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Train A left station T at 3:30 p.m. and traveled on straight

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Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 28 Dec 2013, 05:29
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Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30
[Reveal] Spoiler: OA

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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 28 Dec 2013, 09:38
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mohnish104 wrote:
Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

Dear mohnish104,
I'm happy to help you with this. :-)

First of all, here's a blog you may find helpful:
http://magoosh.com/gmat/2012/gmat-dista ... e-formula/

To start, we need to know how far away Train A was at the moment B started. Train B started at 4:10 pm, 40 minutes after A. Train A moves at 60 mph, which is equivalent to a mile per minute, so in 40 minutes, A has traveled 40 miles. That's how far A is.

Now, B will gain on A at the rate of their difference in speeds --- 75 mph - 60 mph = 15 mph. That's the rate at which the gap between two trains closes. Furthermore, 15 mph is a rate of 1 mile every 4 minutes, for B to close an initial gap of 40 miles will take 40*4 = 160 minutes. That time, 160 minutes is 2 hours and 40 minutes, so this amount of time after 4:10 pm, namely at 6:50 pm, is when the gap has shrunk to zero and B passes A.

Does this make sense?
Mike :-)
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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 25 Jul 2015, 16:13
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Let's see this problem with a diagram:

Train A left 40 mins before Train B at a speed of 60 mph
So distance traveled by Train A in 40 mins= 40 miles
Let's say the total distance at the point of overtaking is x
So, the time taken for covering the remaining distance x-40 by Train A=x-40/60
Time taken for covering the distance x at 75 mph by Train B=x/75

Both Time should be equal which gives us,
x-40/60=x/75
x=160 mins=2 hr 40 mins
Time when train B overtake train A = 3:30+40 mins+ 2 hr 40 mins= 6:50

Correct Answer is D.
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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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mohnish104 wrote:
Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30


Relative Speed = 75 - 60 = 15 miles per hour

Time Difference = 40 minutes = 40/60 = 2/3 hours

Headstart of Train A = Speed x Time = 60*(2/3) = 40 Miles

Time to bridge gap of 40 miles = Relative Distance / Relative Speed = 40/15 = 2 hours and 40 minutes after B started

i.e. 3:30 pm + 40 minutes + 2 hours 40 minutes = 6:50 PM

Answer: option D
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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 04 Aug 2015, 07:15
Train A starts at 3:30 at a speed of 60mph
Train B starts at 4:10 at a speed of 75mph

When train B started train A would have already covered 40 miles, since A travels 60mph which means 1 mile per minute, therefore it will travel 40 miles in 40 minutes

Now lets assume both trains meet in T hours
Therefore in the next T hours
Train A would have travelled 60T
Train B would have travelled 75T+40(extra miles covered by Train A due to starting early)

Since Distance travelled by both trains in T hours will be the same
we can say
60T=75T+40 ---(1)
Solving equation (1) we get T=160 minutes which is 2 hours and 40 minutes
Adding this to 4:10 we get 6:50

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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 14 Nov 2016, 20:28
mohnish104 wrote:
Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30


t*75=(t+2/3)*60
t=8/3 hours=2 hours 40 minutes
4:10+2:40=6:50

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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 15 Nov 2016, 10:46
For the two trains to meet, they must travel the same distance [say 'd']. Let 't' be the time taken by B to cover this distance ['d' miles]. Since A has a head-start of 40 mins. [2/3 hrs], it takes A [t+2/3] hrs. to travel d miles.
Now, from the data we have on B, d=speed*time=75*t
Also, ....................................... A, d=..................=60*[t + 2/3]
So, 75*t=60*[t + 2/3]
or t=8/3 hrs=2 hrs 40 mins
Since B started at 4:10, it catches up with A at 6:50. Ans: D

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Re: Train A left station T at 3:30 p.m. and traveled on straight [#permalink]

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New post 12 Aug 2017, 13:00
60T=(T-40)75
T=200 minutes
3:30 + 200 minutes= 6:50ANS

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Re: Train A left station T at 3:30 p.m. and traveled on straight   [#permalink] 12 Aug 2017, 13:00
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