Discussion
A Train travelled a certain distance at a uniform speed. If the train had been 7 km/hr faster, it would have taken 14 hours less than the scheduled time. If the train were slower by 3 km/hr then it would have taken 10 hours more than the scheduled time. Find the distance covered by the train.
Answer: D
The distance is 900 km.
Let the speed be \(v\) km/h and the distance \(d\) km. The two conditions give
\[ \begin{aligned} \frac{d}{v} - \frac{d}{v + 7} &= 14 \Rightarrow d = 2v(v + 7) \\ \frac{d}{v - 3} - \frac{d}{v} &= 10 \Rightarrow d = \frac{10v(v - 3)}{3} \end{aligned} \]
Equating: \(6(v + 7) = 10(v - 3)\), so \(4v = 72\) and \(v = 18\) km/h. Then \(d = 2 \times 18 \times 25 = 900\) km.
Check: 900/18 = 50 h, 900/25 = 36 h (14 less) and 900/15 = 60 h (10 more).
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