Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{1}{{1 + {2^{a - b}}}} + \frac{1}{{1 + {2^{b - a}}}} \cr & = \frac{1}{{1 + {2^{a - b}}}} + \frac{1}{{1 + {2^{ - \left( {a - b} \right)}}}} \cr & = \frac{1}{{1 + {2^{a - b}}}} + \frac{{{2^{a - b}}}}{{{2^{a - b}} + 1}} \cr & = \frac{{1 + {2^{a - b}}}}{{1 + {2^{a - b}}}} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{3}}\frac{1}{2} - \left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2} \times 1} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \frac{3}{4}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \times \frac{4}{3}} \right] \cr & = \frac{7}{2} - 3 \cr & = \frac{{7 - 6}}{2} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{5}}\frac{1}{3} \div 1\frac{2}{9} \times \frac{1}{4}\left( {10 + \frac{3}{{1 - \frac{1}{5}}}} \right) \cr & = \frac{{16}}{3} \div \frac{{11}}{9} \times \frac{1}{4}\left( {10 + \frac{3}{{\frac{4}{5}}}} \right) \cr & = \frac{{16}}{3} \div \frac{{11}}{9} \times \frac{1}{4}\left( {10 + \frac{{15}}{4}} \right) \cr & = \frac{{16}}{3} \times \frac{9}{{11}} \times \frac{1}{4} \times \frac{{55}}{4} \cr & = \frac{{48}}{{11}} \times \frac{1}{4} \times \frac{{55}}{4} \cr & = 15 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \,\,\,\,\,\,\,\,\,\,\frac{{{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}^2}}}{{{{\left( {\frac{7}{8}} \right)}^2} - {{\left( {\frac{5}{6}} \right)}^2}}} \cr & = \frac{{{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}^2}}}{{\left( {\frac{7}{8} + \frac{5}{6}} \right)\left( {\frac{7}{8} - \frac{5}{6}} \right)}} \cr & = \frac{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}}{{\left( {\frac{7}{8} - \frac{5}{6}} \right)}} \cr & = \left( {\frac{{41}}{{24}} \times 24} \right) \cr & = 41 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{a b}} \cr & = \frac{{4ab}}{{ab}} \cr & = 4\,\,\,\,\,\left\{ {{\text{where }}a = 469,b = 174} \right\} \cr} $$

$$\eqalign{ & 2ab \cr & = \left( {{a^2} + {b^2}} \right) - {\left( {a - b} \right)^2} \cr & = 29 - 9 = 20 \cr & \therefore ab = 10 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\frac{{\sqrt {24} + \sqrt {216} }}{{\sqrt {96} }} \cr & = \frac{{\sqrt {2 \times 2 \times 6} + \sqrt {6 \times 6 \times 6} }}{{\sqrt {4 \times 4 \times 6} }} \cr & = \frac{{2\sqrt 6 + 6\sqrt 6 }}{{4\sqrt 6 }} \cr & = \frac{8}{4} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {3 + \frac{{33}}{{64}}} \div \sqrt {9 + \frac{1}{7}} \times 2\sqrt {3\frac{1}{9}} \cr & = \sqrt {\frac{{225}}{{64}}} \div \sqrt {\frac{{64}}{7}} \times 2\sqrt {\frac{{28}}{9}} \cr & = \sqrt {\frac{{225}}{{64}} \times \frac{7}{{64}} \times \frac{{28}}{9}} \times 2 \cr & = \frac{{5 \times 7}}{{8 \times 4}} \times 2 \cr & = \frac{{35}}{{16}} \cr & = 2\frac{3}{{16}} \cr} $$

$$\eqalign{ & {\text{If }}a = 3\frac{2}{3}, \cr & b = 2\frac{1}{2}, \cr & c = 4\frac{3}{4}, \cr & d = 3\frac{1}{3}, \cr & {\text{then}} \cr & {\text{Given expression ,}} \cr & \,\,\,\,\frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \div \frac{{\left( {a - b} \right)}}{{\left( {c - d} \right)}} \cr & = \frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \times \frac{{\left( {c - d} \right)}}{{\left( {a - b} \right)}} \cr & = \frac{{\left( {a + b} \right)}}{{\left( {c + d} \right)}} \cr & = \frac{{3\frac{2}{3} + 2\frac{1}{2}}}{{4\frac{3}{4} + 3\frac{1}{3}}} \cr & = \frac{{\frac{{11}}{3} + \frac{5}{2}}}{{\frac{{19}}{4} + \frac{{10}}{3}}} \cr & = \frac{{37}}{6} \times \frac{{12}}{{97}} \cr & = \frac{{74}}{{97}} \cr} $$

$$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr} $$