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# M03 #09 PS : Trains

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M03 #09 PS : Trains [#permalink]

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03 Oct 2008, 01:09
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Two trains continuously travel between Washington DC and Baltimore, which is 120 miles away. They start simultaneously, train A at Washington and train B at Baltimore, and run at 30 and 90 mph respectively. The station turnaround times are negligible. What is the distance between the point where the trains meet for the first time and the point where they meet for the second time?

(A) 0
(B) 30 miles
(C) 60 miles
(D) 90 miles
(E) 120 miles

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

what is the best way to approach such problems ? I tried this way , the relative speed between the two trains is 30 + 90 = 120
Total distance between Train A and B is 120. The time taken to meet first time is = 120/120 = 1 hr.

So at the end of 1 hr. the two trains will meet. That is A would have traveled 30 miles. Again they will meet after an hour so train A would have reached 60 miles. Distance between 2 meeting points => 60-30 =30

Is the approach right ? is there a better approach to such problems ?
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Re: PS : Trains [#permalink]

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04 Oct 2008, 07:56
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I got the same answer, 30 but used a different approach. I think it might be coincidence that we got the same answer.

When the trains are heading towards or away from each other, the closing speed is in fact 120mph. After one train turns around however, the faster train is catching up to the slower train at a closing speed of 60mph.

So we agree that the trains first meet 30 miles from DC. It will take train B another 1/3 hour to reach DC and turn around. During that time, train A will have traveled another 10 miles. Now Train B is at the station in DC and Train A is 40 miles away. To find out when they next meet, 90mph * T = 30mph*T+40 miles since train B has a head start. Solving for T gives you 2/3 hour. In 2/3 of an hour, Train B will have gone 60 miles, Train A 20 miles. So they meet 60 miles from DC. 60-30=30 miles between their meeting places.

But this is the answer you got, so it only proves that my method uses more steps. Let's solve the problem assuming that train B is traveling at 80mph.

80mph + 30mph = 110mph
120miles / 110 mph = 12/11 hours
30mph * 12/11 hours = 32 8/11 miles

After another 12/11 hour, train A is now at the 65 5/11 mark. 65 5/11 - 32 8/11 = 32 8/11

Now calculate from train B's point of view. 12/11 hours * 80mph = 87 3/11 miles
87 3/11 + 32 8/11 = 120, so the first point checks out.

Now continue for another 12/11 hours at 80mph, traveling another 87 3/11 miles.
Since the train is only 32 8/11 miles from DC, the end point after that time period is 87 3/11 - 32 8/11 = 54 6/11 miles from DC.

54 6/11 != 65 5/11 so that's not where the trains will meet the second time.

Plugging in to the same method I used above, I found that the second meeting place is actually 72 miles from DC.
72 - 32 8/11 = 39 3/11

After playing around with it in Excel to test a bunch of different numbers, it seems that your method only works when the speed of the faster train is exactly 3 times that of the slower train.
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Re: M03 #09 PS : Trains [#permalink]

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03 Jan 2010, 12:06
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120 m
Washington _____________________________________ Baltimore
Train A = 30mph
Train B = 90mph

1. First meeting point :
As trains are traveling in opposite directions you can consider the sum of both speeds traveling in the same direction, this is:
90 mph + 30 mph = 120 mph
t= d/s = 120/ 120= 1 hour
Then Train A travels 30 miles when the met and Train B 90 miles

Washington ___30m_____|__________90m___________________ Baltimore

1st Meeting point

2. Second meeting point:
Train B then needs 30 more miles to get to Washington. At 90 mph it will take 30m/90 mph = 1/3 hour. At this point Train B advanced 10 more miles

Washington 0 m(Train B)____________|40 m (Train A)_________________________ Baltimore

Here problem becomes an overtaking problem. So we need to find out when Train B and A meet again. From this point we know that time when they meet have to be the same and the distance of train A is 40 more miles:

d=st
sbtb = sata +40

but time is the same when they meet so:

sbt = sat +40

90t = 30t + 40
60t = 40
T= 2/3 hour. They meet after 2/3 hour = 40 minutes

In 2/3 hour Train B travels 60 miles and Train A 20 miles, so they meet at 60 miles

Washington ________| 30m____________| 60 m___________________ Baltimore

1st and 2nd meeting point

Difference 60 -30 = 30 miles
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 06:24
W.DC|<--------------------120km------------------------>|Baltimore
A--> <--B

After 1hr, both trains (trav. in opp. directions) meet at pt C- 1st meeting pt.

30km 90km
W.DC|................>/<.................................................|Baltimore
point C
Train B gets to W.DC in 1/3 hr (30km/90km per hr); same time, Train A will now be 10km (30km x 1/3) away.

