Question : Two stations R and S are 400 km apart from each other. A train leaves from R to S and simultaneously another train leaves from S to R. Both trains meet after 10 hours. If the speed of the first train is 4 km/hr more than the second train, then what is the speed of the slower train?
Option 1: 18 km/hr
Option 2: 26 km/hr
Option 3: 16 km/hr
Option 4: 22 km/hr
Correct Answer: 18 km/hr
Solution :
Let the speed of the slower train be $x$ km/hr. Since the speed of the first train is 4 km/hr more than the second train, its speed is ($x$ + 4) km/hr.
Distance = Speed$\times$Time
For the first train (the faster one), the distance is ($x$ + 4) km/hr × 10 hours = 10($x$ + 4) km.
For the second train (slower one), the distance is $x$ km/hr × 10 hours = 10$x$ km
⇒ 10($x$ + 4) + 10$x$ = 400
⇒ 10$x$ + 40 + 10$x$ = 400
⇒ 20$x$ = 360
⇒ $x$ = 18 km/hr
Hence, the correct answer is 18 km/hr.
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