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# Train A leaves Station 1 and train B leaves station 2, 120

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Train A leaves Station 1 and train B leaves station 2, 120 [#permalink]

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04 Jul 2007, 00:40
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Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

Last edited by GK_Gmat on 04 Jul 2007, 22:01, edited 1 time in total.
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04 Jul 2007, 01:06
Distance is 0. Train B would turn back and end up at the same mark as Train A.

This question is much more easily solved by drawing.
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04 Jul 2007, 01:10
B.

1 -meet: After 1 hour, as 30t +90t=120 t=1

2-meet: After 2 hours, let's say a- total distance traveled by TrainA then a=30T and for TrainB it is 90T=120+a Now sustitute a with 30T and
90T=120+30T T=2.

So 1 -meet was when TrainA traveled 30 miles and 2- meet is when it traveled 60 miles, so DIstance 30.
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04 Jul 2007, 06:04
UMB, please, can you explain the concept behind your calculation, because it seems to be a fairly quick way to solve that problem?
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04 Jul 2007, 06:31
The question is ambiguous. If we think two trains can "meet" when they move in the same direction, then the answer can be B (30km). But if the two trains can "meet" only if they move in opposite directions, the answer must be C (60km).

Explanation for C:
After 3 hours, train A has been driving for 90 kms, while train B has been traveling for 270 kms. Train B has traveled from station Y to station X (90 km), from X back to Y (another 90 km), and from Y again to the "meeting" position (30 km).

Therefore the distance between two meeting points is 90 - 30 = 60.
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04 Jul 2007, 07:32
This question can be easily solved by drwaing. you dont need any complex calcualtions. But make sure to read the last line, it is asked to find the distance b/w there 2 meeting points. The answer is 30.
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04 Jul 2007, 18:03
trahul4 wrote:
This question can be easily solved by drwaing. you dont need any complex calcualtions. But make sure to read the last line, it is asked to find the distance b/w there 2 meeting points. The answer is 30.

Yes, i missed the last line...
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04 Jul 2007, 20:38
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 miles away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

For starters, don't we need to convert km into mi or vice versa? Assuming so, I started like this. Am I on the right track?

120 mi = 193km
a) 30t = d
b) 90t = 193 - d
c) 3d = 193-d
d) d = 48.25, the first time the trains meet is at 3(48.25) = 144.75km.

there are now 48.25 miles left for the two trains together to travel. one train travels 3 times as fast as the other. therefore x + 3x = 48.25. x = 12.06. 3(12.06) - 12.06 = 24.12km distance b/w the trains meeting point on the first run. since the trains are traveling the same speed and distance, on the second run, wouldn't they again meet at the 144.75km mark? and wouldn't the distance b/w the two meeting points be 144.75km + 24.12km = 168.87km? I know this answer isn't even there but this is what made sense to me. where did I screw up?
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05 Jul 2007, 00:23
ggarr wrote:
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 miles away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

For starters, don't we need to convert km into mi or vice versa? Assuming so, I started like this. Am I on the right track?

120 mi = 193km
a) 30t = d
b) 90t = 193 - d
c) 3d = 193-d
d) d = 48.25, the first time the trains meet is at 3(48.25) = 144.75km.

there are now 48.25 miles left for the two trains together to travel. one train travels 3 times as fast as the other. therefore x + 3x = 48.25. x = 12.06. 3(12.06) - 12.06 = 24.12km distance b/w the trains meeting point on the first run. since the trains are traveling the same speed and distance, on the second run, wouldn't they again meet at the 144.75km mark? and wouldn't the distance b/w the two meeting points be 144.75km + 24.12km = 168.87km? I know this answer isn't even there but this is what made sense to me. where did I screw up?

GMAT questions don't require you to convert km to miles unless a conversion factor is given.
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05 Jul 2007, 00:34
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

When they meet the first time, they will have travelled a total of 120 km. Since their combined speed is 120 km/hr, they will meet at a point 30 km from 1. B will reach 1 after 120/90=4/3 hour, when A is at a point 40 km from 1, after which point the speed at which the distance (40 km) between the two decreases at a rate of 90-30=60 km/hour. They will meet for the second time 40/60 =2/3 hour later, when A is 60 km from 1.
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05 Jul 2007, 09:49
ian7777 wrote:
kevincan wrote:
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

When they meet the first time, they will have travelled a total of 120 km. Since their combined speed is 120 km/hr, they will meet at a point 30 km from 1. B will reach 1 after 120/90=4/3 hour, when A is at a point 40 km from 1, after which point the speed at which the distance (40 km) between the two decreases at a rate of 90-30=60 km/hour. They will meet for the second time 40/60 =2/3 hour later, when A is 60 km from 1.

