WholeLottaLove wrote:

Train A leaves New York at 9am Eastern Time on Monday, headed for Los Angeles at a constant rate of 40 miles per hour. On the same 3,000-mile stretch of track, Train B leaves Los Angeles at noon Eastern Time on Monday, traveling to New York at its constant rate of 60 miles per hour, with one exception: Train B is delayed for exactly 2 hours in Las Vegas (approximately 200 miles from Los Angeles) due to track maintenance. Assuming the time it takes for Train B to decelerate and re-accelerate when stopping and resuming in Las Vegas is negligible, at what time (Eastern Time) will the two trains meet?

I think I am being thrown off by Train A starting before train B. How do I figure out how long it takes for them to meet one another if one train starts off before the other?

When the trains travel towards each other, we have to consider the combined speed but only when one of them travel we have to consider only that train's speed.

1. Here, for the first 3 hours i.e., between 9 am and 12 noon only train A travels. The distance covered by it is 3*40 = 120 miles. The remaining distance is 3000-120=2880 miles

2. Also after that for exactly 2 hours train A only travels because train B stops at Las Vegas for 2 hours. The distance covered by train A during this period is 2*40=80 miles. Remaining distance is 2880-80=2800 miles

3. The distance of 2800 miles is covered with both trains moving towards each other. Therefore for this distance combined speed is to be taken. Time taken to cover 2800 miles = 2800/(60+40) = 28 hours.

4. Train A traveled for 5 hours alone and for 28 hours together with train B. The total hours traveled by A is 33 hours. Since it started at 9 am on Monday the two trains will meet 33 hours later i.e., 6 pm on Tuesday.

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Srinivasan Vaidyaraman

Sravna

http://www.sravnatestprep.com

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