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Trains

by kvamsy » Thu Nov 12, 2009 5:36 am
At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.m. the same day. If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour, which of the following represents the speed of the faster train, in miles per hour?

Please provide with solution for the above
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by jaroth04 » Thu Nov 12, 2009 6:25 am
First, I found how quickly two trains moving at the *same* speed could cover that ground:

287 miles / 3.5 hours = 82 miles per hour

That means the trains are moving together at an average of 82 miles per hour. If they were moving at the same speed, that would mean they were each moving 82 / 2 = 41 miles per hour.

But we're given in the problem that the difference between the two trains is 6MPH. To account for that, I add 6 / 2 to the faster train's speed and subtract 6 / 2 from the slower train's speed:

41 + 3 = 44 MPH, 41 - 3 = 38 MPH

So the faster train is moving at 44 MPH.

Back substituting, we get:

44 * 3.5 + 38 * 3.5 = 287

Not sure if there is a trick for the general form of this problem.
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by kvamsy » Thu Nov 12, 2009 6:30 am
jaroth04 wrote:First, I found how quickly two trains moving at the *same* speed could cover that ground:

287 miles / 3.5 hours = 82 miles per hour

That means the trains are moving together at an average of 82 miles per hour. If they were moving at the same speed, that would mean they were each moving 82 / 2 = 41 miles per hour.

But we're given in the problem that the difference between the two trains is 6MPH. To account for that, I add 6 / 2 to the faster train's speed and subtract 6 / 2 from the slower train's speed:

41 + 3 = 44 MPH, 41 - 3 = 38 MPH

So the faster train is moving at 44 MPH.

Back substituting, we get:

44 * 3.5 + 38 * 3.5 = 287

Not sure if there is a trick for the general form of this problem.

Can you please let me know how did you get 3.5 hrs.
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Re: Trains

by palvarez » Thu Nov 12, 2009 9:30 am
kvamsy wrote:At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.m. the same day. If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour, which of the following represents the speed of the faster train, in miles per hour?

Please provide with solution for the above
a > b
a - b = 6
a + b = relative speed = 287/3.5 = 82

10 am to 1:30 pm = 3.5 hrs
what is a?

a = (82+6)/2 = 44
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Re: Trains

by Stuart@KaplanGMAT » Thu Nov 12, 2009 12:50 pm
kvamsy wrote:At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.m. the same day. If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour, which of the following represents the speed of the faster train, in miles per hour?

Please provide with solution for the above
The algebraic solutions are great. Here's a good general rule to remember:

if two objects are moving in opposite directions, always ADD the individual speeds to get total speed;

if two objects are moving in the same direction, always SUBTRACT the individual speeds to get relative speed.

We could have easily solved this question with backsolving had you provided the choices (which is why you should ALWAYS post the full question in these forums).
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Re: Trains

by kvamsy » Fri Nov 13, 2009 12:08 am
Stuart Kovinsky wrote:
kvamsy wrote:At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.m. the same day. If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour, which of the following represents the speed of the faster train, in miles per hour?

Please provide with solution for the above
The algebraic solutions are great. Here's a good general rule to remember:

if two objects are moving in opposite directions, always ADD the individual speeds to get total speed;

if two objects are moving in the same direction, always SUBTRACT the individual speeds to get relative speed.

We could have easily solved this question with backsolving had you provided the choices (which is why you should ALWAYS post the full question in these forums).
Thanks for the above , Solutions are
a)43 b)44 c)45 d) 50 e) 38
Could you please provide how can i back solve the problem here
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Re: Trains

by Stuart@KaplanGMAT » Fri Nov 13, 2009 9:47 am
kvamsy wrote:At 10 a.m. two trains started traveling toward each other from stations 287 miles apart. They passed each other at 1:30 p.m. the same day. If the average speed of the faster train exceeded the average speed of the slower train by 6 miles per hour, which of the following represents the speed of the faster train, in miles per hour?

