Operations on rational numbers - Question 222/OG 13

[gmattesttaker2]

Hello,

I was just wondering if you can please help with this question. This is from OG 13:

OF the 200 students at College T majoring in one or more of th esciences, 130 are majoing 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


OA: D


I was wondering which of the following matrix diagrams are correct? I always had "venn" in mind but an earlier posting of this problem had "venn2". Can you please assist here? Thanks a lot.

Best Regards,
Sri

[spark]

Unfortunately, there are errors in both of the tables you posted. In venn2, the Not Chemistry/Not Biology box should be "at least 30," so to begin, I would write either >=30 or 30+ in that box. Ultimately, the Not Chemistry/Not Biology box must be 30-50, as explained below. In venn, the Chemistry/Not Biology box should be 0-20, and the Not Chemistry/Biology box should be 20-40.

I've attached a new table below with the correct numbers. The key is that every box must satisfy the constraints imposed by its row and by its column.

Since the Not C / Not B box is >=30, the constraint from the 2nd column forces the Not C / B box to be 0-40.

Likewise, since the Not C / Not B box is >=30, the constraint from the 2nd row forces the C / Not B box to be 0-20.

The constraint imposed by the 1st row forces the C / B box to be 110-150, and the constraint imposed by the 1st column forces the C / B box to be 110-130. Since the C / B box must satisfy both constraints, you must go with the more restrictive 110-130.

Also notice that the constraint from the 2nd column forces the Not C / Not B box to be 30-70, and the constraint from the 2nd row forces the Not C / Not B box to be 30-50. Both constraints must be satisfied, so you must go with the more restrictive 30-50.

Likewise, because the Not C / Not B box must actually be 30-50, the Not C / B box could only be 20-40.

[lunarpower]

there's also a more concrete way to deal with this problem, once you've got the diagram in place:
* Find a number that's IN some of the answer choices, and NOT in others.
* Try that number.
* If the diagram works, keep the choices that contain your number.
* If the diagram doesn't work, keep the choices that DON'T contain your number.

e.g.
let's start with 45, which is in choices a/b/c but not d/e.
if you plug 45 into the "both" square, you get 130 - 45 = 85 in the "chemistry only" square, and 150 - 45 = 105 in the "biology only" square.
oops, those three squares already add up to more than 200 students (and that's not even counting the 30+ students majoring in neither subject).
so, eliminate a, b, and c.

now try 140, which is in choice e but not in choice d.
if you plug 140 into the "both" square, then you get -10 (that's negative 10) people for "chemistry only".
that doesn't work either, so eliminate e.

done.

[GMATGuruNY]

Hello,

I was just wondering if you can please help with this question. This is from OG 13:

OF the 200 students at College T majoring in one or more of th esciences, 130 are majoing 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

OA: D

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.

[spark]

A table is not necessarily the most efficient way to evaluate this problem. In my opinion, a table offers the most universally helpful approach to these 2-set overlapping set problems because a table lays out all the information in way that makes it easy not to miss anything. But as in this case, there is definitely a more elegant approach, as long as you don't miss any important constraints.

The general setup can be described as:

TOTAL = B + C - BOTH + NEITHER

...and so

200 = 150 + 130 - BOTH + NEITHER

...and therefore

BOTH = NEITHER + 80

...so since NEITHER >=30, BOTH must be >=110, and C = 130, so the most BOTH could be is 130. This is an important constraint to avoid missing.