Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\,\,\,\frac{{547.527}}{{0.0082}}{\text{ = }}x{\text{ }} \cr & \Rightarrow \frac{{547527}}{{82}} \times 10 = x \cr & \Rightarrow \frac{{547527}}{{82}} = \frac{x}{{10}} \cr} $$

Let the numbers be x and y
According to question,
$$\eqalign{ & \left( {x > y} \right) \cr & {x^2} - {y^2} = 45 \cr & \left( {x + y} \right)\left( {x - y} \right) = 45 \cr & {\text{Make factor of 45}} \cr & {\text{15}} \times 3 \cr & \,\,9 \times 5\,\,\,\,\,\,\,\,\,\,(3\,{\text{pairs}}) \cr & 45 \times 1 \cr} $$
These pairs gives the value of x and y which satisfy the given condition.

According to question,
$$\eqalign{ & \root 3 \of {{3^n}} {\text{ = 27}} \cr & \Rightarrow {\left( {{3^n}} \right)^{\frac{1}{3}}} = {\left( 3 \right)^3} \cr & \Rightarrow {3^{\frac{n}{3}}} = {3^3} \cr & \Rightarrow \frac{n}{3} = 3 \cr & \Rightarrow n = 9 \cr} $$

Let the highest temperature be x degrees
Then, lowest temperature
$$\eqalign{ & {\text{ = }}\left[ {\left( {1 + \frac{1}{3}} \right)\frac{x}{2}} \right]{\text{ degrees }} \cr & = \left( {\frac{4}{3} \times \frac{x}{2}} \right){\text{ degrees}} \cr & = \frac{{2x}}{3}{\text{ degrees}} \cr & \therefore x + \frac{{2x}}{3} = 100 \cr & \Leftrightarrow \frac{{5x}}{3} = 100 \cr & \Leftrightarrow x = \frac{{100 \times 3}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 60 \cr & {\text{So, lowest temperature}} \cr & {\text{ = }}\left( {\frac{2}{3} \times 60} \right){\text{degrees}} \cr & {\text{ = 40 degrees}} \cr} $$

Let the number of hats purchase be x,
$$\eqalign{ & {\text{Then, number of brown hats}} \cr & {\text{ = }}\frac{x}{4} \cr & {\text{Number of hats sold}} \cr & {\text{ = }}\frac{{2x}}{3} \cr & {\text{Number of hats left unsold}} \cr & {\text{ = }}\left( {x - \frac{{2x}}{3}} \right) = \frac{x}{3} \cr & {\text{Number of brown hats sold}} \cr & {\text{ = }}\frac{4}{5}{\text{ of }}\frac{x}{4} = \frac{x}{5} \cr & {\text{Number of brown hats left unsold}} \cr & {\text{ = }}\left( {\frac{x}{4} - \frac{x}{5}} \right) = \frac{x}{{20}} \cr & \therefore {\text{Required fraction}} \cr & {\text{ = }}\frac{{\left( {\frac{x}{{20}}} \right)}}{{\left( {\frac{x}{3}} \right)}} \cr & = \frac{x}{{20}} \times \frac{3}{x} \cr & = \frac{3}{{20}} \cr} $$

Let the number of men in the 1st, 2nd, 3rd and 4th battalions be x, y, z and t respectively.
Then,
$$\eqalign{ & \frac{1}{2}x = \frac{2}{3}y = \frac{3}{4}z = \frac{4}{5}t \cr & \Rightarrow x = \frac{4}{3}y, \cr & \,\,\,\,\,\,\,\,z = \frac{8}{9}y, \cr & \,\,\,\,\,\,\,\,\,t = \frac{5}{6}y{\text{ }} \cr & {\text{Now,}} \cr & x + y + z + t = 7300 \cr & \Rightarrow \frac{4}{3}y + y + \frac{8}{9}y + \frac{5}{6}y = 7300 \cr & \Rightarrow \frac{{24y + 18y + 16y + 15y}}{{18}} = 7300 \cr & \Rightarrow 73y = 7300 \times 18 \cr & \Rightarrow y = 1800 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\left[ {4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{2 + \frac{1}{4}}}}}}}} \right] \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{\frac{9}{4}}}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{4}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{\frac{{31}}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{9}{{31}}}} \cr & = 4 - \frac{5}{{\frac{{40}}{{31}}}} \cr & = 4 - \frac{{5 \times 31}}{{40}} \cr & = 4 - \frac{{31}}{8} \cr & = \frac{1}{8} \cr & {\text{According to question}} \cr & \frac{1}{8}{\text{part = 10 minutes}} \cr & {\text{1 part = 10 minutes}} \cr & \frac{3}{5}{\text{ part = 80}} \times \frac{3}{5} \cr & {\text{ = 48 minutes}} \cr} $$

$$\eqalign{ & {\text{Take first part }} \cr & \frac{{4\frac{1}{7} - 2\frac{1}{4}}}{{3\frac{1}{2} + 1\frac{1}{7}}} \cr & = \frac{{\frac{{29}}{7} - \frac{9}{4}}}{{\frac{7}{2} + \frac{8}{7}}} \cr & = \frac{{\frac{{116 - 63}}{{28}}}}{{\frac{{49 + 16}}{{14}}}} \cr & = \frac{{53}}{{28}} \times \frac{{14}}{{65}} \cr & = \frac{{53}}{{130}} \cr & {\text{The second part}} \cr & \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{5 - \frac{1}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{\frac{{25 - 1}}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{5}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{\frac{{53}}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{{24}}{{53}}}} \cr & = \frac{1}{{\frac{{106 + 24}}{{53}}}} \cr & = \frac{{53}}{{130}} \cr & {\text{According to question,}} \cr & \,\,\,\,\,\,\sqrt {\frac{{53}}{{130}} \div \frac{{53}}{{130}}} \cr & = \sqrt {\frac{{53}}{{130}} \times \frac{{130}}{{53}}} \cr & = \sqrt 1 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{5P + 8C = 145785}} \cr & \Rightarrow 3(5{\text{P + 8C) = 3}} \times {\text{145785}} \cr & \Rightarrow 15{\text{P + 24C = 437355}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \,\,\,\,\,\frac{{{{\left( {75983} \right)}^2} - {{\left( {45983} \right)}^2}}}{{75983 - 45983}} \cr & = \frac{{\left( {75983 - 45983} \right)\left( {75983 + 45983} \right)}}{{(75983 - 45983)}} \cr & = \,75983 + 45983 \cr & = 121966 \cr} $$