Two trains start simultaneously (with uniform speeds) from two stations 270 km apart, each to the opposite station; they reach their destinations in $$6\frac{1}{4}$$ hours and 4 hours after they meet. The rate at which the slower train travels is :
a) 16 km/hr
b) 24 km/hr
c) 25 km/hr
d) 30 km/hr

Answer: b
Explanation:
$$\eqalign{ & {\text{Ratio of speeds}} \cr & {\text{ = }}\sqrt 4 :\sqrt {6\frac{1}{4}} \cr & = \sqrt 4 :\sqrt {\frac{{25}}{4}} \cr & = 2:\frac{5}{2} \cr & = 4:5 \cr }$$
Let the speeds of the two trains be 4x and 5x km/hr respectively
Then time taken by trains to meet each other
$$\eqalign{ & {\text{ = }}\left( {\frac{{270}}{{4x + 5x}}} \right){\text{hr}} \cr & {\text{ = }}\left( {\frac{{270}}{{9x}}} \right){\text{hr = }}\left( {\frac{{30}}{x}} \right){\text{hr}} \cr & {\text{Time taken by slower train to travel}} \cr & {\text{ 270 km = }}\left( {\frac{{270}}{{4x}}} \right){\text{hr}} \cr & \frac{{270}}{{4x}} = \frac{{30}}{x} + 6\frac{1}{4} \cr & \frac{{270}}{{4x}} – \frac{{30}}{x} = \frac{{25}}{4} \cr & \frac{{150}}{{4x}} = \frac{{25}}{4} \cr & 100x = 600 \cr & x = 6 \cr & {\text{Hence speed of slower train}} \cr & {\text{ = 4}}x \cr & = \,24\,{\text{km/hr}} \cr} $$