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Multi-variables & Trains Meeting

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Multi-variables & Trains Meeting

by i_have_no_cool_username » Sat Oct 22, 2011 2:06 pm
Need some help with finding a better, shorter way of working through this:

It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

a) [z(y - x)]/[x + y]
b) [z(x - y)]/[x + y]
c) [z(x + y)]/[y - x]
d) [xy(x - y)]/[x + y]
e) [xy(y - x)]/[x + y]

The many variables could be confusing when setting up equations. The answer key suggested 2 options - one, to plug in numbers for the variables; Two, the algebraic solution (which was what I attempted to do before getting lost amidst the variables in my equations). The first solution seems easy but I'm not confident about picking the right numbers. Part of the algebraic solution is as follows:

Since the two distances sum to the total when the two trains meet, we can set up the following equation:

zt/x + zt/y = z divide both sides of the equation by z
t/x + t/y = 1 multiply both sides of the equation by xy
ty + tx = xy factor out a t on the left side
t(x + y) = xy divide both sides by x + y
t = [xy]/[x + y]

Since it's given that x and y are the respective times taken for the high-speed and regular trains, how can I tell when I should use "t" to set up my equations even when the variables for time have already been given in the prompt?

Also, is there a simpler way to do this algebraically?

Thanks, experts! :)
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by GMATGuruNY » Sat Oct 22, 2011 5:44 pm
I posted a solution here:

https://www.beatthegmat.com/trains-t67410.html
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by rishimaharaj » Sat Oct 22, 2011 6:07 pm
Hi Cool_Username :-),

Here is my attempt at an algebraic solution with comments:

High speed train (H) can travel z miles in x hours. Using Rate = Distance / Time, the general rate of the train is Rate(H) = 1/x.
The regular speed train (R) can travel z miles in y hours. Rate(R) = 1/y.

The diagram would look something like this:

A----------------1/2 way point-------------------B
but the trains would meet not at the half way point; they would be closer to B, because H is faster.

A-----------------------------meeting point------B

So at time T, H would travel Distance(H) = 1/x * T = T/x.
R would travel Distance(R) = 1/y * T = T/y.

Distance(H) + Distance(R) = z miles.
T/x + T/y = z //plugged in the distances from above.
(Ty + Tx) / xy = z //combined fractions on left.
Ty + Tx = xyz //multiplied both sides by xy.
T(y + x) = xyz //factored T from left side.
T = xyz / (y + x) //divided (y+x) from both sides.

What we need to calculate though is how many more miles traveled by H than by R:
Distance(H) - Distance(R) = ?
=T/x - T/y //plugged in distances from above.
=(Ty - Tx) / xy //combined fractions.
=T(y-x) / xy //factored T.
=[xyz / (y+x)] (y-x) / xy //replaced T with the expression we found above.
=[xyz(y-x) / (y+x)] / xy //simplified a bit --- multiplied xyz with (y-x).
=[xyz(y-x)] / (y+x)(xy) //simplified some more --- bringing xy up a level in the denominator.
=z(y-x) / (y+x) //simplified again --- canceled xy in both numerator and denominator.

This is the final answer, which matches with answer choice A.

Hope this helps!

--Rishi
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by i_have_no_cool_username » Sat Oct 22, 2011 9:27 pm
Hi GMATGuru,

Thanks for shedding some light on this - I read your method. I'm wondering why you decided to plug in values for speed, instead of plugging individual values for x,y and z. From there, one would proceed to find the speeds of high-speed and regular train respectively - am I on the wrong track here? I tried doing that, but picking the correct numbers is tough. So, what is a tell that I should be plugging in numbers directly for speed?

Hi Rishi, thanks for acknowledging my laziness in thinking up a creative nickname :) The algebraic method probably comes to you easily, but I anticipate that I'll be "variably-challenged" and might somehow make silly mistakes in the process of simplifying the multiple variables within. Thanks for your help! It's great to read the algebraic solution again and familiarize myself with the alternative method to this question.
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by smackmartine » Sat Oct 22, 2011 10:10 pm
As soon as you see a relation between the rates, try not to indulge too much in individual calculation.

Rf x=z and Rr y=z

Using relativity concept : (Rf+Rr).T = z --> T = z/(Rf+Rr)

Final result :

Rf(T)-Rr(T)= (Rf-Rr)(T)= [(Rf-Rr)/(Rf+Rr)]z = [(Rf/Rr -1)/(Rf/Rr +1)]z=[(y/x-1)/(y/x+1)]z = [spoiler](A)[/spoiler]
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by GMATGuruNY » Sun Oct 23, 2011 2:33 am
i_have_no_cool_username wrote:Hi GMATGuru,

Thanks for shedding some light on this - I read your method. I'm wondering why you decided to plug in values for speed, instead of plugging individual values for x,y and z. From there, one would proceed to find the speeds of high-speed and regular train respectively - am I on the wrong track here?
To determine how long it takes the trains to meet, the value of z (the distance) must be divided by the SUM of the rates of the two trains.
I plugged in values for the RATES in order to ensure that z would be a multiple of all the different rates in the problem -- including the COMBINED RATE of the two trains:
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?


a)

z(y - x)
---------
x + y
Let the high speed rate = 3 miles per hour.
Let the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.

Now we plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.

Only answer choice A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.

The correct answer is A.
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by i_have_no_cool_username » Sun Oct 23, 2011 10:19 am
Thank you, GMATGuru! I think I got it now. I like your method better than the algebraic methods, the only challenge is in picking the right numbers.

Thank you for your time.
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by send2dar » Wed Jul 24, 2013 4:38 pm
I found a very interesting and efficient way of arriving at an answer to this.

As z is the only distance number among x, y, and z, the answer should contain z. Rules out options D and E.
As high-speed train would take lesser time than regular time, x should be less than y. Further, as our answer should be positive, it should not contain x - y term (x - y is negative). Rules out B.
As we are looking at the difference in distances travelled, the numerator should be of the form p - q. Rules out C.

Only Option A is left, which is our answer. In this way, without solving the problem, one can get to the correct answer.

Hope this helps!

If anybody finds loopholes in the above logic, please do let me know.
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