Monday, May 28, 2012

answer to last week's MGRE Math Beast Challenge

The answer to last week's challenge is indeed (A)! That means Charles and I are both right. But here's the thing: I had anticipated that MGRE would take Charles's tack and use the plug-and-chug method, but instead they went full-on algebra, as I did, and offered their own version of plug-and-chug only at the very end as an afterthought, and only as a way to check their algebra. To review, then-- here's what Charles had written in his comment:

Yeah, I got A, too, although my process was not nearly as detailed. I just took 15 as a possible number of women at the party, subtracted 8, and multiplied by 4 to get 28 men originally at the party. Since the question then says that 35 men left the party, I knew that the original number of women had to be greater than 15, so the answer was A.

It took me about five times as long to write the above paragraph as it did to work out the answer. I never did figure out how many women were originally at the party.

Disgustingly simple. My own approach, you may recall, went for the algebra:

Let m = original # of men.

Let w = original # of women.

1st phase: we have m and w.
[All men & women are present.]

2nd phase: we have m and (w - 8).
[Eight women have left.]

3rd phase: we have (m - 35) and (w - 8).
[Thirty-five men have left.]

Given (per what we know of the second phase, and what the word problem tells us):

m = 4(w - 8)

And for the third phase:

(w - 8) = 2(m - 35)

At this point, it's a matter of systems of equations.

m = 4w - 32 (2nd phase)

2m = w + 62 (3rd phase)

Multiply the first equation by 2:

2m = 8w - 64

Match it up with the other equation and solve:

2m = 8w - 64
-(2m = w + 62)


0 = 7w - 126

7w = 126

w = 18

The original number of women was 18, so Quantity A is greater.

I'm going with (A).

Not simple, but definitely thorough. And here, finally, is how MGRE tackled the problem:

This problem can be solved with a system of two equations.

First, “after 8 women leave, there are four times as many men as women.” Thus, once 8 is subtracted from the number of women, there is a 4 to 1 male/female ratio:

m/(w - 8) = 4/1

Cross-multiply and simplify:

m = 4(w – 8)
m = 4w – 32

Then, 35 men leave, and the 8 women don’t come back, resulting in a 1 to 2 male/female ratio:

(m - 35)/(w - 8) = 1/2

Cross-multiply and simplify:

2(m – 35) = (w – 8)
2m – 70 = w – 8

We now have two equations in two variables.
1st equation: m = 4w – 32
2nd equation: 2m – 70 = w – 8

Since the 1st equation is already solved for m, simply plug into the 2nd equation for m:

2(4w – 32) – 70 = w – 8
8w – 64 – 70 = w – 8
8w – 134 = w – 8
8w = w + 126
7w = 126
w = 18

Since 18 is more than 15, the correct answer is A.

Although we are not asked for the number of men, note that we could easily generate it by plugging w = 18 into either equation:

m = 4w – 32
m = 4(18) – 32
m = 40

This would allow us to check our answer. If we begin with 18 women and 40 men, and then 8 women leave, we would have 10 women and 40 men, which indeed would be a 1 to 4 ratio of women to men. If 35 men then leave, we would have 10 women and 5 men, which indeed would be a 2 to 1 ratio of women to men.

The correct answer is A.

1 comment:

Charles said...

Well, obviously it's important to know how to do the algebraic magic. If, for example, you were asked to determine how many women were originally at the party, that would be the way to go.

But since we already have a number for comparison, I figure it would be easiest (and quickest) to plug and chug. I haven't taken a test like this in a really long time, but I do remember that being as efficient as possible was important so that you would have the time you needed to noodle out the hard questions.