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Question

Station A and Station B are 120 Miles apart. A train traveling from station A to station B travels exactly 1/3 of the distance when it is held up and stops for 16 minutes. When the train continues its journey, the engineer increases its average speed by 15 mph so that the train will arrive at station B on schedule. Find the average speed of the train before it stopped for 16 minutes

Answer 1

Without the 16 minute delay, the time would be 120 miles /S mph where S is the usual speed. With the delay the time can be expressed as 120 / 3S + 16 min / 60 min/hr + 2/3 x 120 / (S + 15 mph). This takes into account that 40 miles were at the usual speed, 16 min delay, and 80 miles at S + 15.

As the train was on time despite the delay, we can equate the two times, so

120 / S = 120 / 3S + 16/60 + 2/3 x 120 /(S + 15).

With some algebraic manipulation, this reduces to S² + 15S - 4500 = 0 which can be solved using the quadratic formula.

I got 60 mph as an answer.

Check math carefully!