Students are in clubs as follows: Science-20, Drama-30, and

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

Students are in clubs as follows: Science-20, Drama-30, and

by AAPL » Wed Aug 15, 2018 7:08 am

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Manhattan GMAT

Students are in clubs as follows: Science-20, Drama-30, and Band-12. No student is in all three clubs, but 8 are in both Science and Drama, 6 are in both Science and Band, and 4 are in Drama and Band. How many different students are in at least one of the three clubs?

A. 36
B. 38
C. 42
D. 44
E. 48

OA D.

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Source: Beat The GMAT — Problem Solving | Wed Aug 15, 2018 7:08 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Aug 15, 2018 8:24 am

AAPL wrote:Manhattan GMAT

Students are in clubs as follows: Science-20, Drama-30, and Band-12. No student is in all three clubs, but 8 are in both Science and Drama, 6 are in both Science and Band, and 4 are in Drama and Band. How many different students are in at least one of the three clubs?

A. 36
B. 38
C. 42
D. 44
E. 48

When we count the total in Science, the total in Drama, and the total in Band, any students in exactly two of the clubs are counted TWICE.
Thus, we must SUBTRACT these students so that they are not double counted:
Total = Science + Drama + Band - (exactly 2 clubs) = 20 + 30 + 12 - (8+6+4) = 44.

The correct answer is D.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Apr 19, 2019 9:14 am

AAPL wrote:Manhattan GMAT

Students are in clubs as follows: Science-20, Drama-30, and Band-12. No student is in all three clubs, but 8 are in both Science and Drama, 6 are in both Science and Band, and 4 are in Drama and Band. How many different students are in at least one of the three clubs?

A. 36
B. 38
C. 42
D. 44
E. 48

OA D.

We can use the following formula:

n(At least one set) = n(A) + n(B) + n(C) - n(A and B) - n(A and C) - n(B and C) + n(A and B and C)

n(At least one set) = 20 + 30 + 12 - 8 - 6 - 4 + 0

n(At least one set) = 44

Answer: D

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