Ratio - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \left( {\frac{1}{2}:\frac{2}{3}:\frac{3}{4}} \right) \times 12 \cr & \Rightarrow 6:8:9 \cr & \therefore {1^{{\text{st}}}}{\text{part's}}\,{\text{share}} \cr & = {\text{Rs}}{\text{.}}\,\left( {\frac{6}{{23}} \times 782} \right) \cr & = {\text{Rs}}{\text{.}}\,6 \times 34 \cr & = {\text{Rs}}{\text{.}}\,204 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & A = 2k \cr & B = 3k\,{\text{and}} \cr & C\, = 5k \cr & A's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{115}}{{100}}\,of\,2k = {\frac{{115}}{{100}} \times 2k} = \frac{{23k}}{{10}} \cr & B's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{110}}{{100}}\,of\,3k = {\frac{{110}}{{100}} \times 3k} = \frac{{33k}}{{10}} \cr & C's\,{\text{new}}\,{\text{salary}} \cr & = \frac{{120}}{{100}}\,of\,5k = {\frac{{120}}{{100}} \times 5k} = 6k \cr & \therefore {\text{New}}\,{\text{ratio}} \cr & = {\frac{{23k}}{{10}}:\frac{{33k}}{{10}}:6k} \cr & = 23:33:60 \cr} $$

$$\eqalign{ & {\text{Let}}\,40\% \,{\text{of}}\,A = \frac{2}{3}B \cr & {\text{Then}},\,\frac{{40A}}{{100}} = \frac{{2B}}{3} \cr & \Rightarrow \frac{{2A}}{5} = \frac{{2B}}{3} \cr & \Rightarrow \frac{A}{B} = {\frac{2}{3} \times \frac{5}{2}} = \frac{5}{3} \cr & \therefore A:B = 5:3 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fourth}}\,{\text{proportional}}\,{\text{to}}\,5,\,8,\,15\,{\text{be}}\,x \cr & {\text{Then}},\,5:8:15:x \cr & \Rightarrow 5x = {8 \times 15} \cr & x = \frac{{ {8 \times 15} }}{5} = 24 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{number}}\,{\text{be}}\,3x\,{\text{and}}\,5x \cr & {\text{Then}},\,\frac{{3x - 9}}{{5x - 9}} = \frac{{12}}{{23}} \cr & \Rightarrow 23\left( {3x - 9} \right) = 12\left( {5x - 9} \right) \cr & \Rightarrow 9x = 99 \cr & \Rightarrow x = 11 \cr & \therefore {\text{The}}\,{\text{smaller}}\,{\text{number}} \cr & = {3 \times 11} \cr & = 33 \cr} $$

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively
Then, sum of their values
$$\eqalign{ & = Rs.\,\left( {\frac{{25x}}{{100}} + \frac{{10 \times 2x}}{{100}} + \frac{{5 \times 3x}}{{100}}} \right) \cr & = Rs.\,\frac{{60x}}{{100}} \cr & \therefore \frac{{60x}}{{100}} = 30 \Leftrightarrow x = \frac{{30 \times 100}}{{60}} = 50 \cr & {\text{Hence,}}\,{\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{5p}}\,{\text{coins}} \cr & = \left( {3 \times 50} \right) \cr & = 150 \cr} $$

Rs. 2.80 = 280 paise
∴ Required ration = 280 : 40
= 7 : 1

Maximum possible number of tables = 6
[∵ 1200 × 6 = 7200]
Number of chairs purchased
$$\eqalign{ & {\text{ = }}\frac{{{\text{8100}} - {\text{7200}}}}{{{\text{300}}}}{\text{ = }}\frac{{{\text{900}}}}{{{\text{300}}}}{\text{ = 3}}{\text{}} \cr & {\text{Hence,}} \cr & {\text{Required ratio}} = {\text{3}}:{\text{6}} \cr & {\text{ = 1}}:{\text{2}} \cr} $$

$$\eqalign{ & x = \frac{1}{3}y\,\,{\text{and }}y = \frac{1}{2}z \cr & \frac{x}{y} = \frac{1}{3}\,{\text{and }}\frac{y}{z} = \frac{1}{2} \cr & x:y = 1:3 \cr & y:z = 1:2 \times 3\,({\text{multiply}}) \cr & i.e.\,y:z = 3:6 \cr & \therefore x:y:z = 1:3:6 \cr} $$

$$\eqalign{ & x:y = 3:1 \cr & \therefore \frac{x}{y} = \frac{3}{1} \cr & \therefore \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} \cr & \Rightarrow \frac{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} - 1} \right)}}{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} + 1} \right)}} \cr & {\text{taking }}{{\text{y}}^3}{\text{ common}} \cr & {\text{ = }}\frac{{\frac{{{x^3}}}{{{y^3}}} - 1}}{{\frac{{{x^3}}}{{{y^3}}} + 1}} \cr & \Rightarrow \frac{{27 - 1}}{{27 + 1}} \cr & \Rightarrow \frac{{26}}{{28}} \cr & \Rightarrow \frac{{13}}{{14}} \cr} $$