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# PS - 700 level - trains

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PS - 700 level - trains [#permalink]  12 Aug 2011, 15:26
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Train A left Centerville station, heading towards Dale City Station, at 3PM. Train B left Dale City Station, heading toward Centerville Station, at 3:20PM on the same day. The trains rode on straight tracks that were parallel to each other. If Train A traveled at a constant speed of 30 miles per hour and Train B traveled at a constant speed of 10 miles per hour, and the distance between the Centerville Station and Dale City Station is 90 miles, when did the trains meet each other?
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Re: PS - 700 level - trains [#permalink]  12 Aug 2011, 17:59
bschool83 wrote:
Train A left Centerville station, heading towards Dale City Station, at 3PM. Train B left Dale City Station, heading toward Centerville Station, at 3:20PM on the same day. The trains rode on straight tracks that were parallel to each other. If Train A traveled at a constant speed of 30 miles per hour and Train B traveled at a constant speed of 10 miles per hour, and the distance between the Centerville Station and Dale City Station is 90 miles, when did the trains meet each other?

From 3 Pm to 3:20 PM , train A will have travelled a distance of 20/60 * 30 = 10 miles. So at 3:20 distance remaining between train A and train B =80 miles.

Now let train A and train B meet x hours after 3:20. In x hours both trains will have travelled a total distance of 80 kms.

so, 30 X + 10X = 80

40x =80. there X =2

so trains will meet each other at 5:20.
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Re: PS - 700 level - trains [#permalink]  13 Aug 2011, 00:15
this is how i did it ...

let time be "t"
distance = speed * time
distance travelled by A = 30t
distance traveled by B = 10 ( t - 20/60) --- 20 mins later

distance traveled by A + distance traveled by B = 90
30t+ 10 (t-20/60) = 90

solving for t we'd get t= 7/3 = 2 and 1/3 = 2 h 20 mins
this when added to the original time 1500 hours would give the time reqd 1720
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Re: PS - 700 level - trains [#permalink]  13 Aug 2011, 07:25
For those who want to learn a more strategic way to solve such problems:

If two objects start moving towards each other in straight line at the same time, their speeds add up. If moving away, subtract their speeds.

Therefore index the time to 3:20. In 20 minutes, A has moved 10 miles already.

So at 3:20, train A and train B are 80 miles apart. Their speeds together is 40 mph.
Therefore, they meet in 80/40 hrs = 2 hours.

Therefore at 5:20pm

Last edited by bschool83 on 13 Aug 2011, 07:46, edited 1 time in total.
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Re: PS - 700 level - trains [#permalink]  13 Aug 2011, 07:39
bschool83 wrote:

So at 3:20, train A and train B are 80 miles apart. Their speeds together is 40 mph.
Therefore, they meet in 80/20 hrs = 2 hours.

Therefore at 5:20pm

Did you mean 80/40??
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Re: PS - 700 level - trains [#permalink]  13 Aug 2011, 07:46
yes thanks for the correction. it's been changed.
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Re: PS - 700 level - trains [#permalink]  13 Aug 2011, 07:59
Train A equation: Xa = Xao + Sa·t

Train A has 20 minutes in advance, that's 1/3 hours, so 1/3 · 30 = 10.

Starting position, Xao = 10.

Xa = 10 + 30t

Train B equation: Xb = Xbo + Sb·t

Speed is negative since it's going in the opposite direction and starting position is 90 (starts at the end of the track).

Xb = 90 + (-10t)

Condition for meeting: Xa = Xb

10 + 30t = 90 - 10t

40t = 80

t = 2 hours (from 3:20) = 5:20.
Re: PS - 700 level - trains   [#permalink] 13 Aug 2011, 07:59
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