Two trains continuously travel between Washington DC and New York, which is 226 miles away. They start simultaneously, train A at Washington and train B at New York, and run at 60 and 40 mph respectively. The station turnaround times are negligible, however, each train stops at destination for 15 minutes. These two train will meet 10th time after
A) 36.11 hours
B) 37.11 hours
C) 38.11 hours
D) 39.11 hours
E) None of the above
OA will be given later.
Note: The question can be solved using some logic and little algebra.
Journey of Two Trains
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As nobody has posted any solution, I am giving first hint.
Hint 1: As both trains meet 10th time, the faster train has to complete 9 one-side trip to meet the another train 10th time.
Note: I hope, many viewers will be able to find solution with the above hint. If I don't get solution in another 12 hours, I shall post second hint.
Hint 1: As both trains meet 10th time, the faster train has to complete 9 one-side trip to meet the another train 10th time.
Note: I hope, many viewers will be able to find solution with the above hint. If I don't get solution in another 12 hours, I shall post second hint.
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Somehow I missed this one! Here goes nothing:
Extend the tracks so that A has a 15 mile turnaround at each city and that B has a 10 mile turnaround at each city. Now force each train to run constantly, without stops, in Gomez Addams fashion.
Call a full trip the time it takes a train to go from newly facing one city to newly facing the other city (after the turnaround). A's full trip = 241 miles and B's full trip = 236 miles.
To meet the slower train B ten times, the faster train A will complete 9 full trips, then make his tenth encounter with B somewhere in the midst of his tenth full trip. This is not a trivial deduction, but we can unpack it in another post.
When A finishes his ninth full trip, he's traveled 9*241 miles, which will take him 9*241/60 hours, or 36.15 hours.
As A begins his tenth full trip, we know that B has traveled for the same time as A, or 36.15 hours. B has traveled at 40 mph, so B's total distance = 36.15 * 40 = 1446. Each of B's trips is 236 miles, and 1446/236 has remainder 30, so B is 30 miles into his seventh trip.
Since B is 30 miles into the 226 miles that separate the cities and A has just begun his tenth full trip, the trains have (226 - 30) miles between them. Their joint rate is 100 mph, so they'll meet in (226 - 30)/100 hours or 1.96 hours. 36.15 + 1.96 = 38.11, so that looks like our answer.
Extend the tracks so that A has a 15 mile turnaround at each city and that B has a 10 mile turnaround at each city. Now force each train to run constantly, without stops, in Gomez Addams fashion.
Call a full trip the time it takes a train to go from newly facing one city to newly facing the other city (after the turnaround). A's full trip = 241 miles and B's full trip = 236 miles.
To meet the slower train B ten times, the faster train A will complete 9 full trips, then make his tenth encounter with B somewhere in the midst of his tenth full trip. This is not a trivial deduction, but we can unpack it in another post.
When A finishes his ninth full trip, he's traveled 9*241 miles, which will take him 9*241/60 hours, or 36.15 hours.
As A begins his tenth full trip, we know that B has traveled for the same time as A, or 36.15 hours. B has traveled at 40 mph, so B's total distance = 36.15 * 40 = 1446. Each of B's trips is 236 miles, and 1446/236 has remainder 30, so B is 30 miles into his seventh trip.
Since B is 30 miles into the 226 miles that separate the cities and A has just begun his tenth full trip, the trains have (226 - 30) miles between them. Their joint rate is 100 mph, so they'll meet in (226 - 30)/100 hours or 1.96 hours. 36.15 + 1.96 = 38.11, so that looks like our answer.