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og rates 119

by resilient » Sat Mar 08, 2008 8:30 pm
Two trains, x and y, started simultaneously from oppsoite ends of a 100 mile route and traveleed toward each other. Train x traveling constant rate, completed the 100 mile trip in 5 hours: train y, completed the 100 mile trip in 3 hours. how many miles had train x traveled when it met rain y? (I now underline all questions).

a.37.5
b. 40
c.60
d.62.5
e.77.5

qa is a. I see that adding both rates to see the time in which they will finish up the distance of 100 miles and then applying that time # to the time X rate of train x. IS there a different or more simple way to look at this?
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by musicdaemon » Sun Mar 16, 2008 11:01 pm
The concept you have used is called "Relative Velocity" This is the simplest method possible for this problem.

However, you can also use proportions to solve this

Velocity of train A = 100/5 = 20 mph
Velocity of train B = 100/3 mph

Suppose A covers distance x, when it meets B
Then the distance covered by B on 100 mile track = 100-x miles

Since, both trains started simultaneously, the time taken while they meet is equal for A & B

Thus,
Velocity is proportional to the distance covered only (Since Velocity=distance/time), i.e.

Velocity of A/Velocity of B = Distance covered byA/Distance covered by B

=> 20/(100/3) = x/(100-x)
=> x= 37.5 miles ........................................ your answer


hope you get the point
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by camitava » Sun Mar 16, 2008 11:21 pm
Enginpasa1, a good and mathematical approach is given by musicdaemon!
Actually stmt made by musicdaemon -
Thus,
Velocity is proportional to the distance covered only (Since Velocity=distance/time), i.e.

Velocity of A/Velocity of B = Distance covered byA/Distance covered by B
u can take it in this way also -
Time taken to cover the distance from the starting point to meeting point by x = Time taken to cover the distance from the starting point to meeting point by y.
Now the rests are quite similar to the explanation given by musicdaemon.
Correct me If I am wrong


Regards,

Amitava
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rates

by resilient » Mon Mar 17, 2008 11:31 pm
Thanks a lot friends. THis has helped a lot! I am trying to apply the rules learned from manhattan gmat, but they are hard to apply from case to case. This will be great during the exam.

NOte: backsolving is a good measure here also!
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one more note

by resilient » Mon Mar 17, 2008 11:36 pm
THis is a great rule and I am running out of practice questions for this topic. ANy ideas?
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by Stacey Koprince » Tue Mar 18, 2008 11:45 am
Nice, musicdaemon!

You can (and should) also use logic / estimation to know what your answer should be close to - make this as real-world as possible. Imagine the trains running on the tracks! Draw a diagram and pretend you're sitting on the train!

If it takes X 5 hours and it takes Y 3 hours, then Y is going faster than X. Therefore, Y will have covered more than half the distance when they meet and X will have covered less than half the distance. So C, D, and E are all out as possible answers for X's distance traveled when they met. If you notice this, it can help you to know what to expect (and to know whether you made a mistake) when doing the actual math. This also makes it much more efficient to use a technique such as backsolving b/c you know you should only try answers A and B.

Generally speaking, you should always look at the answer choices before you decide how to solve a particular math problem, because the form and spread of the specific answers may change how you want to do the problem!
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ok

by resilient » Wed Mar 19, 2008 8:09 am
perfect! I see your points. THis is a great way to solve. Back solving always works for me and I can easily answer the question. Do you recommend that I back solve and answer the question or should I try to make the variable charts using drt formula? Also how would you even set that up here? The manhattan way helped me answer many questions I normally would not be able to answer and I just want to nail it down. This way I will have two ways of being able to answer these questions!
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by Stacey Koprince » Wed Mar 19, 2008 8:17 am
For d=rt, this is what you'd do (though, note, I can't actually draw a chart here):

Train X:
distance = d
rate = 100/5 = 20mph
time = t

Train Y:
distance = 100-d (the two distances have to add up to 100)
rate = 100/3mph
time = t (the start simultaneously and meet at a specific time, so t is the same)

This yields two formulas:
d = 20t
100-d = (100/3)t

I want to find d, so manipulate first equation to substitute in for t:
d/20 = t
substitute into second equation:
100-d = (100/3)(d/20)
100-d = 5d/3
300 - 3d = 5d
300 = 8d
37.5 = d

As to whether you should solve with d=rt or with backsolving - that's a personal choice. Sometimes you'll want to do it one way, sometimes the other. The key is to do it both ways when you're studying and then ask yourself which way you think is the better way for this problem and WHY it's better for this particular problem (so that you can replicate the behavior when you see another similar problem in future). You might decide, for example, that since you could tell it was either A or B using logic, you'd prefer to use backsolving (since there are only 2 options).
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rates

by resilient » Fri Mar 21, 2008 9:06 pm
I originally was having a tough time with rate questions that dealt with two object and a question dealing with when they will meet. WIth the help of the pros here I think I have a good breakthrough here.

Sample NEMESIS question.

Two airplanes are 300 miles a part and flying directly toward each other. ONe is flying at 200 miles per hour and the at 160 miles per hour. How long will it take for the two planes to meet.

qa is 50 minutes.

The logic is the same as two painters working at two different rates, while dealing with the time that they will BOTH finish the job.

Similiarly, this is question is both rates combined ( 360 mph) plugged into the DRT formula. D=RT but rearranged to T= D/R

So, Time = 300/360 THis simplifies to 5/6 hours or 50 minutes.

Does anyone know of more questions like this for me to work on. I have come a long way and need to keep the momentum!

THanks for all the help. BTG is the best
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