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Expressed in minutes, on day one, the distance to P is 120*s (120mins=2 hrs, s=speed)

Then for bus 1 on the second day, distance covered is (120-24)*s =96s

So shouldn't distance to P (120*s) minus 24 miles equal 96s? As in 120s-24=96s, which would calculate to s=1mile/min=60mph

The point is that we cannot say that bus 1 traveled 96 minutes on the second day.

BELOW IS REVISED VERSION OF THIS QUESTION:

A bus leaves city M and travels to city N at a constant speed, at the same time another bus leaves city N and travels to city M at the same constant speed. After driving for 2 hours they meet at point P. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance between the two cities?

A. 48 B. 72 C. 96 D. 120 E. 192

Say the distance between the cities is \(d\) miles.

Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point P, is \(\frac{d}{2}\) miles away from M (and N).

Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours.

Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover.

The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway.

Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours.

On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover.

They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

You destroyed that problem and took its lunch money too. So this is a great opportunity for me to ask a question about a concept that I've never fully understood. When you create this equation:

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

What makes you subtract 24 instead of add 24? I feel like either one signifies "distance from", right?

First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway.

Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours.

On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover.

They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

You destroyed that problem and took its lunch money too. So this is a great opportunity for me to ask a question about a concept that I've never fully understood. When you create this equation:

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

What makes you subtract 24 instead of add 24? I feel like either one signifies "distance from", right?

The simple fact that d/2 is greater than 0.75(d/2).

First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway.

Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours.

On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover.

They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

I am not sure whether this is a good question...but can we do this problem by plugging in the answer choices...for eg. assumin 'd' to be 192,120...etc??? ..?? is it possible in any way??

First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway.

Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours.

On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover.

They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

I am not sure whether this is a good question...but can we do this problem by plugging in the answer choices...for eg. assumin 'd' to be 192,120...etc??? ..?? is it possible in any way??

The latest diagnostic test version (v6.2) does not have the exact wording of the original post nor of Bunuel's revision.

I skipped the question when I took the test because I did not know what they were asking. Upon reading the question in this thread, I solved it fairly quickly.

gmatclubot

Re: GMAT Diagnostic Test Question 24
[#permalink]
10 Mar 2014, 13:02