Please provide with solution for the above

a)38
b)43
c)44
d)45
e)50
OK.. backsolving.

The correct answer is the speed of the faster train and we know that the slower train is 6 mph less.

When we backsolve, we generally start with B or D. Here, D looks like a simpler number, so let's work with 45.

If the fast speed is 45, then the slow speed is 39. Since the trains are travelling directly towards each other, we add their speeds together to get a combined rate of 84.

We know that the time is 3.5 hours (10am to 1:30pm).

So, since d = r*t, we can solve for distance:

d = 84 * (7/2)
d = 42 * 7 = 294

Is that how the story ends? No - according to the question, they only travelled 287 miles. Therefore, our speed is too high: eliminate D and E.

Now we have two choices: we can either plug in B or we can try to reason out the correct answer.

If you can come up with a quick way to get from your result to the correct result, do it; if you don't see a quick rationale, then plug in.

Plugging in: next up we try B. If the fast speed is 43, then the slow speed is 37, giving us a combined speed of 80.

d = 80 * (7/2)
d = 40 * 7 = 280

Now our distance is too low, which means that our speed is too low: eliminate B and A.

Only C is left, choose C.

If you backsolve and you can determine whether the answer you plugged in is either too big or too small, you'll never have to test more than 2 choices (and 40% of the time you'll only have to test 1 choice).

We also could have reasoned it out after just 1 plug-in. Our result (294) was 7 over what we wanted (287). If we're travelling for 7/2 hours, we need to go 2mph to make up 7 miles (7=r*(7/2)). So, if we subtract 1 mph from each train's rate, we win. Accordingly, choose C (44-1=43).
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by kvamsy » Mon Nov 16, 2009 1:35 am
Nice Solution, This makes solutions easier
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by ershovici » Mon Nov 16, 2009 6:46 am
I think this is an easier solution:
We know time 3.5 h and distance 287 miles - so average speed is 82m/h
Since faster train is mooving 6 miles faster we have
2x+6 = 82
2x = 76
x = 38
so teh faster train is moving 38+6 or 44 m/h
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by Testluv » Mon Nov 16, 2009 3:54 pm
ershovici wrote:I think this is an easier solution:
We know time 3.5 h and distance 287 miles - so average speed is 82m/h
Since faster train is mooving 6 miles faster we have
2x+6 = 82
2x = 76
x = 38
so teh faster train is moving 38+6 or 44 m/h
Actually, this was the first algebraic solution posted (by jaroth04), and it was endorsed by Stuart.

But it is not the case that the average speed is 82 m/h. Instead, 82 is the sum of their speeds. So, if we let the slower speed be "x" the sum of their speeds is: x + (x+6) = 82. And that is where you get your equation of: 2x + 6 = 82.

When two objects are approaching each other, the rate at which they are mutually closing the distance between them is the sum of their speeds. When approaching each other, they are moving in opposite directions. In fact, whenever two objects are moving in opposite directions you can just add their rates (so if they are moving away from each other, also add their rates).

But if the objects are moving in the same direction--one object "catching up" to another--then you can take the difference of their rates. In these problems, you will have an earlier-departing but slower moving object that accumulates a "headstart" distance on a later-departing but faster moving object. The rate at which the second object catches up to the first is the difference between their rates.
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by ershovici » Tue Nov 17, 2009 12:48 am
Testluv wrote: But it is not the case that the average speed is 82 m/h. Instead, 82 is the sum of their speeds. So, if we let the slower speed be "x" the sum of their speeds is: x + (x+6) = 82. And that is where you get your equation of: 2x + 6 = 82.

When two objects are approaching each other, the rate at which they are mutually closing the distance between them is the sum of their speeds.
You right, it is the summ of their speeds, my foult, but the solution is still right
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by [email protected] » Sun Jun 27, 2010 8:00 pm
how yo are saying that trains get avg speed as 82km/hr? because 1.30 pm is the time taken oly to cross each other.but all the answers taken that as the time taken to cover all 287miles completly.how cum it ll be?
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