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3/8 of all students at Social High are in all three of the

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3/8 of all students at Social High are in all three of the  [#permalink]

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New post 21 Aug 2010, 06:07
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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
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 21 Aug 2010, 06:33
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zisis wrote:
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


Let's # of students at Social High be 8 (I picked 8 as in this case 3/8 of total and 5/8 of total will be an integer).

3/8 of all students at Social High are in all three clubs --> 3/8*8=3 people are in exactly 3 clubs;

1/2 of all students are in Albanian club --> 1/2*8=4 people are in Albanian club;
5/8 of all students are in Bardic club --> 5/8*8=5 people are in Bardic club;
3/4 of all students are in Checkmate club --> 3/4*8=6 people are in Checkmate club;

Also as every student is in at least one club then # of students in neither of clubs is 0;

Total=A+B+C-{# of students in exactly 2 clubs}-2*{# of students in exactly 3 clubs}+{# of students in neither of clubs};

8=4+5+6-{# of students in exactly 2 clubs}-2*3+0 --> {# of students in exactly 2 clubs}=1, so fraction is 1/8.

Answer: A.

For more about the formula used check my post at: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups or overlapping-sets-problems-87628.html?hilit=exactly%20members

Hope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 22:30
If you plot this data on a venn diagram, the problem would be much clearer .
General formula for 3 intersecting sets of data:
n(A u B u C) = n(A) + n(B) + n(C) - n(A int B) - n(B int C) - n(A int C) + n(A int B int C).
Assuming all students were part of atleast 1.
My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 22:43
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vicksikand wrote:
If you plot this data on a venn diagram, the problem would be much clearer .
General formula for 3 intersecting sets of data:
n(A u B u C) = n(A) + n(B) + n(C) - n(A int B) - n(B int C) - n(A int C) + n(A int B int C).
Assuming all students were part of atleast 1.
My solution is giving me an absurd result and it was definitely not 1/8. Please explain, I might have missed the point.


It's not clear how you applied the formula you wrote and what answer did you get, so it's hard to tell where you went wrong.

But the problem might be in formula you apply. Actually there are 2 formulas for 3 overlapping sets and the second one is better for this particular problem. Check the solution above and also this link about the mentioned 2 formulas: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups

Hope it helps.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 22:57
1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 23:02
vicksikand wrote:
1. For 3 sets A, B, and C: P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
Thats the one I have used all along and has worked fine for me. I would think that this formula would apply to this problem.


It depends how you apply the formula. Again it's not the only formula for 3-overlapping sets, there is a second one and it fits this problem better. You can check the link I gave if you are interested to know the difference and to study how to apply either of them.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 23:08
5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

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?

It doesnt say 1/2 of students are in Albanian only - In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.

We have to find (P(A n B) + P(A n C) + P(B n C))/2
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 20 Sep 2010, 23:23
vicksikand wrote:
5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)

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?

It doesnt say 1/2 of students are in Albanian only - In my opinion 1/2 includes the number that are common with Bardic and Checkmate. Formula 2 is for finding no. of persons in one set. I kind of see how the author arrived at that formula, but I wouldnt apply it to our case.

We have to find (P(A n B) + P(A n C) + P(B n C))/2


You are right it doesn't say that 1/2 of students are in Albanian only and nowhere in the solution above it's mentioned. Also this is not the formula I meant.

Refer to my post at: formulae-for-3-overlapping-sets-69014.html?hilit=exactly%20groups

And refer to the solution above to see how the second formula is applied to this problem.
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Re: The Quest for 700: Weekly GMAT Challenge  [#permalink]

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New post 21 Sep 2010, 00:07
I agree. Good explanation on the proof.
(a n b n c) was bothering me and you did a good job explaining why we need to subtract it twice and let that 1 remain so that it is counted at least once.

I remember having seen 1 question based on this in the MGMAT test series.

Thanks!
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 11 Jun 2013, 08:23
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Overlapping Sets:
advanced-overlapping-sets-problems-144260.html
how-to-draw-a-venn-diagram-for-problems-98036.html

All DS Overlapping Sets Problems to practice: search.php?search_id=tag&tag_id=45
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 30 Oct 2014, 09:58
Why did you multiply {# of students in exactly 3 clubs} by 2?
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 30 Oct 2014, 10:01
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3/8 of all students at Social High are in all three of the  [#permalink]

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New post 01 Nov 2014, 22:40
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Basic formula is:

Total=A+B+C - exactly two - 2*all three, so

Let pick 16 as total number of students

16=8+10+12 - exactly two - 2*6
16=30-x-12
x=2

2/16=1/8

with fractions:

8/8=1/2+5/8+3/4 - x - 3/4
8/8=15/8-x-3/4
x=1/8

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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 05 Jul 2018, 16:45
zisis wrote:
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


We can let the total number of students = 8, and thus:

Since 3/8 of all students are in all 3 clubs, 3 students are in all 3 clubs

Since 1/2 of all students are in Albanian, 4 students are in Albanian

Since 5/8 of all students are in Bardic, 5 students are in Bardic

Since 3/4 of all students are in Checkmate, 6 students are in Checkmate.