2nd meeting point:

W.DC|------------/----------A-->------------------------------|Baltimore
B--> 40km
Pt B to A = 40km; 2nd meeting pt requires that both trains spend equal time, t, to travel distance x.
t = d/s; Train A: t = d1/s1. Train B: t = d2/s2. Where d1 = 40+x; d2 = x
From d1/s1 = d2/s2 => (40+x)/90 = x/30
40+x = 3x => x = 20km

so pt D is 30km (10+20) away from pt C.
Correct response is 30km
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 06:33
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let X be distance towards Baltimore between first meeting point and 2nd time meeting point
then,
time take by train A to travel distance X = time taken by train B to cover (120-X) distance
==> X/30 = (120-X)/90 ==>X = 30

hence difference between two meeting points = 30miles
and this difference between any two crossings will be constant till the trains velocities are constant ...

Thanks

Last edited by einstein10 on 22 Mar 2010, 21:14, edited 2 times in total.
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 08:19
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amitdgr wrote:
Two trains continuously travel between Washington DC and Baltimore, which is 120 miles away. They start simultaneously, train A at Washington and train B at Baltimore, and run at 30 and 90 mph respectively. The station turnaround times are negligible. What is the distance between the point where the trains meet for the first time and the point where they meet for the second time?

(A) 0
(B) 30 miles
(C) 60 miles
(D) 90 miles
(E) 120 miles

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

what is the best way to approach such problems ? I tried this way , the relative speed between the two trains is 30 + 90 = 120
Total distance between Train A and B is 120. The time taken to meet first time is = 120/120 = 1 hr.

So at the end of 1 hr. the two trains will meet. That is A would have traveled 30 miles. Again they will meet after an hour so train A would have reached 60 miles. Distance between 2 meeting points => 60-30 =30

Is the approach right ? is there a better approach to such problems ?

I think this approach is the best. In case trains are travelling from the same station say washington and we have to find their meet point, we will take
speed as 90-30 = 60 miles/hr. Time taken will be 120/60 = 2 hrs
in 2 hrs first train has travelled 180 or 120 from W to B and 60 from B to W and first train has travelled 60 from W to B.
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 08:33
2
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amitdgr wrote:
Two trains continuously travel between Washington DC and Baltimore, which is 120 miles away. They start simultaneously, train A at Washington and train B at Baltimore, and run at 30 and 90 mph respectively. The station turnaround times are negligible. What is the distance between the point where the trains meet for the first time and the point where they meet for the second time?
(A) 0
(B) 30 miles
(C) 60 miles
(D) 90 miles
(E) 120 miles
[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions
what is the best way to approach such problems ? I tried this way , the relative speed between the two trains is 30 + 90 = 120
Total distance between Train A and B is 120. The time taken to meet first time is = 120/120 = 1 hr.
So at the end of 1 hr. the two trains will meet. That is A would have traveled 30 miles. Again they will meet after an hour so train A would have reached 60 miles. Distance between 2 meeting points => 60-30 =30
Is the approach right ? is there a better approach to such problems ?

The approach is good enough and very solvavble in 2 mins.
Trains will cross each other every hour. So distance b/w first cross and second cross will be whatever B will travel in 1 hour.
B travel first hour = 30
B travel second hour = 60 km.
Difference = 30 hence B.
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 08:52
The first meeting point is after 1 hour ( 120/(30+90)
The second m. point is after 2 hours
=>
2∙30 - 1∙30 = 30m
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 09:54
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sandeep25398 wrote:
let x be the distance from DC when trains meet first time :
then,
time taken to cover x distance for train A = time taken to cover distance (120-x) by B
==> x/30 = (120-x)/90 ==> x= 30miles

now, let Y be distance towards Baltimore between first meeting point and 2nd time meeting point
then,
time take by train A to travel distance Y = time taken by train B to cover (X+X+Y) distance
==> Y/30 = (2X+Y)/90 ==>X=Y = 30

hence difference between two meeting points = 30miles

Thanks

Sandeep i am not sure ur approach will give the correct answer if we change the speed of train to 80mph frm 90mph as previously pointed by csvobo.
How would you proceed then ??????
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Re: M03 #09 PS : Trains [#permalink]

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22 Mar 2010, 12:41
hardnstrong wrote:

Sandeep i am not sure ur approach will give the correct answer if we change the speed of train to 80mph frm 90mph as previously pointed by csvobo.
How would you proceed then ??????

you are correct, i did some mistake, now i have done the corrections. thanks for informing
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Re: M03 #09 PS : Trains [#permalink]

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23 Mar 2010, 17:00
For the first time they meet:

(30)t1 + (90)t1 = 120
=> t1 = 1 hr
=> distance from Washington DC is
d1 = 30 x t1 = 30 miles (1)

For the second time they meet:
Because train B is moving with speed of 90mph, B will get to Wash D.C and then turn around to meet train A.
Hence:

(30)t2 = (90)t2 - 120 (Minus 120 to subtract turn around distance)
=> t2 = 2 hr
=> distance from Washington DC is
d2 = 30 x t2 = 60 miles

From (1)&(2), distance between two meeting points is:
|d2 - d1| = |60 - 30| = 30 miles

Correct me if my logic is incorrect.Thanks
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Re: M03 #09 PS : Trains [#permalink]

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25 Mar 2010, 12:52
The quickest way to solve this will be

First meeting point.

