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# An express train leaves Detroit at 3:00 PM and reaches Niles

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Manager
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An express train leaves Detroit at 3:00 PM and reaches Niles [#permalink]

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06 Nov 2004, 10:44
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An express train leaves Detroit at 3:00 PM and reaches Niles at 6:00 PM. The slow train leaves Niles at 1:30 PM and arrives at Detroit at 6:00 PM. If both trains travel at constant speeds, at what time do they meet?

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06 Nov 2004, 10:56
Rapidly sketched, I have 4:10PM
Let's say distance b/w 2 cities is 300km
Speed of A(fast train) = 100km/h
Speed of B(slow train) = approx 66km/h since it completes 300km in 4.5 hours.
When train A leaves, train B has done 1.5 hours thus 100km. There is now only 200km b/w A and B
Time = Distance/Speed
The sum of their speed: 166km/h
Time = 200/166 = 1 1/6 hour --> 1h10 min
Add it to time at which A left and you get 4:10PM which is time at which they cross. Lots of words for something that can be done within 2 min.
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Paul

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06 Nov 2004, 11:05
Let the distance between Detroit and Nile be D

so, D = V1*3
and D = V2*4.5 , time taken = 4.5 hrs

Now the slow train starts earlier at 1:30 pm, so distance covered during 1:30pm and 3:00 pm = d = V2 * 1.5

So distance between the trains at 3:00 pm = D - d . From here we can use relative speed.

we need to find the time that is taken with a relative speed, V1+V2

D-d = (V1+V2) . T

or V2 * 4.5 - V2 * 1.5 = (V1+V2)T

or V2*3 = (V1+V2)T

or D*3/4.5 = (D/3 + D/4.5) T

or 30/45 = 25/45T

or T = 30/25 = 6/5 hr = 1 hr 12 min

so the trains must have met at 4:12 pm.

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Director
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06 Nov 2004, 11:16
let speed of express train =x and the speed of slow train =y

we have x*3 = y*(4.5)
=> x= (3/2)*y

Relative speed to x+y = y + y(3/2) = (5/2)*y

distance y*(t-1.5). (t-1.5) is because, the slower train leaves 1.5hrs before the express train. Also the time taken (T) to cover this distance would then be

y*3 = (5/2)y * T

T= 3/2.5 = 1.2 hrs.

Meaning they will meet at 4:12 PM.

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06 Nov 2004, 11:24
Hehe venksune, I was trying to figure out how you got 4:20 at first Was about to point out the flaw but you were lightning fast at editing it. My calculation is a bit off because of the rounding of slower train's speed to 66km/h instead of 66.67km/h
4h12 should be right. Given the answer choices, it would have been an easy pick though
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Paul

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Manager
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07 Nov 2004, 00:16
Another way to do it...

They move in the 270:180 min. ratio, or 3:2

In a given time, the faster train will be moving 2/5 * 180 = 72 min.

That is, they will meet at 4:12 PM.

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07 Nov 2004, 00:16
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