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In a race over m yards

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In a race over m yards

by Brent@GMATPrepNow » Sun Jan 25, 2009 4:58 pm
In a race over m yards, Sharon would finish 15 yards ahead of Rick; Rick would finish 20 yards ahead of Christine; and Sharon would finish 31 yards ahead of Christine. What is the value of m?
(A) 75
(B) 100
(C) 125
(D) 150
(E) 175

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by gaggleofgirls » Sun Jan 25, 2009 5:21 pm
I am assuming that the three run at constant rates (isn't that necessary to solve this?)

Over that last 15 yards of the race, Rick went from being 16 yards ahead of Christine (31-15) to being 20 yards ahead of her, so he is gaining 4 yards every 15 yards.

Since everyone in a race starts at the same time and Rick finished 20 yards ahead, then he rust have run 20/4 * 15 yards = 75 yards.

Answer = A

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by awesomeusername » Sun Jan 25, 2009 5:28 pm
Yes, I got 75 too. If Rick gains 4 yards over Christine for every 15 yards he runs, and he beats Christine by 20 yards, then he must have run (20/4)*15 yards, or 75 yards. m=75.
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by Brent@GMATPrepNow » Sun Jan 25, 2009 6:08 pm
Nice work - the answer is A
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by 4meonly » Wed Jan 28, 2009 8:47 am
I do not understand how you got this equation...
gaggleofgirls wrote: then he rust have run 20/4 * 15 yards
Please help
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by Brent@GMATPrepNow » Wed Jan 28, 2009 9:39 am
4meonly wrote:I do not understand how you got this equation...
gaggleofgirls wrote: then he rust have run 20/4 * 15 yards
Please help
I'll let gaggleofgirls explain her method. In the meantime, you might consider another (albeit slightly longer) approach to this question:

The important thing to recognize here is that the ratio of speeds between different pairs of people will be the same as the ratio of the distances they cover over a certain time. So, for example, in the race between Sharon and Rick, we see that Sharon runs m yards in the same time that Rick runs m-15 yards. So, we can conclude that the ratio of Sharon’s running speed to Rick’s running speed is m : m-15

Using this idea, we can conclude the following ratios of speeds:
(1) Sharon : Rick = m : m-15
(2) Rick : Christine = m : m-20
(3) Sharon : Christine = m : m-31

It will be useful to create an equation with this information. We can use ratios (1) and (2) to create a ratio between Sharon’s and Christine’s speeds. This new ratio must equal ratio (3). This will be our equation to solve.

We get (1) Sharon : Rick = m : m-15 = m^2 : (m-15)m
(2) Rick : Christine = m : m-20 = (m-15)m : (m-20)(m-15)
From here, we can conclude that Sharon : Christine = m^2 : (m-20)(m-15)

But this must equal (3)
So we get: m^2 : (m-20)(m-15) = m : m-31
We can solve this for m to get m=75 (answer choice A)
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by gaggleofgirls » Wed Jan 28, 2009 3:46 pm
4meonly wrote:I do not understand how you got this equation...
gaggleofgirls wrote: then he rust have run 20/4 * 15 yards
Please help
Here is how I thought of it...

For every 15 yards of the race, Rick gains 4 yards on Christine.

Since they started together (both at 0) and Rick finishes 20 yards ahead of Christine, there must be 20/4 sets of 15 yards in the race. Hence 20/4*15 is the length of the race.

Here is a chart that might help:

Race length | Rick Ahead
------------------------------
0 | 0
15 | 4
30 | 8
45 | 12
60 | 16
75 | 20 *****
From the chart, you can see that there are 5 sets of 15 yards in the race to give Rick the time to get 20 yards ahead.

That formula would work fine also if the race length were not a multiple of 15.

Lets say that Rick finished 22 yards ahead of Christine,
Then there were 22/4 sets of 15 yards in the race, so 22/4 * 15 = race length.
5 1/2 * 15 = race length
82 1/5 yards = race length.

HTH.

-Carrie
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