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Weird?!?

by Deepthi Subbu » Wed Aug 03, 2011 4:16 am
Is this question right ?

Of the 250 students in a certain class, each student who majors
in mathematics also majors in computer science, and 90 of the
students major only in biology. If no student majors in all
three subjects, how many of the students major in two of the
three subjects?
(1) 50 of the students major only in computer science.
(2) 40 of the students do not major in any of the three
subjects.


In the question stem it says - no student majors in all
three subjects . But in statement B it says 40 of the students do not major in any of the three
subjects . Arent they contrary to each other ?

Also nothing is mentioned about students taking only math . Does it mean we dont have the necessary info or it means none has taken only math ?

To my surprise the answer is C .
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by Frankenstein » Wed Aug 03, 2011 5:58 am
Hi,
In the question stem it says - no student majors in all
three subjects . But in statement B it says 40 of the students do not major in any of the three
subjects . Arent they contrary to each other ?
No, they are not contrary. The question is just fine.
If we represent majors in Maths,Computer Science and Biology as M,C,B then
no student majors in all three subjects -> n(M n C n B) = 0
40 of the students do not major in any of the three subjects -> n( M U C U B) = 250 - 40.
Coming to the problem,
Each student who majors in mathematics also majors in computer science -> M is a subset of C i.e. set M lies entirely in C, when we draw a Venn diagram.
So, n(M n C) = n(M) and n(M U C) = n(C)
So, n(M U C U B) = n(C U B)
As no student majors in all three subjects, there should be any elements common to M and B.
So, n(M n B) = 0
Let n(M n B) = x.
Number of students student who major in 2 of the 3 subjects is given by
n(M n B) + n(M n C) + n(B n C) = 0 + n(M) + n(B n C)
So, we need to find n(M) + n(B n C)
Given that 90 of the students major only in biology -> n(only B) = 90
From(1):
n(Only C) = 50
This means n(C) - n(M) - n(C n B) = 50.
So, n(M) + n(B n C) = n(C) - 50.

From(2):
n(C U B U M) = 210.
So, n(C) + n(only B) = 210.
So, n (C) = 210 - 90 = 120.

From(1) and (2):
n(M) + n(B n C) = n(C) - 50 = 120-50 = 70

Hence, C

Hey I couldn't draw this as a Venn diagram and upload this. Please draw a Venn diagram, then it will be easy to understand.
Draw it in such a way, M lies entirely in C and there will be no common part for M and B although C and B can overlap.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by gmatboost » Wed Aug 03, 2011 8:57 am
A good starting diagram of the situation could look like this

Image

We must find x + y.

You must remember that "None" is a possibility.
The 0 in the center and the 90 come from the question prompt directly, and the other two 0's come from the fact that every Math major MUST also be a CS major. So, no one can be Math only, and no one can be just Math and Bio.

To answer the original question: The prompt says that no one is in the very center, but Statement 2 is addressing the "None" region.

The answer is C because we need both the CS-only number and the "None" number in order to get everything but x+y. Once we have that total of everything else (90 + 50 + 40 = 180), we can get x + y = 250 - 180 = 70.

Hope this helps.
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