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by zaarathelab » Sat Oct 15, 2011 1:29 am
In a sample of college students, 40 percent are third-year students and 70 percent are not second-year students. What fraction of those students who are not third-year students are second-year students?

(A) 3/4
(B) 2/3
(C) 4/7
(D) 1/2
(E) 3/7




I setup the double matrix correctly (SEE ATTACHMENT), but i got lost in the final line of the problem. Can someone explain the last line? Don't we have to find a?
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by sl750 » Sat Oct 15, 2011 1:50 am
Going by your double matrix diagram a=30 . So 30/60 = 1/2. The case of a student being both in the second and third year is not possible, hence it is zero

Percentage of students who are NOT third year = 60
Percentage of students who are second year = 30

So 30/60 is the fraction who are NOT third year but second year students
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by jayavignesh » Sat Oct 15, 2011 3:23 am
zaarathelab wrote:In a sample of college students, 40 percent are third-year students and 70 percent are not second-year students. What fraction of those students who are not third-year students are second-year students?

(A) 3/4
(B) 2/3
(C) 4/7
(D) 1/2
(E) 3/7




I setup the double matrix correctly (SEE ATTACHMENT), but i got lost in the final line of the problem. Can someone explain the last line? Don't we have to find a?
Please forget about the matrix .Go by percentages.
Look for information that can be withdrawn from the statements given.
statement 1: 60% are not third year students.
statement 2: 30% may or may not be second year students from the pool of 60 %

So it is 30% out of 60% i.e., (30/60)=(1/2)


Ans:D
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by rohit_gmat » Sat Oct 15, 2011 5:07 am
plug-in....

100 is a good number.
so # of 3rd yr students = 40
# of NOT 2nd yr students = 70

& so # of 2nd yr students = 30
# of students who are not 3rd students = 100 - 40 = 60
fraction = 30/60 = 1/2


D
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by zaarathelab » Sat Oct 15, 2011 7:17 am
OA is D
Success = Max(Hardwork) + Min(Luck)
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