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Group problem

by talaangoshtari » Tue Mar 10, 2015 1:58 am
Hi,

How can we solve this problem by using the double-matrix method?

OF the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from

A - 20 to 50
B 40 to 70
C - 50 to 130
D - 110 to 130
E - 110 to 150

Thanks!
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by DavidG@VeritasPrep » Tue Mar 10, 2015 3:51 am
Who doesn't love a good matrix problem?

Start by inserting the concrete values. If there are 200 people total and 130 take Chemistry, that means 70 don't take Chemistry. Similarly, if there are 150 people taking Biology, 50 don't take Biology. So here's our starting matrix:


Image

If we know that at least 30 take neither, we know that the No/Bio/No Chem cell must be AT LEAST 30. So that scenario gives us the following:

Image

This gives us a low end of our range who take both, 110.

Next we want to maximize that No Bio/Bo Chem cell. Notice that we're limited by that total of 50 who don't take Bio. Therefore 50 is the largest possible number of students who take neither Biology nor Chemistry. So now we have this:

Now we have the high end of our range who take both, 130.

Answer is D


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by GMATGuruNY » Tue Mar 10, 2015 5:22 am
Of the 200 students at College T majoring in one or
more of the sciences, 130 are majoring in chemistry
and 150 are majoring in biology. If at least 30 of the
students are not majoring in either chemistry or
biology, then the number of students majoring in both
chemistry and biology could be any number from

(A) 20 to 50
(B) 40 to 70
(C) 50 to 130
(D) 110 to 130
(E) 110 to 150
Chemistry = 130 and Biology = 150.
Since 130+150 = 280 -- exceeding the total number of students by 80 -- at least 80 students must major in both subjects.
Eliminate A, B and C.

Since only 130 students major in chemistry, the number of students majoring in both subjects cannot exceed 130.
Eliminate E.

The correct answer is D.
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by Brent@GMATPrepNow » Tue Mar 10, 2015 7:51 am
It's worth noting that David's Double Matrix approach can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of students, and the two characteristics are:
- majoring in chemistry or not majoring in chemistry
- majoring in biology or not majoring in biology

This question type is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch our free video: https://www.gmatprepnow.com/module/gmat- ... ems?id=919

Once you're familiar with this technique, you can attempt these additional practice questions:

Easy Problem Solving questions
- https://www.beatthegmat.com/the-aam-aadm ... 72242.html
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Difficult Problem Solving questions
- https://www.beatthegmat.com/ratio-problem-t268339.html
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Easy Data Sufficiency questions
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Medium Data Sufficiency questions
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Difficult Data Sufficiency questions
- https://www.beatthegmat.com/double-set-m ... 71423.html
- https://www.beatthegmat.com/sets-t269449.html
- https://www.beatthegmat.com/mba/2011/05/ ... question-3

Cheers,
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by Matt@VeritasPrep » Tue Mar 10, 2015 11:57 pm
talaangoshtari wrote:Hi,

How can we solve this problem by using the double-matrix method?

OF the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from

A - 20 to 50
B 40 to 70
C - 50 to 130
D - 110 to 130
E - 110 to 150

Thanks!
Another approach here. Since we have fewer people in chemistry than in biology, our MAXIMUM # of double majors comes if we assume that EVERY chemistry major is a double major. So we could have 130 double majors.

The other case is the MINIMUM, which is given in the problem (if exactly 30 students major in neither subject). That would give us 130 + 150 - (overlap) = (200 - 30), or an overlap of 110.

Hence it's 110 to 130, D, and we saved a lot of time!
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by gmatbeater1989 » Sat Oct 10, 2015 8:59 am
talaangoshtari wrote:Hi,

How can we solve this problem by using the double-matrix method?

OF the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from

A - 20 to 50
B 40 to 70
C - 50 to 130
D - 110 to 130
E - 110 to 150

Thanks!
Can I use the formua Group 1 + Group 2 + Neither - Both = Total here?
Is there a way to know when this formula works? I see it used some times and not others.
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by [email protected] » Sat Oct 10, 2015 9:08 am
Hi gmatbeater1989,

Yes, you can use the Overlapping Sets Formula here (although it won't be applicable on every Overlapping Sets question that you might see).

This prompt comes with a couple of 'twists' to it:
1) The number of students who study 'neither' is NOT a fixed value - it's a range.
2) The question asks for the RANGE of students who could study both Chemistry and Biology.

Here's how you can use the Formula though...

Total = Gp.1 + Gp.2 - Both + Neither

200 = 130 + 150 - B + (>=30)
200 = 280 - B + (>=30)
200 = (>=310) - B
B = >=110

This gives you the 'lower end' of the range, but does not immediately give you the 'upper end.' To find that, you have to think about the numbers involved. Since 130 students study Chemistry and 150 study Biology, the MAXIMUM number who could study both would be 130 (and that's if EVERY Chemistry student also studied Biology). Thus, the range is 110 to 130, inclusive.

Final Answer: D

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by gmatbeater1989 » Sat Oct 10, 2015 9:18 am
[email protected] wrote:
Yes, you can use the Overlapping Sets Formula here (although it won't be applicable on every Overlapping Sets question that you might see).
Thanks Rich.
So how do I know when the formula will work?
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