Races And Games - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

$$\eqalign{ & A:B = 100:90 \cr & A:C = 100:72 \cr & B:C = {B \over A} \times {A \over C} = {{90} \over {100}} \times {{100} \over {72}} = {{90} \over {72}} \cr} $$
When B runs 90 m, C runs 72 m.
When B runs 100m, C run
$$\left( {{{72} \over {90}} \times 100} \right)m = 80\,m$$
∴ B can give C 20 m

Answer: Option B

$$\eqalign{ & {\text{A's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {5 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{18}}{\text{m/sec}} \cr & {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{A}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 100 m}} \cr & = \left( {100 \times \frac{{18}}{{25}}} \right)\sec \cr & = 72\sec \cr & \therefore {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{B}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 92 m}} \cr & = \left( {72 + 8} \right) = 80\sec \cr & \therefore {\text{B's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right){\text{kmph}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.14{\text{ kmph}} \cr} $$

Answer: Option C

To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.
While A covers 3 m, B covers 4 m.
While A covers 360 m, B covers $$\left( {\frac{4}{3} \times 360} \right)$$  m = 480 m.
Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.

Answer: Option D

$$\eqalign{ & A:B = 100:90 \cr & A:C = 100:87 \cr & \frac{B}{C} = \frac{B}{A} \times \frac{A}{C} \cr & \,\,\,\,\,\,\,\, = \frac{{90}}{{100}} \times \frac{{100}}{{87}} \cr & \,\,\,\,\,\,\,\, = \frac{{30}}{{29}} \cr} $$
When B runs180m, C runs $$ \left( {\frac{{29}}{{30}} \times 180} \right){\text{m}} = 174{\text{m}}$$
∴ B beats C by (180 - 174) m = 6 m

Answer: Option C

$$\eqalign{ & A:B = 60:45 \cr & A:C = 60:40 \cr & \therefore \frac{B}{C} = {\frac{B}{A} \times \frac{A}{C}} = {\frac{{45}}{{60}} \times \frac{{60}}{{40}}} \cr & = \frac{{45}}{{40}} = \frac{{90}}{{80}} = 90:80 \cr & \therefore {\text{B}}\,{\text{can}}\,{\text{give}}\,{\text{C}}\,{\text{10}}\,{\text{points}}\,{\text{in}}\,{\text{a}}\,{\text{game}}\,{\text{of}}\,{\text{90}} \cr} $$