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Each of the 59 members in a high school class is required [#permalink]

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25 Nov 2011, 14:59

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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?

Union_3sets.gif [ 11.63 KiB | Viewed 15796 times ]

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? A. 2 B. 5 C. 6 D. 8 E. 9

Translating: "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" Total=59; Neither=0 (as members are required to sign up for a minimum of one); "22 students sign up for the poetry club": P=22; "27 students for the history club": H=27; "28 students for the writing club": W=28; "6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs)

"How many students sign up for all three clubs": question is \(PnHnW=x\). Or section 4 =?

Apply formula: \(Total=P+H+W -\){Sum of Exactly 2 groups members}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

Re: Each of the 59 members in a high school class is required [#permalink]

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26 May 2012, 17:50

Hey Bunuel,

Dude your awesome! Just a quick question, as I like to attack problems from different methods. The forumula you used is perfect but I also learned another cool formula from another problem which you can see here:

The formula involves finding the minimum value for the intersection of all the sets (A and B and C) in a three overlapping set problem. Here it is: According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?

The way it is attacked is by finding the # of people that arent in the set

So the solution to the above problem I posted would be : A and B and C = Total - [(Total-A) + (Total-B) + (Total-C)]

My question is if I were to use this method to find the solution I get weird numbers.

Total=59 A=Poetry=22 B=History=27 C=Writing=39

A and B and C =59-[(59-22)+(59-32)+(59-28)]=49. Actual answer is 6.

Why does this formula work on the problem I posted but not on this question?

Re: Each of the 59 members in a high school class is required [#permalink]

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09 Nov 2014, 02:02

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Re: Each of the 59 members in a high school class is required [#permalink]

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08 Jun 2015, 03:10

Bunuel wrote:

manalq8 wrote:

QUESTION

why did u multiply 2*(two clubs)?

Attachment:

Union_3sets.gif

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? A. 2 B. 5 C. 6 D. 8 E. 9

Translating: "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" Total=59; Neither=0 (as members are required to sign up for a minimum of one); "22 students sign up for the poetry club": P=22; "27 students for the history club": H=27; "28 students for the writing club": W=28; "6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs)

"How many students sign up for all three clubs": question is \(PnHnW=x\). Or section 4 =?

Apply formula: \(Total=P+H+W -\){Sum of Exactly 2 groups members}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

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? A. 2 B. 5 C. 6 D. 8 E. 9

Translating: "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" Total=59; Neither=0 (as members are required to sign up for a minimum of one); "22 students sign up for the poetry club": P=22; "27 students for the history club": H=27; "28 students for the writing club": W=28; "6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs)

"How many students sign up for all three clubs": question is \(PnHnW=x\). Or section 4 =?

Apply formula: \(Total=P+H+W -\){Sum of Exactly 2 groups members}\(-2*PnHnW + Neither\) --> \(59=22+27+28-6-2*x+0\) --> \(x=6\).

Re: Each of the 59 members in a high school class is required [#permalink]

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27 Aug 2015, 13:19

The following may be helpful to those struggling with the logic of the formula Total=P+H+W -{Sum of Exactly 2 groups members}-2*PnHnW + Neither .

First, forget this formula for a moment and rather recall the first "easier" formula: Total = A + B + C - (all cases of 2 group overlap, which equals AnB + AnC + BnC) + (the single 3 group overlap, which is AnBnC, or g) + (neither)

Recall that when we subtract all cases of group overlap, AnB + AnC + BnC, then we also subtract the 3 group overlap, g, 3 times. Since we add g 3 times when we add A +B + C, we are back at 0 regarding g, hence why we add it back following this subtraction, as you can see up there in the first formula.

In the gmat question we have above, the "simpler" first formula rendition would therefore go: 59 = 22 + 27 + 28 - (AnB + AnC + BnC) + g + 0

We know that the total number of members in exactly two groups (so NOT the ones in all three) is 6.

Now, can we make AnB + AnC + BnC just equal 6? No, since AnB, and the rest of them, also include those members of all three groups, and thus may be larger than 6. The work around is that we simply need to account for g in each case, by including it explicitly in the formula and then set the part that is NOT g equal to 6.

Hence, AnB = (some part of the total 6) + g, and the same for the others. Just to give 'the part of the total 6' on A and B's side a name, let's substitute 'anb'. Don't get confused here: anb is just AnB with g subtracted.

But what does anb equal? We don't know, but what we DO know that anb + anc + bnc, all the sets that have ONLY members of two and no more or less, is equal to 6.

Hence anb + anc + bnc = 6

With this understood we just replace anb, anc, and bnc with 6.

Hence, (AnB + AnC + BnC) = (6 + g + g + g)

Back to the first formula rendition, we now have 59 = 22 + 27 + 28 - (6 + g + g +g) + g + 0 Simplified (a little): 59 = 22 + 27 + 28 - 6 - g - g - g + g + 0 Simplified (a little more): 59 = 22 + 27 + 28 - 6 - 2g + 0

But wait, this just is the second formula!

Second formula: Total = P+H+W -{Sum of Exactly 2 groups members}-2*PnHnW + Neither

Where we left off: 59 = 22 + 27 + 28 - 6 - 2g + 0

You can now hopefully see WHY 6, the sum of those in exactly 2 groups, is subtracted, as well as WHY 2g is subtracted.

Re: Each of the 59 members in a high school class is required [#permalink]

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30 Sep 2016, 03:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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