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Train X is travelling at a constant speed of 30 miles per hour and Tra

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Train X is travelling at a constant speed of 30 miles per hour and Tra  [#permalink]

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New post 19 Oct 2016, 08:17
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Train X is travelling at a constant speed of 30 miles per hour and Train Y is travelling at a constant speed of 40 miles per hour. If the two trains are travelling in the same direction along the same route but Train X is 25 miles ahead of Train Y, how many hours will it be until Train Y is 10 miles ahead of Train X?


A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5
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Re: Train X is travelling at a constant speed of 30 miles per hour and Tra  [#permalink]

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New post 19 Oct 2016, 08:26
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Train x is 25 miles ahead and we want to calculate the time by which Train y will be 10 miles ahead.

So total distance to be covered by Y is = 25+10 = 35 miles

Relative speed of Train y = 40-30 = 10 miles per hour (Relative speed concept)

Total time taken = Total Distance/Speed = 35/10 = 3.5 hrs.

Option E.

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Train X is travelling at a constant speed of 30 miles per hour and Tra  [#permalink]

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New post 03 Feb 2018, 16:38
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Manonamission wrote:
Train X is travelling at a constant speed of 30 miles per hour and Train Y is travelling at a constant speed of 40 miles per hour. If the two trains are travelling in the same direction along the same route but Train X is 25 miles ahead of Train Y, how many hours will it be until Train Y is 10 miles ahead of Train X?


A. 1.5
B. 2.0
C. 2.5
D. 3.0
E. 3.5

This is a "chase" question. One approach is the "gap" method.

The distance between Trains X and Y is the gap.

In "chase" problems (one traveler is behind another and needs to catch up), the relative rate, R used is (Faster - Slower)

Distance here?
Y is 25 miles behind X, and
Y needs to get 10 miles ahead of X
25 + 10 = 35 miles (= gap)

Relative rate (speed)?

Y is faster than X. Y catches and passes X.
Use relative speed (Faster - Slower) = (Y - X)
Rate = (40 mph - 30mph) = 10 mph = \(R\)
That's the rate (speed) at which the gap closes.*

Time required to cover that distance/close the gap?

\(R * T = D\), and \(T=\frac{D}{R}\)
\(Time = \frac{D}{R}=\frac{35mi}{10mph}=3.5\) hours

Answer E

*Y actually closes the gap. X and Y both move. Y moves faster. The rate at which Y closes the gap is Y's speed - X's speed.
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Re: Train X is travelling at a constant speed of 30 miles per hour and Tra  [#permalink]

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Re: Train X is travelling at a constant speed of 30 miles per hour and Tra   [#permalink] 11 Apr 2019, 12:10
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