In a certain senior class, 72 percent of

[GGUY]

In a certain senior class, 72 percent of the male students and 80 percent of the female students have applied to college. What fraction of the students in the senior class are male?

(1) There are 840 students in the senior class
(2) 75 percent of the students in the senior class have applied to college.


I'm trying to solve it using the double set matrix approach, will appreciate an example

[StaceyKoprince]

[Edited by Stacey b/c Ron pointed out that we can of course solve this via double-set matrix! See, even we make mistakes sometimes... :)]

This is one way to solve this, using weighted averages:
72% male apply; therefore 28% male don't apply
80% female apply; therefore 20% female don't apply

I want to know what fraction of students is male. I could get this one of two ways:
a) the actual number of males in the class and the actual total number of students in the class (# male / # total)
b) the proportion or percentage of male students to female (which, together, comprise the total - so I don't necessarily need to know the actual numbers in order to know a percentage)

(1) 840 = m + f. I don't know actual numbers for male and female. I also don't know proportions for male / female. Can't do it.

(2) 75% of all students (m+f) have applied. Don't know actual numbers for male and female. I can, however, figure out proportions. If 72% of males apply, and 80% of females apply, and 75% overall apply, then I can figure out the proportion of males to females in the class - this is a weighted average problem.

Here's how: 8 percentage points separate the males and the females (80-72). The overall average, 75, is 3 points away from the males (at 72) and 5 points away from the females (at 80). This means there are more males than females and, further, it tells me the actual proportion. Of every 8 students, 5 are male and 3 are female. (If there were equal numbers of males and females, the overall average would be 76%, or 4 points away from each. Because the overall number is closer to the male percentage, the males have more of an impact on the final number. Further, the proportion of males to females is equal to that skew - a 50/50 weighting would mean that each group "pulled" the average by 4 points. Since the males "pulled" the average one point closer to them, they have a weighting of 5, and the females lose that point for a weighting of 3.)

So, 5/8 of the students are male and 3/8 are female.

This is another way to solve this, using double-set matrix:

----------------male-------fem-------t
apply----------.72x-------.8y-------
not apply-----.28x--------.2y------
t----------------x-----------y-------x+y
The above is from the question stem

Statement 1:
----------------male-------fem-------t
apply----------.72x-------.8y-------
not apply-----.28x--------.2y------
t----------------x-----------y-------840
Can't use the above to find out x.

Statement 2:
----------------male-------fem-------t
apply----------.72x-------.8y-------.75(x+y)
not apply-----.28x--------.2y------.25(x+y)
t----------------x-----------y-------x+y

.72x + .8y = .75(x+y) = .75x + .75y
I'm trying to find x. I can simplify the equation to .5y = .3x or 5y = 3x. I would get the same proportion if I wrote out and simplified the second equation (from the "not apply" line). So, basically, I'd need to multiply y (females) by 5 and x (males) by 3 to make them equal. This means that x (male) is bigger than y (female) and that they are in the proportion 5 males to 3 females.

The question is what proportion of all students are males, and I know that for every 5 males there are 3 females. 5+3 = 8 total, so 5/8 of the students are males. Sufficient.

[nikhil.mysore]

Very good explanation

[RonPurewal]

Very good explanation

sweet.

[ameya]

This out of the world classic. Never had thought about weighted averages here. I struggled on this Q and consequently solved incorrectly. Great explanation Stacey! Kudos.

Ameya

[RonPurewal]

stacey is pretty darn smart.

[domgluck]

Initially I got this wrong, but when I went back and looked at it I immediately started drawing a matrix. I must have not seen the puzzle pieces correctly.

I go with Ron's method. If you have the right pieces its very easy to solve the problem using the matrix, but it's good to have multiple routes to take.

male/female, apply/not apply

draw matrix exactly as stacey has
#1, cant find x no good
#2, use to make equation -->

.72x + .80y = .75(x+y)
.72x + .80y = .75x + .75y
.05y = .03x
5y = 3x
5/3 = x/y
5/8 = x/(x+y) looking at x relative to whole

so #2 is good

[tim]

thanks for sharing. let us know if anyone has any further questions on this one..

[rakshaki]

Hi Stacey/Ron,

I have a problem with the following statement:

.72x + .80y = .75(x+y)
.72x + .80y = .75x + .75y

Please refer to this thread of a similar GMATPrep problem: female-students-at-college-c-ds-t13977.html

In that problem, choice B was wrong because we understood that 2/5(m+f) = 2/5th of m + 2/5th of f. This is why equating 2/5(f) = 200 was wrong and B was insufficient to answer the question.

I understand that we are performing a very similar step here:
0.75(x+y) = 0.75(x) + ).75(y). Why is this step correct in this problem and wrong in the former?

After getting the first problem wrong, I noted down that 2/5(whole) does not mean it is the sum of 2/5(induvidual groups).

What are the take aways from this problem? What am I missing?

[RonPurewal]

You can't argue with the fact that (2/5)(a + b) = 2a/5 + 2b/5. That's just distributing, as you learned in first-year algebra.

In other words, If you have the quantity (2/5)(m + f) -- for the 2/5 of all students who are business majors -- then that's algebraically equivalent to 2m/5 + 2f/5.
However, this does not necessarily mean that two-fifths of the people in each separate group are business majors. You cannot separate the 2m/5 and the 2f/5 and assume that they represent individual numbers of business majors of each sex. They are purely algebraic expressions.

[RonPurewal]

For instance, let's say there are 50 male students and 50 female students. Say 30 of the 50 male students are business majors, but only 10 of the female students are business majors.

Check the above:
* (2/5)(m + f) correctly gives 40, the total number of business majors.
* 2m/5 + 2f/5 also correctly gives 40.
* 2m/5 and 2f/5 do not separately yield 30 and 10. These quantities are not meaningful individually; they are algebraic artifacts.