What is the ratio of men to women enrolled in a certain class?

Hi MWithrock ,

M and W must be positive integers therefore you don´t have "all freedom" in statement (1).

On the other hand, you are right in statement (2), although W*(5/2)=M, of course. I explain:
Once M/W is unique, the fact that it must be possible to be written as a ratio of positive integers is the examiner´s burden (not yours).

Regards,
Fabio.

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Typo edited. Than you Fabio.

Statement 1) The number of women enrolled in the class is 3 less than half the number of men enrolled.

w = \(\frac{m}{2} - 3\)

\(\frac{m}{w} = \frac{m}{\frac{m}{2}-3}\)

Clearly INSUFFICIENT.

Statement 2) The number of women enrolled in the class is 2/5 of the number of men enrolled.

w = \(\frac{2}{5}m\)

\(\frac{m}{w} = \frac{m}{\frac{2}{5}m}\)

\(\frac{m}{w} = \frac{5}{2}\)

SUFFICIENT.

OPTION: B

Hi, I know this is a relatively simple question, but I got this wrong because I defaulted to thinking "there is a clear relationship between men and women presented here, therefore statement 1 must be sufficient." Does it not work because the "-3" means that it is not a directly proportional relationship but rather one that is slightly skewed? Would the best approach be to test out different numbers for Statement 1 to see if it still holds the same relationship, or would any ratio with such addition/subtraction provide different values?

fskilnik Bunuel

Yeah I guess there's two ways to think about this. The first is to look for the intuitive / general idea. The second is to prove it's insufficient by counterexample.

Your intuition outlined here is right. It's the "3 less" thing that makes it insufficient. As m and n get large, the significance of that "3 less" shrinks, eventually approaching zero asymptotically.

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Starting with the general idea
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Let m be the count of men enrolled; let w be the count of women enrolled.

We want to know the ratio of men to women enrolled, i.e. we want a unique solution for m / w

(1) Tells us w = 0.5m - 3.

We can rearrange this a little bit:
---> w = 0.5m - 3
---> 0.5m = w + 3
---> 0.5m/w = 1 + 3 / w
---> m/w = 2 + 6 / w

We're forced to notice that the ratio of men to women depends on how many women there are (we could do it the same way to show that the ratio of men to women depends on how many men there are; it's the same thing).

It's not sufficient because we could create many different ratios.

(Actually, I'm reasonably sure we could identify a countably infinite number of pairings here. Note that w = 0.5m -3 resembles a y = mx + b equation (although well defined in this context only on a subset of integers),

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Just doing it by counterexample
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Again let m be the count of men enrolled; let w be the count of women enrolled.

If we can show two different ratios for two viable combinations of m and w, then we prove the statement is insufficient.

Let's start with m = 10, which means w = 2. Then the ratio of men to women is 5:1.

But we could also have m = 12, in which case w = 3. Then the ratio of men to women is 4:1.

Because we can show two distinct ratios, the information provided is insufficient to identify a unique ratio.

Beautiful explanation, thank you!