Each of the 59 members in a high school class is required

Question

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

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

Ritterberkeley · Accepted Answer

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.

Hence AnB + AnC + BnC = (anb + g) + (anc + g) + (bnc + g)

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.

Bunuel · Answer

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 . Answer: C. For more check ADVANCED OVERLAPPING SETS PROBLEMS Similar problem at: ps-question-94457.html#p728852 Hope it helps.

kostyan5 · Answer

59 - poetry - history - writing + 2*(two clubs) + three clubs = 0
59 - 22 - 27 - 28 + 12 + three clubs = 0
three clubs = 6

C

manalq8 · Answer

QUESTION

why did u multiply 2*(two clubs)?

alphabeta1234 · Answer

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:

overlapping-sets-84100.html
ps-venn-diagrams-77473.html

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?

Your my hero dude. Thank you!

irishiraj87 · Answer

Hi,
the question states that -A total of 22 students sign up for the poetry club which means 22 is the whole of poetry club including the intersections with other clubs....Can you please explain how do we know if poetry is exclusive of the intersections?

chetan2u · Answer

it is inclusive.... 
P+N+W-(sum of two clubs)..... Means that P and N and W are inclusive and you are subtracting repeated elements from the SUM

2asc · Answer

I wasn't all that familiar with the overlapping concept, but I solved it through drawing a diagram -- posting in case anyone is looking for a different logic.

Attachment:

IMG_0282.JPG [ 1.45 MiB | Viewed 39087 times ]

**The shaded areas cover the 6 students who joined exactly 2 clubs.

Total # of students = (3club) + (2club) + (1club).
3-club: X
2-club: 6
1-club: P+H+W-(2clubs)-(3clubs)
Note that when we add the given P H W values together, we have added each of the shaded 2-club areas twice, and the center 3-club area three times, so be sure to account for that in your equation.

59=X+6+(22+27+28-(2*6) - 3X)
2X=6+77-12-59
2X=12
X=6.

Adedami · Answer

Hi house, i just went through the advanced overlapping set topic and it is very well helpful. God bless the hands responsible for putting it all together. However i have a question pls and it goes thus; How do i know when to use the first formular i.e TOTAL= A+B+C-(SUM OF 2-GROUP OVERLAPS)+ALL THREE+NEITHER and when to use second formular i.e TOTAL=A+B+C-(SUM OF EXACTLY 2-GROUP OVERLAPS)+2(ALL THREE)+NEITHER. 
 i need a quick response its just a few days to my test.

Nik11 · Answer

I saw that many don't understand why sometimes someone multiply *2 in the middle of the formula.
Despite it's trivial I believe this might help.
When you should use the second formula, instead of using it if you subtract twice you can come back to the first type formula.. that's it
So instead of resolve for 2*g
You can resolve for g
But the concept is the same.
Moreover, you know when you have to use the second type when they give you the data:"exactly 2", because in this case it's already "net of g"

Posted from my mobile device

willreedy10 · Answer

Is anyone familiar with the "Double Set Matrix" strategy to solve overlapping set questions? If so, is that possible here/ can you walk through that?

Vubar · Answer

Hi Will, I am not an expert by any stretch, but I don't believe you can use the standard double set matrix table for more than 2 sets. That said, GMATNinja (specifically Harry at The GMAT Ninja org), ran through a method I think is pretty great (because I never learned the formulas mentioned above). He taught the method in this 24-hour fundraiser stream: https://youtu.be/FJa6Dz5p1Ko?t=23573 - Check it out, and support the cause if you can (or another cause depending when you see this!). But using GMAT Ninja's table method, my working looked like this: https://media.discordapp.net/attachments/924831114869235722/974509842523889684/unknown.png I can give more explanation if anyone needs it. I just realised that I didn't add column headings for the other two columns. If I did, the column headings would have been: # clubs | # students | # with overlaps (club members)

ayakik · Answer

Hello  KarishmaB   avigutman  I have gone through the thread twice and I'm still having trouble understanding this concept.

I got the correct answer through the following process:
P+H+W = 77 - 59 = 18 
18 - 2*6 = 6 (since 6 people are in 2 groups)
So remaining 6 must be in 3 groups

Is this reasoning correct? I'm a little confused about how to understand the general formula, Total=P+H+W -{Sum of Exactly 2 groups members}-2*PnHnW + Neither.

I always find both of your explanations so intuitive and so much easier to understand than general formulas so it would be really great if I could get your take on this. Thank you.

avigutman · Answer

My advice is to ignore this problem,  ayakik . I haven't seen a question like this on the GMAT in at least 15 years.

KarishmaB · Answer

Just about but not quite correct. You are thinking in terms of people and instances - which is great! There are total 59 people and 77 instances (number of choices made by these 59 people. Some people choose only one, some choose two and some choose 3 so we have more total choices than number of people) Since everyone chooses at least one, 77 - 59 = 18 instances are left. These are of those people who choose two or three clubs. Since we have already removed one club for each person, now the people who chose 2 clubs are counted in 18 once. So we are left with 18 - 6 = 12 instances. These 12 instances represent the people who chose 3 clubs. They will have two representations each in these 12 instances. So number of people who chose 3 clubs = 12/2 = 6. Alternatively, use the Venn diagram. Screenshot 2022-11-14 at 10.19.27 AM.png Say each letter represents the region of a single colour only. So (d + f+ e) represents people who chose exactly two clubs. 59 = a + b + c + d + e + f + k (each person lies somewhere in the region) 22 + 27 + 28 = (a + d + f + k) + (b + d + k + e) + (c + f + k + e) 77 = a + b + c + 2(d+f+e) + 3k 77 = 59 + (d + e + f) + 2k 77 = 59 + 6 + 2k k = 6

ScottTargetTestPrep · Answer

We can use the following formula:

#Total = #Group A + #Group B + #Group C - #Exactly two - 2 * #Exactly three + #Neither

Since we are told that every student is a member of at least one group, #Neither = 0. Let's substitute #Total = 59, #Group A = 22, #Group B = 27, #Group C = 28, #Exactly two = 6, and #Neither = 0 in the above formula:

59 = 22 + 27 + 28 - 6 - 2 * #Exactly three + 0

59 = 71 - 2 * #Exactly three

-12 = -2 * #Exactly three

6 = #Exactly three

Answer: C

GobaChatro · Answer

Is there a way to solve this problem using only Venn Diagrams?

KarishmaB · Answer

Yes, but you need to understand the Venn diagram properly. It is explained here: https://anaprep.com/sets-statistics-thr ... ping-sets/ Overall you have 59 people. In 22+27+28 = 77, those belonging to exactly two sets are counted twice and to exactly 3 sets are counted thrice. So from 77 you can go to 59 by subtracting 'those belonging to Exactly two sets (given = 6)' once and 'those belonging to 3 sets (say = x)' twice. 77 - 6 - 2*x = 59 So x = 6 Answer (C)

findingmyself · Answer

Exactly two=∣A∩B∣+∣B∩C∣+∣C∩A∣−3∣A∩B∩C∣ 
6+ 3∣A∩B∩C∣=A∩B∣+∣B∩C∣+∣C∩A∣

∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣B∩C∣−∣C∩A∣+∣A∩B∩C∣ 
 59=22+27+28-6-3∣A∩B∩C∣+∣A∩B∩C∣ 
 2A∩B∩C=12 
 A∩B∩C=6

Each of the 59 members in a high school class is required

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Each of the 59 members in a high school class is required

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards