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Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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10 Oct 2012, 04:04

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37% (01:02) wrong based on 155 sessions

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Train X departs Rockburg for Williamsland at 1:00 p.m. and train Y departs Williamsland for Rockburg at 2:00 p.m. If Rockburg and Williamsland are 200 miles apart, how much faster is train X than train Y ?

(1) Train X arrives at Williamsland at 5:00 p.m. (2) Train X and train Y pass each other at 3:30 p.m.

Statement 1 can give us the rate for train x which is 200/4=50, so its not sufficient.

Statement 2 is not sufficient beccause it doesn't give a clue about either rates, Not sufficient.

I realize its C, but in order to use it I think the equation should be like this:

2.5*50= 3/2*Ry I just need the confirmation whether my construction for the equation from statements 1 and 2 is right , because I saw an OA stating that the equation constructed from both A and B would be:

(2.5 hours)(50 mph)+ (1.5 hours)(Y’s rate) = 200 miles

I don't understand where that equation came from since the meeting distance when train x and y pass by each other is not 200 miles yet, so how is the equation above was constructed???

I would appreciate it if I can get some hint about the right equation from A and B

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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01 Aug 2013, 13:25

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This post received KUDOS

Train X departs R for W at 1:00 p.m. and train Y departs W for R at 2:00 p.m. If R and W are 200 miles apart, how much faster is train X than train Y ? We are looking for the speed of train X vs. train Y.

Speed = distance/time Distance = rate*time

(1) Train X arrives at Williamsland at 5:00 p.m. This tells us that train X traveled for four hours. The average speed of X is therefore 50 miles/hour. Still, we know nothing about the speed of Y. INSUFFICIENT

(2) Train X and train Y pass each other at 3:30 p.m. Depending on the speed of X and Y, they could have passed one another at the midpoint or at some other point along the tracks. We know nothing of their speed. INSUFFICIENT

Train X travels 2.5 hours and train Y travels 1.5 hours when they pass each other. When they do pass one another, they will have covered a total of 200 miles. d1+d2=200 distance = r*t d1: (50mph*2.5 hours) d2: (s*1.5 hours+ total distance = 200 (50*2.5)+(s*1.5)=200 125+1.5s = 200 1.5s = 75 s = 50

The speed for both trains is the same, 50 miles/hour. SUFFICIENT

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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10 Oct 2012, 10:24

madimo wrote:

Train X departs Rockburg for Williamsland at 1:00 p.m. and train Y departs Williamsland for Rockburg at 2:00 p.m. If Rockburg and Williamsland are 200 miles apart, how much faster is train X than train Y ?

(1) Train X arrives at Williamsland at 5:00 p.m. (2) Train X and train Y pass each other at 3:30 p.m.

Statement 1 can give us the rate for train x which is 200/4=50, so its not sufficient.

Statement 2 is not sufficient beccause it doesn't give a clue about either rates, Not sufficient.

I realize its C, but in order to use it I think the equation should be like this:

2.5*50= 3/2*Ry I just need the confirmation whether my construction for the equation from statements 1 and 2 is right , because I saw an OA stating that the equation constructed from both A and B would be:

(2.5 hours)(50 mph)+ (1.5 hours)(Y’s rate) = 200 miles

I don't understand where that equation came from since the meeting distance when train x and y pass by each other is not 200 miles yet, so how is the equation above was constructed???

I would appreciate it if I can get some hint about the right equation from A and B

Thank you very much

(1) Not sufficient, as we can only deduce the speed of train X. We have no information on train Y.

(2) Now we know that until they pass each other, the two trains traveled 2.5 and 2.5 hours, respectively. Again not sufficient. Either train can be faster than the other one, we just know for how long each traveled.

(1) and (2): sufficient, as we have now the speed of train X, the time it travelled, so we can find where did they meet (distance covered by each one), so we can find Y's speed and we can compare the two speeds. Of course, sufficient, and no need to compute anything.

Answer C. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 11 Oct 2012, 02:27, edited 1 time in total.

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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11 Oct 2012, 00:04

piyatiwari wrote:

Asked : how much faster is train X than train Y ?

say X's speed is X and Y's speed is Y We need to find speeds of both trains OR difference between speeds OR X-Y

(1) Train X arrives at Williamsland at 5:00 p.m.

So X travels 200 miles in 4 hours (1 PM to 5PM) X's speed : 50 mph

Can't get Y's speed from statemet (1) - Insufficient.

(2) Train X and train Y pass each other at 3:30 p.m.

At 3:30 pm, X has travelled for 2.5 hours Y has travelled for 1.5 hours

when they cross each other, they both have completed 200 miles together.

