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The figure above shows the railway tracks between three cities A, B, and C. The distance between city A and city B is 100 kilometers and city A and city C is 240 kilometers. A train can travel from city A to city B via city C only and vice versa. Two trains are scheduled to start from city A and city B respectively at 9 AM and travel to city B and city A respectively. The speeds of the train starting from city A and city B are 20 kilometers per hour and 25 kilometers per hour respectively. If the train from city A starts half an hour early and the train from city B starts half an hour late, at what time will the two trains meet?

A. 7:10 PM
B. 8:10 PM
C. 8:23 PM
D. 8:40 PM
E. 8:53 PM

The OA is B

Route of the train from A to B: ACB = AC + CB = 240 + √(100^2 + 240^2) = 240 + 260 = 500 km
Route of the train from A to B: BCA = ACB = BC + CA = 500 km

Since from the scheduled time of their departure, 9 am, the train from A starts 1/2 an hour early (8:30 am) and train B starts 1/2 and later (9:30 am), Train from A would have traveled 1 hour before the train from B starts.

Distance covered by the train from A in 1 hr = 1 hr *20 kmph = 20 km

Total distance to be covered by both the trains = 500 - 20 = 480 kms

Since the trains are moving in the opposite direction, their speeds will be added.

Thus, relative speed = 20 + 25 = 45 kmph

Time taken to cover 480 km = 480 / 45 = 10 hrs 40 min.

Time when they meet = 8:30 am + 1 hr + 10 hrs 40 min = 8:10 pm

The correct answer: B

Hope this helps!

-Jay
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