This is EXACTLY how I did it, too.

ok, I understand that the first time the trains meet is after each has traveled 30km. I solved like this:
30t = d; 90t = 120-d
3d=120-d
4d=120
d=30.

now, on the return trip isn't each train traveling at the same speed? won't they meet at 30 km again. wouldn't that make the answer 0 (30 - 30 = 0)?

If possible, I'd like to stick with the above method to solve. This is the first thing that came to mind and will more than likely be the first thing that comes to mind on G day.
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05 Jul 2007, 09:57
ggarr wrote:
ian7777 wrote:
kevincan wrote:
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

When they meet the first time, they will have travelled a total of 120 km. Since their combined speed is 120 km/hr, they will meet at a point 30 km from 1. B will reach 1 after 120/90=4/3 hour, when A is at a point 40 km from 1, after which point the speed at which the distance (40 km) between the two decreases at a rate of 90-30=60 km/hour. They will meet for the second time 40/60 =2/3 hour later, when A is 60 km from 1.

This is EXACTLY how I did it, too.

ok, I understand that the first time the trains meet is after each has traveled 30km. I solved like this:
30t = d; 90t = 120-d
3d=120-d
4d=120
d=30.

now, on the return trip isn't each train traveling at the same speed? won't they meet at 30 km again. wouldn't that make the answer 0 (30 - 30 = 0)?

If possible, I'd like to stick with the above method to solve. This is the first thing that came to mind and will more than likely be the first thing that comes to mind on G day.

The question says that they continue after they meet, turn around at the end, and then meet again. It doesn't say that they each get to their destinations, wait until they both leave at the same time, and then meet again. I think that is how you're interpretting it.

The slow train is REALLY slow. When they meet after 1 hour, it has only gone 30 km. In the next hour, it's only gone 60 km. Meanwhile, the fast train is plugging along at 90 each hour, so it's really kickin butt.

So we look to the fast train to see what's going on. It's travelled already 90 when they meet, and has only 30 more to go before the end. That takes it all of 20 min.

In that same 20 min, the slow train has gone only 10 km.

Now the fast train turns around. It's going 90 while the slow one is going 30, but now they're in the SAME direction. There is 40 km between them, and essentially, the fast train has to close that gap at 60 km/hour (because they're going the same direction, it's travelling 90 - 30 = 60 km/hour FASTER). So it takes the fast train another 40 minutes to travel 40 km.

in that 40 min, the slow train has gone another 20 km, just slowly chugging along. All together, an hour has passed, and the slow train's travelled 30 km since the last time they met.

So that's 30 KM.
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05 Jul 2007, 14:10
ian7777 wrote:
ggarr wrote:
ian7777 wrote:
kevincan wrote:
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

When they meet the first time, they will have travelled a total of 120 km. Since their combined speed is 120 km/hr, they will meet at a point 30 km from 1. B will reach 1 after 120/90=4/3 hour, when A is at a point 40 km from 1, after which point the speed at which the distance (40 km) between the two decreases at a rate of 90-30=60 km/hour. They will meet for the second time 40/60 =2/3 hour later, when A is 60 km from 1.

This is EXACTLY how I did it, too.

ok, I understand that the first time the trains meet is after each has traveled 30km. I solved like this:
30t = d; 90t = 120-d
3d=120-d
4d=120
d=30.

now, on the return trip isn't each train traveling at the same speed? won't they meet at 30 km again. wouldn't that make the answer 0 (30 - 30 = 0)?

If possible, I'd like to stick with the above method to solve. This is the first thing that came to mind and will more than likely be the first thing that comes to mind on G day.

The question says that they continue after they meet, turn around at the end, and then meet again. It doesn't say that they each get to their destinations, wait until they both leave at the same time, and then meet again. I think that is how you're interpretting it.

