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RATE/DISTANCE QUESTION

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RATE/DISTANCE QUESTION

by Mclaughlin » Mon Jul 21, 2008 8:53 pm
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
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by egybs » Tue Jul 22, 2008 12:07 am
I'm exhausted and about to go to bed.. but here's the solution... There is very likely a quicker way to do this, but here's an idea... I'm too tired to throw in full explanations for what's going on.. so if you can't figure out what i've done, lemme know and i'll revisit it.


Let's call time when they meet t.. so:
t = z/(z/x +z/y)
t = 1/(1/x +1/y)
t= (xy)/(x+y)

We're looking for:
tz/x - tz/y
tz (1/x - 1/y)
tz (y-x)/(xy)

put it all together:
z(xy)(y-x)/((x+y)(xy))
z(y-x)/(x+y)
which is A.
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by pepeprepa » Tue Jul 22, 2008 12:59 am
I think, first I try something theoric but after I would choose examples.

z=120
Train X is 40m/h so x=3h
Train is 30m/h so y=4h
The result we look for is (40/70)*120-(30/70)*120=17 and things

So now we can delete all answers with (x-y) (3-4<0) because our answer cannot be negative --> B and D out

Now let's try A)
z(y – x)/x + y
(120(4-3))(3+4)=120/7=17,14......
So it seems good

Now you can try C) E) to confirm you are right.
If you do not make fool errors it can take less than 2' but it is true it is not a fast method...
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