Question
A train which is moving at an average speed of 40 km/h reaches its destination on time. When its average speed reduces to 35 km/h, then it reaches its destination 15 minutes late. The distance traveled by the train is?
Answer: Option A
$$\eqalign{
& {\text{Average speed of train}} \cr
& {\text{ = 40 km/hr}} \cr
& {\text{Reach at its destination at on time }} \cr
& {\text{New average speed of train}} \cr
& {\text{ = 35 km/h}} \cr
& {\text{Time = 15 minutes}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr
& {\text{Then distance travelled}} \cr
& {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = 70}}\,{\text{km}} \cr} $$
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$$\eqalign{
& {\text{Average speed of train}} \cr
& {\text{ = 40 km/hr}} \cr
& {\text{Reach at its destination at on time }} \cr
& {\text{New average speed of train}} \cr
& {\text{ = 35 km/h}} \cr
& {\text{Time = 15 minutes}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr
& {\text{Then distance travelled}} \cr
& {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = 70}}\,{\text{km}} \cr} $$
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