The slow train is REALLY slow. When they meet after 1 hour, it has only gone 30 km. In the next hour, it's only gone 60 km. Meanwhile, the fast train is plugging along at 90 each hour, so it's really kickin butt.

So we look to the fast train to see what's going on. It's travelled already 90 when they meet, and has only 30 more to go before the end. That takes it all of 20 min.

In that same 20 min, the slow train has gone only 10 km.

Now the fast train turns around. It's going 90 while the slow one is going 30, but now they're in the SAME direction. There is 40 km between them, and essentially, the fast train has to close that gap at 60 km/hour (because they're going the same direction, it's travelling 90 - 30 = 60 km/hour FASTER). So it takes the fast train another 40 minutes to travel 40 km.

in that 40 min, the slow train has gone another 20 km, just slowly chugging along. All together, an hour has passed, and the slow train's travelled 30 km since the last time they met.

So that's 30 KM.
thanks for your post. and yes, that is how I was interpreting the question. here is my solution:

30t = d; 90t = 120-d
3d=120-d
4d=120
d=30

the trains meet at the 30km mark the 1st time around.

train B finishes it's initial run and turns around. it took train B, 80 min to complete its first run (90t=120; t=4/3; 60(4/3) = 80)

once train B turns around and heads back, train A would've been traveling for 80 min. we can set up 2 equations illustrating this:

B: 90t = 120-d
A: 30(4/3t) = d; 120t/3 = d; 40t = d

(90/40)40t=d(90/40) ==> 90t =90d/40

120-d = 90d/40 ==> 90d = 4800 - 4d ==> d = 60km (2nd meeting point)

60 - 30 = 30km
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05 Jul 2007, 21:37
ian7777 wrote:
ggarr wrote:
ian7777 wrote:
kevincan wrote:
GK_Gmat wrote:
Train A leaves Station 1 and train B leaves station 2, 120 km away. The two trains are traveling by parallel routes at 30 km/hr and 90 km/hr respectively. The trains meet and continue moving. Once a train reaches any station, it turns immediately around and moves in the opposite direction. The trains meet for the second time. What is the distance between their two meeting points.

a) 0
b) 30
c) 60
d) 90
e) 120

When they meet the first time, they will have travelled a total of 120 km. Since their combined speed is 120 km/hr, they will meet at a point 30 km from 1. B will reach 1 after 120/90=4/3 hour, when A is at a point 40 km from 1, after which point the speed at which the distance (40 km) between the two decreases at a rate of 90-30=60 km/hour. They will meet for the second time 40/60 =2/3 hour later, when A is 60 km from 1.

This is EXACTLY how I did it, too.

ok, I understand that the first time the trains meet is after each has traveled 30km. I solved like this:
30t = d; 90t = 120-d
3d=120-d
4d=120
d=30.

now, on the return trip isn't each train traveling at the same speed? won't they meet at 30 km again. wouldn't that make the answer 0 (30 - 30 = 0)?

If possible, I'd like to stick with the above method to solve. This is the first thing that came to mind and will more than likely be the first thing that comes to mind on G day.

The question says that they continue after they meet, turn around at the end, and then meet again. It doesn't say that they each get to their destinations, wait until they both leave at the same time, and then meet again. I think that is how you're interpretting it.

The slow train is REALLY slow. When they meet after 1 hour, it has only gone 30 km. In the next hour, it's only gone 60 km. Meanwhile, the fast train is plugging along at 90 each hour, so it's really kickin butt.

So we look to the fast train to see what's going on. It's travelled already 90 when they meet, and has only 30 more to go before the end. That takes it all of 20 min.

In that same 20 min, the slow train has gone only 10 km.

Now the fast train turns around. It's going 90 while the slow one is going 30, but now they're in the SAME direction. There is 40 km between them, and essentially, the fast train has to close that gap at 60 km/hour (because they're going the same direction, it's travelling 90 - 30 = 60 km/hour FASTER). So it takes the fast train another 40 minutes to travel 40 km.

in that 40 min, the slow train has gone another 20 km, just slowly chugging along. All together, an hour has passed, and the slow train's travelled 30 km since the last time they met.

So that's 30 KM.

Great explanation. Thanks.

OA is B
Re: PS: Moving Trains   [#permalink] 05 Jul 2007, 21:37
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