Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Given equation is :}} \cr & \frac{{\left( {5x + 1} \right)}}{x} \times \frac{{\left( {4y + 3} \right)}}{4} = 20 \cr & \Leftrightarrow \left( {5x + 1} \right)\left( {4y + 3} \right) = 80x \cr & {\text{Clearly, }}x = 3{\text{ and}} \cr & {\text{ }}y = 3{\text{ satisfy}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {248 + \sqrt {52 + \sqrt {144} } } \cr & \Rightarrow \sqrt {248 + \sqrt {52 + 12} } \cr & \Rightarrow \sqrt {248 + \sqrt {64} } \cr & \Rightarrow \sqrt {256} \cr & \Rightarrow \pm 16 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} \cr & \Rightarrow \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} \cr & \Rightarrow \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}} \cr & \Rightarrow 6.4 + 0.64 + 0.064 + 0.0064 \cr & \Rightarrow 7.11 \cr} $$

$$\eqalign{ & {\text{Required differnce}} \cr & {\text{ = }}\frac{{19}}{{16}} - \frac{{16}}{{19}} = \frac{{{{19}^2} - {{16}^2}}}{{304}} \cr & = \frac{{\left( {19 + 16} \right)\left( {19 - 16} \right)}}{{304}} \cr & = \frac{{35 \times 3}}{{304}} \cr & = \frac{{105}}{{304}} \cr} $$

$$\eqalign{ & \,a = \left( {4 \div 3} \right) \div {\text{3}} \div {\text{4}} \cr & \,\,\,\,\,\,{\text{ = }}\frac{4}{3} \times \frac{1}{3} \times \frac{1}{4} \cr & \,\,\,\,\,\, = \frac{1}{9} \cr & b = 4 \div \left( {3 \div 3} \right) \div {\text{4}} \cr & \,\,\,\,\,{\text{ = 4}} \div 1 \div 4 \cr & \,\,\,\,\, = 1 \cr & c = 4 \div 3 \div \left( {3 \div 4} \right) \cr & \,\,\,\,\, = 4 \div 3 \div \frac{3}{4} \cr & \,\,\,\,\, = \frac{4}{3} \times \frac{4}{3} \cr & \,\,\,\,\, = \frac{{16}}{9} \cr & {\text{Clearly, c is the greatest}}{\text{.}} \cr} $$

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

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

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

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

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