Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{{{{\left( {0.1} \right)}^2} + {{\left( {0.01} \right)}^2} + {{\left( {0.009} \right)}^2} + 2\left[ {0.1 \times 0.01 + 0.01 \times 0.009 + 0.009 \times 0.1} \right]}}{{0.01 + 0.001 + 0.0009}} \cr & = \frac{{10{{\left( {0.1 + 0.01 + 0.009} \right)}^2}}}{{\left( {0.1 + 0.01 + 0.009} \right)}} \cr & = 10 \times 0.119 \cr & = 1.19{\text{ Answer}} \cr} $$

$$\eqalign{ & \frac{4}{5} \div 3\frac{1}{4}{\text{ of }}\frac{8}{{13}} - \frac{{\frac{1}{5} - \frac{1}{8}}}{{\frac{1}{5} + \frac{1}{8}}} \times 5\frac{1}{5} + \frac{5}{6} \cr & = \frac{4}{5} \div \frac{{13}}{4} \times \frac{8}{{13}} - \frac{{\frac{1}{5} - \frac{1}{8}}}{{\frac{1}{5} + \frac{1}{8}}} \times 5\frac{1}{5} + \frac{5}{6} \cr & = \frac{2}{5} - \frac{3}{{13}} \times \frac{{26}}{5} + \frac{5}{6} \cr & = \frac{2}{5} - \frac{6}{5} + \frac{5}{6} \cr & = \frac{{12 - 36 + 25}}{{30}} \cr & = \frac{1}{{30}} \cr} $$

$$\text{I}\text{. }\underset{}{\underbracket{{100}^2-{99}^2}}+{98}^2-{97}^2+....+2^2-1^2$$
$$\eqalign{ & = \left( {100 + 99} \right)\left( {100 - 99} \right) + \left( {98 + 97} \right)\left( {98 - 97} \right) + \left( {96 + 95} \right)\left( {96 - 95} \right) + \,.\,.\,.\,.\, + \left( {2 + 1} \right)\left( {2 - 1} \right) \cr & \Rightarrow \frac{{n\left( {n + 1} \right)}}{2} = {\text{ sum of }}n{\text{ terms}} \cr & = \frac{{100 \times 101}}{2} \cr & = 5050 \cr & {\text{Statement I is true}} \cr & {\text{II}}{\text{. }}8x + \frac{8}{x} = - 16 \cr & {\text{Put }}x = - 1 \cr & {\text{Then }}{x^{197}} + {x^{ - 197}} \cr & = {\left( { - 1} \right)^{197}} + {\left( { - 1} \right)^{ - 197}} \cr & = - 1 - 1 \cr & = - 2 \cr & {\text{Statement II is incorrect}} \cr} $$

$$\eqalign{ & \frac{3}{4} \times 2\frac{2}{3} \div \frac{5}{9}{\text{ of }}1\frac{1}{5} + \frac{2}{{23}} \times 3\frac{5}{6} \div \frac{2}{7}{\text{ of }}2\frac{1}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \div \frac{5}{9}{\text{ of }}\frac{6}{5} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{7}{\text{ of }}\frac{7}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \div \frac{2}{3} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{3} \cr & = \frac{3}{4} \times \frac{8}{3} \times \frac{3}{2} + \frac{2}{{23}} \times \frac{{23}}{6} \div \frac{2}{3} \cr & = 3 + \frac{1}{2} \cr & = 3\frac{1}{2} \cr} $$

$$\eqalign{ & 3\frac{1}{3} \div 2\frac{1}{2}{\text{ of }}1\frac{3}{5} + \left( {\frac{3}{8} + \frac{1}{7} \times 1\frac{3}{4}} \right) \cr & = \frac{{10}}{3} \div \frac{5}{2} \times \frac{8}{5} + \left( {\frac{3}{8} + \frac{1}{7} \times \frac{7}{4}} \right) \cr & = \frac{{10}}{3} \times \frac{1}{4} + \frac{5}{8} \cr & = \frac{5}{6} + \frac{5}{8} \cr & = \frac{{35}}{{24}} \cr} $$

[7 + 7 × (7 + 7 ÷ 7)] + 7 ÷ 7
= [7 + 7 × (7 + 1)] + 1
= [7 + 56] + 1
= 64

$$\eqalign{ & \frac{{6.25 - 1.96}}{{1.1}} \cr & = \frac{{{{\left( {2.5} \right)}^2} - {{\left( {1.4} \right)}^2}}}{{\left( {1.1} \right)}} \cr & = \frac{{\left( {1.1} \right) \times 3.9}}{{\left( {1.1} \right)}} \cr & = 3.9{\text{ Answer}} \cr} $$

$$\eqalign{ & 225 - \left[ {42 - \left\{ {25 - \left( {18 - \overline {18 + 13} } \right)} \right\}} \right] \cr & = 225 - \left[ {42 - \left\{ {25 - \left( {18 - 18 - 13} \right)} \right\}} \right] \cr & = 225 - \left[ {42 - \left\{ {25 + 13} \right\}} \right] \cr & = 225 - \left[ {42 - 38} \right] \cr & = 221 \cr} $$

$$\eqalign{ & \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} - \frac{1}{5} \times \left( {\frac{7}{8} - \frac{5}{4}} \right)} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} - \frac{1}{5} \times \left( {\frac{{7 - 10}}{8}} \right)} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} + \frac{1}{5} \times \frac{3}{8}} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{4}{5} + \frac{3}{{40}}} \right] \cr & = \frac{{33}}{{40}} + \frac{1}{5}\left[ {\frac{{32 + 3}}{{40}}} \right] \cr & = \frac{{33}}{{40}} + \frac{7}{{40}} \cr & = \frac{{40}}{{40}} \cr & = 1 \cr} $$