2.5X + 1.5Y = 200

we can not get X-Y from this statemet (2) - Insufficient.

combining

X : 50 mph -> from statemet (1) 2.5X + 1.5Y = 200 -> statemet (2)

We can get Y's value and X-Y.

Together they are sufficient.

Hence C.

Thank you for your explain, actually I was wondering about the extracted equation from the two statements, I didn't get why when they cross each other, they both have completed 200 miles together??

shouldn't the equation be like this

2.5*50 = 1.5y------ because when they cross each other, then they both will have the same distance, but it also means that they didn't complete the whole distance, which is 200

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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11 Oct 2012, 01:06

Expert's post

madimo wrote:

piyatiwari wrote:

Asked : how much faster is train X than train Y ?

say X's speed is X and Y's speed is Y We need to find speeds of both trains OR difference between speeds OR X-Y

(1) Train X arrives at Williamsland at 5:00 p.m.

So X travels 200 miles in 4 hours (1 PM to 5PM) X's speed : 50 mph

Can't get Y's speed from statemet (1) - Insufficient.

(2) Train X and train Y pass each other at 3:30 p.m.

At 3:30 pm, X has travelled for 2.5 hours Y has travelled for 1.5 hours

when they cross each other, they both have completed 200 miles together.

2.5X + 1.5Y = 200

we can not get X-Y from this statemet (2) - Insufficient.

combining

X : 50 mph -> from statemet (1) 2.5X + 1.5Y = 200 -> statemet (2)

We can get Y's value and X-Y.

Together they are sufficient.

Hence C.

Thank you for your explain, actually I was wondering about the extracted equation from the two statements, I didn't get why when they cross each other, they both have completed 200 miles together??

shouldn't the equation be like this

2.5*50 = 1.5y------ because when they cross each other, then they both will have the same distance, but it also means that they didn't complete the whole distance, which is 200

X started out from R and Y started out from W. The distance between R and W is 200 miles. They met each other at 3:30. HOw much distance do they need to cover together to meet? They need to cover the original distance between them. What is the original distance between them? 200 miles. So they together covered 200 miles. That's how you get the equation.

Distance covered by X + Distance covered by Y = 200 (time of X * speed of X) + (time of Y * speed of Y) = 200 2.5*50 + 1.5 * Y = 200

Think, if you are your friend are 10 m apart and you want to meet, how many meters do you need to cover together? 10 M of course. It doesn't matter whether you cover 9 and he covers 1 or you cover 2 and he covers 8 etc. But together, you need to cover 10. It is not essential that you both need to cover equal distances. If you start at different times and/or if your speed is different, you will likely cover different distances. This is the same case.

You can actually do this equation using a diagram and without making any equation. I will explain it in a while. _________________

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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11 Oct 2012, 02:34

Expert's post

madimo wrote:

Train X departs Rockburg for Williamsland at 1:00 p.m. and train Y departs Williamsland for Rockburg at 2:00 p.m. If Rockburg and Williamsland are 200 miles apart, how much faster is train X than train Y ?

(1) Train X arrives at Williamsland at 5:00 p.m. (2) Train X and train Y pass each other at 3:30 p.m.

This is what we know: Distance between X and Y initially is 200 miles. X starts at 1:00 pm and Y at 2:00 pm. We need to know the difference between speeds of X and Y.

(1) Train X arrives at Williamsland at 5:00 p.m. We know the time taken by X to cover 200 miles but not the time taken by Y. So we cannot say what the difference between their speeds is.

(2) Train X and train Y pass each other at 3:30 p.m. The two trains cross each other at 3:30. So X has traveled for 2.5 hrs and Y for 1.5 hrs and they have together covered 200 miles. But we don't know the individual distances covered by them and hence, we don't know their speeds. Think of it this way: say X is very slow. It could still start at 1:00 and meet at 3:30 if Y is very fast. On the other hand, if X is very fast, it could start at 1:00, meet at 3:30 (Y is slow) and reach the other side.

Now the problem is to figure out if together the two statements are sufficient. Note that they meet at 3:30 and X reaches the other side at 5:00 i.e. it takes 1.5 hrs to cover the remaining distance. This 'remaining distance' is the distance covered by Y when it started at 2:00 and met at 3:30 i.e. in 1.5 hrs. Basically, V=X and Y both covered the same distance in 1.5 hrs and hence their speeds are the same. So X has the same speed as Y.

I have made a diagram below to reinforce this.

Attachment:

Ques3.jpg [ 8.4 KiB | Viewed 1948 times ]

Notice that Y travels on the green line and takes 1.5 hrs to do it. X travels from the meeting point to the other end (same as the distance of the green line) in 1.5 hrs. SO we can conclude that their speeds are the same. _________________

Re: Train X departs Rockburg for Williamsland at 1:00 p.m. and [#permalink]

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25 Jun 2015, 08:16

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