near to the value of 63522.
∴ 63522 - x = 63504
⇒ x = 18

Answer: Option C

$$\eqalign{ & {\text{According to question,}} \cr & \frac{5}{{3 + \frac{3}{{1 - \frac{2}{3}}}}}\,\, \cr & \Rightarrow \frac{5}{{3 + \frac{3}{{\frac{1}{3}}}}}\, \cr & \Rightarrow \frac{5}{{3 + 9}} \cr & \Rightarrow \frac{5}{{12}} \cr} $$

Answer: Option C

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}}{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}} \cr & {\text{let ,}}1 + \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{111}}{{101}} = a \cr & \,\,\,\,\,\,\,\,1 - \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{91}}{{101}} = b \cr & \Rightarrow \frac{{{a^2} - {b^2}}}{{a + b}}\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{a + b}} \cr & \Rightarrow \left( {a - b} \right) \cr & \Rightarrow \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow \frac{{20}}{{101}} \cr} $$

Answer: Option C

$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{5}{2}of\frac{3}{4} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \frac{3}{2}\left[ {\frac{2}{3} - \frac{1}{3}} \right] \cr & = \frac{5}{2}of\frac{3}{4} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \left( {\frac{3}{2} \times \frac{1}{3}} \right) \cr & = \frac{{15}}{8} \times \frac{1}{2} \div \frac{3}{2} + \frac{1}{2} \div \frac{1}{2} \cr & = \frac{{15}}{8} \times \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times 2 \cr & = \frac{5}{8} + 1 \cr & = \,1\frac{5}{8} \cr} $$