Data Sufficiency

In a certain class 72% of male and 80% of females have applied to college. What fraction of students are male.

1. There are 840 students in the class

2. 75% of the students have applied to college

STM1. 840 does not do anything.

SMT2.


(0.72M + 0.8F)/(M+F) = 0.75

0.72M + 0.8F = 0.75M + 0.75 F

3M = 5F

3/8 are male!! SUFFICIENT!!

sakurle wrote:In a certain class 72% of male and 80% of females have applied to college. What fraction of students are male.

1. There are 840 students in the class

2. 75% of the students have applied to college

B

s1) insufficient. we don't need to know total number of students in class.

s2)

let x=total number of students
.75x=students applied therefore (.72)(.75)x=fraction of students who are male.

however, i will add caveat. this is with assumption that question is asking "what fraction of students(that applied) are male?" You may have typo. Because if that's not explicit, this might be an E. In that, there might be males that did not apply that we are not taking into account when we generate fraction.

OA plz.

shahdevine wrote: however, i will add caveat. this is with assumption that question is asking "what fraction of students(that applied) are male?" You may have typo.

wow. It makes total sense!! If answer is E, then I will have to spend more time with reading Q.

Hope its typo.

OA is B

When I have comparable averages or percentages (where weights influence the percentage or average), I use what I call the "number line" method. I think its quick.

Statement 1 is clearly insufficient.

72% of males applied to college, 80% of females did and 75% of all students applied to college.

M-------->A-------->F
72%--->75%--->80%

Distance between M and A = 3%
Distance between F and A = 5%


As A is closer to M, there must be more males.

M:F = 5:3

The answer is B.

Wow this is really fast, but will it work everytime? If yes then nothing like it.