Time taken to meet for the first time

120/90+30 (Approaching)
=1 hr
So train B would have covered 90 miles.

(Wash) 0---------30--------60--------90-------120(Baltimore)

Now since both the trains are at 30 mile pt, for train B to meet train A, train B must travel from 30 to 0 and then back. So the hypothetical distance between A and B is the distance from 30 to 0 and back to 30 will 60 miles.

For 2nd point.
60/90-60 (Moving in same direction)
= 1 hr.
ie 30 to 0 to 60.
So the 2nd point is at 60 miles. Diff is 30 miles.
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Re: M03 #09 PS : Trains [#permalink]

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24 May 2010, 12:28
punzoo wrote:
For 2nd point.
60/90-60 (Moving in same direction)
= 1 hr.
ie 30 to 0 to 60.
So the 2nd point is at 60 miles. Diff is 30 miles.

Why 60/90-60 ?
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Re: M03 #09 PS : Trains [#permalink]

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24 Mar 2011, 05:52
W---------x----------y------------------B

They meet first at x after 120/120 hr = 1 hr

So x = (Distance from W) = 30 Miles.

Train A reaches Baltimore in 4 hrs and restarts

Train B reaches Washington in 4/3 hr and restarts

So Train B is travelling in same direction as Train A now

And in 4/3 hrs train A reaches 4/3 * 30 = 40 Km

Hence time interval to meet again = 40/90-30 = 40/60 = 2/3 hrs from the time train B reaches Washington

So B travels 2/3* 90 miles = 60 miles then

Hence distance = 60 - 30 = 30

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Re: M03 #09 PS : Trains [#permalink]

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09 Apr 2011, 12:12
I found the most efficient way to solve this problem was drawing out a number line. It took less than 20 sec to draw out position 0, 1, and 2 and arrive at the correct answer.
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Re: M03 #09 PS : Trains [#permalink]

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28 Mar 2012, 05:51
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after 1 Hr....30mph train >>30 miles from A
after 1 Hr....90mph train >>90 miles from B

>>> this means after 1 hr they will meet for 1st time as (90+30 = 120 miles)

after 2 Hr....30mph train >>60 miles from A
after 2 Hr....90mph train >>180 miles from B - (120 from B + 60 from A)

>>> 2nd time they meet

Ans = 60-30 = 30 miles (OB)

Last edited by Bazzinga on 29 Mar 2012, 05:49, edited 1 time in total.
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Re: M03 #09 PS : Trains [#permalink]

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28 Mar 2012, 06:10
Quite Simple....
Washington to Baltimore - 120kms
A takes 30mph
B takes 90mph
A at Washington and B at Baltimore at same time

at 1st hr, they meet for the first time, A would have covered 30kms from Washington, B 90kms from Baltimore
at 2nd hr, they meet again, A would have covered 60kms from Washington, B would have covered 30kms from their first meeting point to Washington and then travels return journey of 60kms to meet at 60Kms point from Washington.

So first was at 30kms from Washington and 2nd was at 60kms from Washington...hence the difference is 30kms

Hope this is a simple method...but my worry is the more complex problems of this nature....can somebody share such problems?
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Re: M03 #09 PS : Trains [#permalink]

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28 Mar 2012, 13:33
All great answer. Helped me answer the question better. You must take note that train A has traveled an extra 10 miles after they meet. which is what i forgot.
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Re: M03 #09 PS : Trains [#permalink]

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28 Mar 2012, 22:00
Sure it is c.....

1 meeting point x/ Speed of A = 120-x /speed of B
gives x=30 miles.

now while B trvaels 30 miles to reach end point with in that 20 min A travels 10 miles.

2. Relative speed in same direction= 90-30=60
now 40/60 = 2/3 hr i.e 40 min

in 40 min B travels 60 Km i.e second meeting pt at 60 miles

so 60-30= 30 miles apart.
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Re: M03 #09 PS : Trains [#permalink]

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29 Mar 2012, 03:26
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Easy question!!!

Train speeds: Train A 30mph, Train B 90mph. Total distance is 120 miles.
in 1st hour, train A would have travelled 30 miles, train B would have travelled 90 miles.

I st hour:
Train A
--------->
30 miles
<----------------------------- Train B
90 miles
Train A and B meet at 30 miles.

2nd hour: train A would have another travelled 30miles, train B would have travelled another 90 miles.(remaining 30 miles and 60 miles on return)

Train A
--------->------------>
60miles
----------------------> Train B
60miles
second time, they meet at 60 miles.
Difference in miles between the point they met at first time and second time is 60-30= 30 miles

So the answer is B.
Re: M03 #09 PS : Trains   [#permalink] 29 Mar 2012, 03:26

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