Of the 1400 College teachers surveyed 42% said they considered engaging in research an essential goal. How many of the college teachers surveyed were women?
1. In the survey 36% of the men and 50% of the women said that they considered engaging in research an essential goal.
2. In the survey, 288 men said they considered engaging in research an essential goal.

The ans for this is A
Can you please explain why it is not C?

stem: 42/100 x 1400 said that they considered engaging..., so 588 are considered engaging, 812 are not considered engaging...

2) of the 588, 288 are men, so 300 are women are engaging -- but how many women are NOT considered engaging...?
INSUFFICIENT

harder of the 2 statements:

1) 36% of Men and 50% of Women are considered engaging...so 36/100 M + 50/100 W = 588 and M + W = 1400

2 distinct linear equations, 2 variables
SUFFICIENT

answer is A

this is good. remember that you don't have to actually solve a system, if you know that it can be solved for a unique solution.

note that this is a WEIGHTED AVERAGES problem.
on weighted averages problems, if you know any 2 of the following 3, then you can find the third one:
1. the ratio of 'weights' of the different quantities
2. the values of the quantities
3. the weighted average

in this problem, we have the values of the quantities (36 and 50), as well as the weighted average (42). that's enough information to find the ratio of the 'weights' (which happens to be 4:3 m:f, although you don't particularly care because it's data sufficiency).
because you have a total for everybody (1400), that ratio is enough to give you a hard number for the # of women.
sufficient.

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if you know the relationships posited above, you can get through weighted averages on data sufficiency very quickly - although it's still helpful to know the algebraic method so you can break it out on problem solving.

this question can also be easily solved using a double set matrix. statement 1 gives you two linear equations with two variables that can be simultaneously solved for unique solutions for each variable. just a thought...
Wtd average is such a dominant topic on GMAT!

yep -- correct.

by the way, if you're going to make posts like this one, please give the specifics
i.e., don't just state that "there are two linear equations" -- give the equations! if you don't, then there is very little point in posting this.

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i'm assuming that you're talking about the following equations:
let m = # of male teachers, w = # of female teachers
then
m + w = 1400
0.36m + 0.50w = 0.42(1400)

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by the way, if we change statement (1) to say
In the survey 42% of the men and 42% of the women said that they considered engaging in research an essential goal
... then it's now insufficient.
make sure you know why.

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Is it because it just repeats information already given in the question stem? .42M+.42W = .42(1400) --->
.42(M+W) =.42(1400)
but M+W = 1400 ---> .42(1400)=.42(1400)
Does not give us any useful new information.

no ... that would definitely give you new information. the fact that 42% is the individual percentage of both men and women is much, much more specific than the fact that 42% is a composite weighted average.
(you'll be able to think about this more intuitively if you think with "half" instead of "42%". i.e., knowing that half of all men *and* half of all women support something is much more specific information than just knowing that half of everyone supports it.)

the reason why 42% and 42% wouldn't help is that you would no longer be able to determine the "weights" in the weighted average. i.e., if 42% of one population and 42% of another population support something, then the weighted average will *always* be 42%, regardless of the relative sizes of the two populations.
on the other hand, a weighted average of two different figures will vary according to the relative population/sample sizes. for instance, a weighted average of 40% men / 50% women can only be 42% of the ratio of men to women is exactly four to one.