From the given information, we know that 0 students are in none

Letting t = the number of students in exactly two clubs, we can now use the equation:

Total = A + B + C - (exactly two clubs) -2(all three clubs) + none

8 = 4 + 5 + 6 - t - 2(3) + 0

8 = 15 - t - 6

t = 1

So, 1/8 are in exactly two clubs.

Answer: A
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 11 Jul 2018, 22:19
ScottTargetTestPrep wrote:
zisis wrote:
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


We can let the total number of students = 8, and thus:

Since 3/8 of all students are in all 3 clubs, 3 students are in all 3 clubs

Since 1/2 of all students are in Albanian, 4 students are in Albanian

Since 5/8 of all students are in Bardic, 5 students are in Bardic

Since 3/4 of all students are in Checkmate, 6 students are in Checkmate.

From the given information, we know that 0 students are in none

Letting t = the number of students in exactly two clubs, we can now use the equation:

Total = A + B + C - (exactly two clubs) -2(all three clubs) + none

8 = 4 + 5 + 6 - t - 2(3) + 0

8 = 15 - t - 6

t = 1

So, 1/8 are in exactly two clubs.

Answer: A
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 11 Jul 2018, 22:26
Let the total number of students be 96, a multiple of 8. Then P(A)=48, P(B)=60, P(C)=72 and P(An BnC)= 36,APPLYING FORMULA
P(AUBUC)= P(A)+P(B)+P(C)-P(An b)-P(BnC)-P(AnC)+ P(AnBnC)
We get P(AnB) +P(BnC)+P(AnC)= 120
So, number of students in exactly two clubs will be 120- 3*36= 12, which is 1/8th of 96. Answer: A
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 30 Jan 2019, 13:27
zisis wrote:
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



When they ask "what fraction of some population is category x?" Pick an easy to use number. 8, 4, and 2 are divisible by 64, so let the total number of students = 64. then the number of students in all three groups = 3/8 * 64 = 24

The number of students in Albanian = 32, and the number in Bardic = 40, and the number in checkmake = 48


The formula for 3 set overlaps is Group A + Group B + Group C - (sum of 2 group overlaps) - 2*(3 group overlaps) - neither group = total Since every student is in one of the clubs, we don’t have to worry about the neither group

So then by the formula 32+40 +48-(2overlap) - 2* 24=64

Then 120 - (2 group overlap) -48 = 64

and 72 - 2 group overlap - 64 so 2 group overlap =8. Since we chose 64 to equal the total student body, then 8/64 = 1/8 of the student body is in exactly 2 clubs.
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 05 Feb 2019, 12:26
zisis wrote:
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


My reasoning:

\(\frac{1}{2} + \frac{5}{8} + \frac{3}{4} - BOTH - 2*\frac{3}{8} = 1\)

BOTH = \(\frac{1}{8}\)
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 01 Jul 2019, 11:14
zisis wrote:
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


let total = 64
so all three = 3/8*64 ; 24
A=32
B=40
C=48
64=32+40+48-(2*24)-exactly 2
exactly 2 = 8 i/e 1/8 of 64
IMO A
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Re: 3/8 of all students at Social High are in all three of the  [#permalink]

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New post 19 Nov 2019, 17:34
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zisis wrote:
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


A student asked me to solve this without using a formula, so here we go....

Let's assign a nice value to the TOTAL number of students.
So we need a number that will work well with all of the fractions in the question.

Let's say there are 16 students in total

3/8 of 16 = 6. So there are 6 students and all three clubs.
1/2 of 16 = 8. So, there are 8 students in in Albanian
5/8 of 16 = 10. So, there are 10 students in in Bardic
3/4 of 16 = 8. So, there are 12 students in in Checkmate

We get the following:
Image


What fraction of the student body is in exactly 2 clubs?
Let a, b and c = the number of students in exactly 2 clubs
Image
So, our goal is to find the SUM a + b + c


Since there is a total of 8 students in Albanian, we know that 2 - a - b = the number of students that are in Albanian only (this way the total number of students in Albanian adds to 8)
Since there is a total of 10 students in Bardic, we know that 4 - b - c = the number of students that are in Bardic only (this way the total number of students in Bardic adds to 10)
Since there is a total of 12 students in Checkmate, we know that 6 - a - c = the number of students that are in Checkmate only (this way the total number of students in Checkmate adds to 10)
We get:
Image


Since there are 16 students all together, the sum of all the expressions in our diagram must add to 16.
In other words: (2 - a - b) + b + (4 - b - c) + 6 + a + c + (6 - a - c) = 16
Simplify to get: 18 - a - b - c = 16
Subtract 16 from both sides to get: 2 - a - b - c = 0
Rearrange to get: a + b + c = 2

In other words, 2 of the 16 students are in exactly two clubs.

2/16 = 1/8

Answer: A

Cheers,
Brent
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Re: 3/8 of all students at Social High are in all three of the   [#permalink] 19 Nov 2019, 17:34

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