Manhattan GMAT Challenge Problem of the Week – 19 Aug 2010

by , Aug 19, 2010

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As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

(A) 1/8

(B) 1/4

(C) 3/8

(D) 1/2

(E) 5/8

Answer

Since we are only dealing in fractions, lets pick a simple number for the total number of students: 8. Thus, 3 students are in all 3 clubs. 4 students are in A, 5 students are in B, and 6 students (3/4 of 8 ) are in C. Since every student is in at least one club, then we know that the total number of club members, appropriately counting people in two or three clubs, must be 8.

We can use the following idea: suppose there were no overlap in the club memberships at all. Then we could just add up the club memberships, and there would have to be 4 + 5 + 6 = 15 students. Now, we know that this is overcounting; we counted 3 students 3 times (for all 3 clubs), when we should have only counted them once. So we overcounted them twice, meaning that we should subtract 23 = 6 students from the 15, leaving 9. We still have an extra student in the count, since we really only have 8 people. Now, lets think about the people in exactly 2 clubs: we counted those people twice, when we should have counted them just once. So we overcounted them once, and we should subtract their number once from the total. Representing this number by x, we get 9 x = 8, so x = 1. The fraction of people in exactly 2 clubs is 1/8.

An easier formula for this process that captures the same idea is this:

Total people in at least one club

= Total of separate memberships of each club,

MINUS the people in exactly two clubs,

MINUS TWICE the people in all three clubs.

Plugging in numbers, we get

8 = 4 + 5 + 6 x 2(3)

8 = 15 x 6

8 = 9 x

x = 1

Again, the fraction is 1/8.

This can also be solved by a Venn diagram with three overlapping circles and placing numbers to ensure that the circles total properly (A = 4, B = 5, C = 6), that the central overlap has 3 people, and that the overall total is 8. Trial and error reveals that we will need to place 1 person in one of the leaves (indicating membership in exactly 2 clubs).

The correct answer is (A).

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