In a certain senior class, 72 percent of the male students and 80 perc

Question

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

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Official Answer

B

Bunuel · Accepted Answer

Check this: 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? This is weighted average question. Q: \frac{m}{m+f}=? (1) There are 840 students in the senior class --> m+f=840 . Stem info is useless. Not sufficient. (2) 75 percent of the students in the senior class have applied to college --> 0.72m+0.8f=0.75(m+f) --> 3m=5f --> \frac{m}{f}=\frac{5}{3} --> \frac{m}{m+f}=\frac{5}{8} . Sufficient. Answer: B. Also discussed here: percentage-students-101231.html Hope it helps.

esledge · Answer

I think you must have! Sorry if this repeats what others have said, but here's a solution that illustrates that pitfall, with a little additional commentary on Weighted Averages. With just the info from the question stem, the overlapping sets chart looks like this. The question is “What is m/Total?” (or, “What is m/(m+f)?”) ------Have-----Haven’t----Total M----0.72m-----0.28m-----m F-----0.8f--------0.2f-------f Total------------------------m+f (1) Tells us that m+f = 840, but nothing about how the 840 splits between m and f. (2) If 75% of students have applied to college, let’s put that expression in the chart: ---------Have--------Haven’t-------Total M-------0.72m--------0.28m--------m F--------0.8f-----------0.2f----------f Total--0.75(m+f)---0.25(m+f)----m+f We can pull two new equations out of the vertical Have and Haven’t columns: 0.72m + 0.8f = 0.75(m+f) 0.28m + 0.2f = 0.25(m+f) Multiply everything by 100 to avoid dealing with decimals. 72m + 80f = 75(m+f) 28m + 20f = 25(m+f) Move all m’s and f’s to one side: -3m + 5f = 0 3m + -5f = 0 When we add the equations we get 0=0. Yikes! This is why you would conclude that (2) is insufficient! All the math we’ve done so far is right, but there are two things we have overlooked: a. We haven’t made full use of the fact that m and f are tightly related to one another, because there’s not a third gender (in this problem): m = total –f and f = total –m. b. We were trying to solve for m and f. But our question is really about m/total. For combo problems (where you are solving for not one variable, but an expression with 2 or more), the “trick” is to solve for the combo as directly as you can. So let’s back up to this: ---------Have--------Haven’t-------Total M-------0.72m--------0.28m--------m F--------0.8f-----------0.2f----------f Total--0.75(m+f)---0.25(m+f)----m+f Pull out that vertical sum from the “Have” column, and rewrite the f’s in terms of m. 0.72m +0.8f = 0.75(Total students) 0.72m + 0.8(Total students – m) = 0.75(Total students) 72m + 80(Total students) – 80m = 75(Total students) 5(Total students) = 8m 5/8 = m/Total students 5/8 of the students are male. As others have noted, there is a “weighted average” insight we could have exploited. The “weighted average” formula here is: (Fraction of the class that is male)*(male application rate) + (Fraction of the class that is female)*(female application rate = overall application rate. Here’s another post I wrote on the subject: weighted-average-problems-93260.html#p717555

walker · Answer

B

First of all, we have a red flag in the problem: it is a %-problem and it does not tell us about any absolute value, only about % and fraction. So, if we have one DS-statement only with one absolute value, the statement is useless for solving the problem.

1) Yeah, it stands here to catch us.

2) How to determine sufficiency of the statement for a few seconds? Just try slightly to increase or decrease 75% in your mind, do you need to change a fraction? Of course, as you need to have more/less male/female ratio to decrease/increase 75%. 
If it is difficult, the traditional solution:

75% = (72%*male + 80%*female)/(male + female) = (72% male/female+ 80%) / (male/female + 1) - we can find male/female fraction.

Now, male/(male + female) = 1 / (1 + female/male) - using information from the statement above, we can find the fraction of the students who are male.

KieuPhuong · Answer

The answer is B

If X is the FRACTION of the male students in the senior class, then 1-X is the fraction of female in class.

1) is not sufficient to solve the problem
2) from the provided information, we have:

0.72*X + 0.80*(1-X) = 0.75 <=> X = 5/8

marcusaurelius · Answer

Great, thank you. When I was reading through the problem I think I assumed they were asking for the number of males rather than the percentage.

KarishmaB · Answer

(I am assuming that 'senior college' in question stem is the same as 'senior class')
This is a question of weighted averages.

Statement 1: Total students is 840 but we do not know how many of them are male and how many are female.

Statement 2: 75% of total have applied to college. 72% of males have applied and 80% of females have applied. The weighted average of 72% and 80% is known to be 75%.

_______________________
72% 75% 80%

Diff between 72 and 75 is 3 and that between 75 and 80 is 5 hence males:females ratio is 5:3.

Hence 5/8 of the students in senior class are males.
Statement 2 is sufficient. Answer is (B).

thesfactor · Answer

I got this problem wrong as usual...I HATE DS!

Anyway I checked the answer and it dawned on me that B is sufficient. However, I still cant seem to get 5/8, what am I doing wrong?
X = m + f where X is total number of students in college.
So from B) 75% of all students applied to college. Therefore we have 75% (X)
Now we know 72% male students applied to college, therefore the male number will be 72% of 75% of X
The fraction should be:
( 72% of 75% of X) / X
This gives me around 54/100 males attending college.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

This question has a logic, which is frequently given on GMAT Math exam lately. The logic is there are 4 variables, and a question includes percentage and the number of percentage should be 3, which makes the con sufficient.
That is, this is a frequently given question, which is "2 by 2" question like the table below. There are 4 variables.

Attachment:

GCDS tejal777 In a certain senior class 72 percent (20160217).jpg [ 35.6 KiB | Viewed 29656 times ]

The question asks percentage and the number of percentage should be 3. In the original condition, the number of percentage is 2(72%, 80%) and 75% in 2), which makes the number of percentage 3 in total. Thus, it is sufficient and the answer is B.

 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

 from con 1) and con 2), if one of the conditions is given by numbers and the other is given by ratio (percent,fraction), then the condition with ratio (percent,fraction) value has higher chance of being the answer.

GMATinsight · Answer

Answer: Option B

Video solution by  GMATinsight

https://www.youtube.com/watch?v=EtE2ql71nkw

RastogiSarthak99 · Answer

Weighted average question but would suggest using the alligation method here avoiding algebra.

Basically --> 75-2 and 80-75 will give us a ratio of M/F of 5/8. No need to solve further % since its a DS question. Sufficient. B

A tells us total qty. Irrelevant (since we are only talking in %'s and question has asked for fractions)

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In a certain senior class, 72 percent of the male students and 80 perc

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In a certain senior class, 72 percent of the male students and 80 perc

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