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Question:

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Explanation:

Let distance between X and Y be ‘d’ km and train T travels at speed ‘s’ kmph.

Therefore, speed of train S = $\frac{3}{4}$ s kmph

At 4 pm, T must have covered ‘s’ km.

Therefore, distance between the two

= (d – s) km

Time taken for the two to cover this distance = $\frac{d - s}{s + \frac{3}{4} s}$ = $\frac{4 (d - s)}{7 s}$

Distance covered by train T by now = s + $\frac{4 (d - s)}{7 s} \times s$ = $\frac{3 s + 4 d}{7}$

By the given condition, $\frac{3 s + 4 d}{7}$ = $\frac{3}{5} d$

Solving this, we get d = 15s

To travel from X to Y, train T takes 15s/s = 15 hours

Hence, 15.

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