* I understand everything about this problem. I even got 12 as an answer but I didn't divide by 2.
* Can you please explain why the following is true and how I know when to do this:
o For each student who signs up for three clubs, there are two extra sign-ups. Therefore, there must be 6 students who sign up for three clubs:
12 duplicates / (2 duplicates/student) = 6 students
THANKS!
GMAT Forum
6 postsPage 1 of 1
- Anne1276
Original Problem
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
- Anne1276
It is MGMAT CAT #1 Question 32:
Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
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Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?
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- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
MGMAT CAT PS
Tricky problem. I think your question comes from the "intuitive" explanation example. And do remember that there are multiple ways to do any problem - if you found a way that works for you, feel free to use it (as long as you make sure it is really a valid method and you didn't just get lucky - but sounds like you understand this one).
If you add up the numbers for each of the three clubs (22 + 27 + 28 = 77) you are overcounting the students, because some students are in 2 clubs and some are in 3 clubs. So, the 77 figure includes the double-counts and triple-counts. The problem also tells you that there are 59 members in the class, so we know how many overcounts we have: 77 - 59 = 18. Those 18 "extras" represent not unique students but those who have been counted either twice or three times, instead of just once.
If a student has been counted twice, one of those "counts" is in the first 59 (which is fine, because we want the student to be counted once) but the second count is part of the 18 overcounts. The problem tells us that 6 students have been counted twice, so I can subtract out the second count for each one (which adds up to 6 overcounts) from the 18 total overcounts. I now have 12 overcounts left.
If a student has been counted three times, one of those counts is part of the first 59 (which is fine, because we want the student to be counted once), but the second AND third counts are part of the 12 remaining overcounts. So, for each student in 3 clubs, that student gets overcounted twice - which is why I divide 12 by 2 to get 6 students who have been triple-counted.
Essentially, you use the above method whenever you have something double or triple (or more) counted. You just have to adjust the method depending upon how many extra times (above the first time) the student has been counted.
If you add up the numbers for each of the three clubs (22 + 27 + 28 = 77) you are overcounting the students, because some students are in 2 clubs and some are in 3 clubs. So, the 77 figure includes the double-counts and triple-counts. The problem also tells you that there are 59 members in the class, so we know how many overcounts we have: 77 - 59 = 18. Those 18 "extras" represent not unique students but those who have been counted either twice or three times, instead of just once.
If a student has been counted twice, one of those "counts" is in the first 59 (which is fine, because we want the student to be counted once) but the second count is part of the 18 overcounts. The problem tells us that 6 students have been counted twice, so I can subtract out the second count for each one (which adds up to 6 overcounts) from the 18 total overcounts. I now have 12 overcounts left.
If a student has been counted three times, one of those counts is part of the first 59 (which is fine, because we want the student to be counted once), but the second AND third counts are part of the 12 remaining overcounts. So, for each student in 3 clubs, that student gets overcounted twice - which is why I divide 12 by 2 to get 6 students who have been triple-counted.
Essentially, you use the above method whenever you have something double or triple (or more) counted. You just have to adjust the method depending upon how many extra times (above the first time) the student has been counted.
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
6 posts Page 1 of 1