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CS103 and M101

Expert replies
by 4meonly » Mon Jan 26, 2009 12:47 pm
If 53 students are enrolled in both the CS103 (Algorithms and Data Structures) and the M101 (Mathematics for Computer Science) classes. How many of the CS103 students are not enrolled in M101?

(1) 72 students are taking the M101
(2) 59 students are taking the CS103

OA after
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Source: — Data Sufficiency |

by bluementor » Mon Jan 26, 2009 2:52 pm
Stmt 1: No info on CS103 participation. Insufficient.

Stmt 2: CS103_total = 59 = CS103_only + Both. Both is given, so statement 2 is suffcient.

Choose B.

-BM-
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by 4meonly » Wed Jan 28, 2009 3:07 am
why we chould not take into count those students who are NOT involved in any class?
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by sdjomou » Wed Jan 28, 2009 3:44 am
Statement 2 is definitely sufficient to answer the question.
if 59 students are enrolled in arithmetic and 53 are enrolled in both courses, that means 53 of the 59 are enrolled in Mathematics, thus 6 are not.
ready to take a challenged GMAT with a score target of 750
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by 4meonly » Wed Jan 28, 2009 11:49 pm
4meonly wrote:why we chould not take into count those students who are NOT involved in any class?
Guys,
I know the OA. The question is why answer is B?
We do not know the total number of students and the number of students that are enrolled only in M101. That's why I think answer is E.
What do you think?
Can anybody post exhaustive reasoning?
Thank you!

Image
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by Zipper » Wed Jan 28, 2009 11:57 pm
4meonly wrote:why we chould not take into count those students who are NOT involved in any class?

Because in the question you are asked "How many of the CS103 students are not enrolled in M101?" To answer this question you need only CS103 size and how many students are in both CS103 and M101 classes. The second is given in the stem, so you are looking only for CS103 size.

In 1 you don't know how many people are in CS103, so even when you know that M101 has 72 students you have no way of knowing how many of the CS103 students are not enrolled. so 1 //insuff

In 2 we know that 59 students are taking the CS103 class. From the stem we also know that 53 students take both CS103 and M101 classes. So 6 students from CS103 are not taking M101, hence //sufficient

Answer B.
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by 4meonly » Thu Jan 29, 2009 12:09 am
Ohhhh, yeeeees..... Thank you!
Stupid mistake. :?
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by Zipper » Thu Jan 29, 2009 12:24 am
U r welcome :)
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by sanju09 » Thu Jan 29, 2009 6:30 am
4meonly wrote:
4meonly wrote:why we chould not take into count those students who are NOT involved in any class?
Guys,
I know the OA. The question is why answer is B?
We do not know the total number of students and the number of students that are enrolled only in M101. That's why I think answer is E.
What do you think?
Can anybody post exhaustive reasoning?
Thank you!

Image
Where did you get this image from, friend? 72 should be the total number of students taking M101, not M101 only. Though it doesn't help much.

By the way IMO B.
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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Re: CS103 and M101

by maihuna » Thu Jan 29, 2009 6:38 am
4meonly wrote:If 53 students are enrolled in both the CS103 (Algorithms and Data Structures) and the M101 (Mathematics for Computer Science) classes. How many of the CS103 students are not enrolled in M101?

(1) 72 students are taking the M101
(2) 59 students are taking the CS103

OA after
c and m 53
c total 59 which includes the 53 taking m as well, so 6 not enrolled for m
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