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.

got another one right!

I consider myself quite capable of teaching the Quantitative, Verbal, and Analytical Writing sections of the GRE, but I still need to keep myself on my toes. That's one reason why I do the weekly Math Beast Challenges that Manhattan GRE throws us. The challenges appear on Monday; I splash them up on this blog on Tuesday (since Tuesday is the designated math day here) and offer my own answers to the questions. When the solutions to the problems appear the following week, I recopy and display them here for your edification. This way, Dear Reader, you get to see two approaches to the problem, which allows you to decide what works best for you.

As it turns out, I got last week's problem right: the answer is, in fact, (C): the quantities are equal. Here's MGRE's explanation:

We can set up and simplify two equations, one for the calories consumed and the other for cost, using variables A and B for the number of servings of Snacks A and B, respectively.

Calories:
200A + 350B = 3250
20A + 35B = 325 {divided by 10}
4A + 7B = 65 {divided by 5}

Cost:
$1.70A + $0.60B = $11.00
17A + 6B = 110 {multiply by 10 to eliminate decimals}

So we now have a system of two equations and two variables, which could be solved for A and B. But, even in their simplified form, these two equations have awkward coefficients that will make solving messy.

Since this is a Quantitative Comparison question, it would be smarter to “cheat off of the easy statement.” That means we plug in the 4 from Quantity B as a possible number of servings of Snack A, and see what that tells us.

We’ll plug A = 4 in to each equation.

Calories: 4(4) + 7B = 65, so 7B = 65 – 16 = 49. Therefore B = 7.
Cost: 17(4) + 6B = 110, so 6B = 110 – 68 = 42. Therefore B = 7.

We have effectively shown that A = 4 and B = 7 is the solution we would have found had we solved this system of equations ourselves.

Thus, Quantity A and Quantity B are both 4.

The correct answer is C.

I'd say that's not so different from my own approach, until we get to the next-to-final step. MGRE and I certainly agreed on the fundamental issue: this was indeed a "systems of equations" problem. Where we disagreed was in how to reach the very end. I did the typical thing, and simply solved all the algebra. MGRE, by contrast, "cheated" by plugging in data from the chart to solve for one variable. This method is arguably quicker, although I think the difference between MGRE's and my respective methods amounts to just a few seconds. On the actual test, you've got a little more than a minute to solve each math problem, so there's a bit of a time buffer there, as long as you're fast on your feet.


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