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# S96-17

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Math Expert
Joined: 02 Sep 2009
Posts: 54544

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16 Sep 2014, 01:51
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85% (hard)

Question Stats:

51% (02:28) correct 49% (02:25) wrong based on 116 sessions

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$$\frac{3}{8}$$ of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. $$\frac{1}{2}$$ of all students are in Albanian, $$\frac{5}{8}$$ are in Bardic, and $$\frac{3}{4}$$ are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

A. $$\frac{1}{8}$$
B. $$\frac{1}{4}$$
C. $$\frac{3}{8}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{8}$$

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16 Sep 2014, 01:51
Official Solution:

$$\frac{3}{8}$$ of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. $$\frac{1}{2}$$ of all students are in Albanian, $$\frac{5}{8}$$ are in Bardic, and $$\frac{3}{4}$$ are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

A. $$\frac{1}{8}$$
B. $$\frac{1}{4}$$
C. $$\frac{3}{8}$$
D. $$\frac{1}{2}$$
E. $$\frac{5}{8}$$

Since we are only dealing in fractions, let’s pick a simple number for the total number of students: 8. Thus, 3 students are in all 3 clubs. 4 students are in A, 5 students are in B, and 6 students ($$\frac{3}{4}$$ of 8) are in C. Since every student is in at least one club, then we know that the total number of club members, appropriately counting people in two or three clubs, must be 8.

We can use the following idea: suppose there were no overlap in the club memberships at all. Then we could just add up the club memberships, and there would have to be $$4 + 5 + 6 = 15$$ students. Now, we know that this is overcounting; we counted 3 students 3 times (for all 3 clubs), when we should have only counted them once. So we overcounted them twice, meaning that we should subtract $$2 \times 3 = 6$$ students from the 15, leaving 9. We still have an extra student in the count, since we really only have 8 people. Now, let’s think about the people in exactly 2 clubs: we counted those people twice, when we should have counted them just once. So we overcounted them once, and we should subtract their number once from the total. Representing this number by $$x$$, we get $$9 - x = 8$$, so $$x = 1$$. The fraction of people in exactly 2 clubs is $$\frac{1}{8}$$.

An easier formula for this process that captures the same idea is this:

Total people in at least one club = Total of separate memberships of each club, MINUS the people in exactly two clubs, MINUS TWICE the people in all three clubs.

Plugging in numbers, we get:
$$8 = 4 + 5 + 6 - x - 2(3)$$
$$8 = 15 - x - 6$$
$$8 = 9 - x$$
$$x = 1$$

Again, the fraction is $$\frac{1}{8}$$.

This can also be solved by a Venn diagram with three overlapping circles and placing numbers to ensure that the circles total properly ($$A = 4$$, $$B = 5$$, $$C = 6$$), that the central overlap has 3 people, and that the overall total is 8. Trial and error reveals that we will need to place 1 person in one of the "leaves" (indicating membership in exactly 2 clubs).

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30 Dec 2014, 13:49
Hi Bunuel,
It's not clear to me since I think that there could be another answer:

Considering that:
A=4
B=5
C=6

IN A VENN DIAGRAM:

A ∩ B ∩ C = 3 (Info)

A ∩ B = 3 (3 in ABC + 0 in AB but no C)
A ∩ C = 4 (3 in ABC + 1 in AC but no in B)
B ∩ C = 5 (3 in ABC + 2 in BC but no in A)

Estudents only in A = 0
Estudents only in B = 0
Estudents only in C = 0

This way the correct answer should be 3/8.

I just can find out what I did wrong. Help pls.
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31 Dec 2014, 04:56
TheRunawayFound wrote:
Hi Bunuel,
It's not clear to me since I think that there could be another answer:

Considering that:
A=4
B=5
C=6

IN A VENN DIAGRAM:

A ∩ B ∩ C = 3 (Info)

A ∩ B = 3 (3 in ABC + 0 in AB but no C)
A ∩ C = 4 (3 in ABC + 1 in AC but no in B)
B ∩ C = 5 (3 in ABC + 2 in BC but no in A)

Estudents only in A = 0
Estudents only in B = 0
Estudents only in C = 0

This way the correct answer should be 3/8.

I just can find out what I did wrong. Help pls.

Check alternative solution below. Hope it helps.

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.

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
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10 Apr 2015, 13:37
I was struggling for a while with the Venn diagram. I kept coming up with 9 people instead of 8. I finally figured it out, but in the final solution, it is not clear if there is either 0 people who are in club A and ONLY in one club, or if there are 0 people who are ONLY In club b.

in other words, the solution will in fact tell us that there is 1 person who is in two clubs, and thus there is only one person in the class who is only a member of either A or B. there could be multiple answers for the question of whether A or B has a single member. it is possible to get the right answer and still not know if only one person is in A or B. or perhaps the one person in two clubs is in C+B or C+A? the way that I drew my venn diagram, the person that was in exactly two clubs was a member of A+B.
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07 Jul 2015, 02:21
1
Hey. I sort of used the min/max to solve this one, as an alternative way to the formula, which is what I usually do (no thinking involved... )

To begin with, I used 16 as the total number of students. This, based on the fractions we are given, dials down to:
A = 8
B = 10
C = 12
All three clubls = 3.

Looking at the venn diagram, we add 6 in the middle, as 3/8 are in all three clubs.
Focusing on A, we know that there should be 8 people there. 6 are already in A (in the union of the 3 clubs), so we are left with 2 people. Let's put 1 person only in A (the min possible number, as we know that 1 person is in at least one club) and 1 person in the union of A and B. This automatically means that there are 0 people in the union of A and C.

Moving on to B. There should be 10 people in B. Now, we already have 7 people in B. So, we need 3 more. Again using the min possible number, we add 1 person in B alone, which means that there should be 2 people in the union of B and C.

Finally, we need 12 people for C. We already have 8 people in C, so we still need to add 4 people. As the union of A and C includes 0 people, all 4 remaining people will be in C alone.

The question asks for the fraction of those in exactly 2 clubs (the unions between 2 different clubs), relative to the total number of people. This is 3/16 = 1/8. So, ANS A
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18 Jan 2016, 20:00
I agree with explanation.
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21 Jan 2016, 05:52
Number of students = number in club a + number in club b + number in club c - number in 2 - 2*number in 3
1 = 1/2 + 5/8 + 3/4 - x - 2*3/8
x = 1/8
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14 May 2016, 06:13
Quote:
Since we are only dealing in fractions, let’s pick a simple number for the total number of students: 8. Thus, 3 students are in all 3 clubs. 4 students are in A, 5 students are in B, and 6 students (3434 of 8) are in C. Since every student is in at least one club, then we know that the total number of club members, appropriately counting people in two or three clubs, must be 8.

We can use the following idea: suppose there were no overlap in the club memberships at all. Then we could just add up the club memberships, and there would have to be 4+5+6=154+5+6=15 students. Now, we know that this is overcounting; we counted 3 students 3 times (for all 3 clubs), when we should have only counted them once. So we overcounted them twice, meaning that we should subtract 2×3=62×3=6 students from the 15, leaving 9. We still have an extra student in the count, since we really only have 8 people. Now, let’s think about the people in exactly 2 clubs: we counted those people twice, when we should have counted them just once. So we overcounted them once, and we should subtract their number once from the total. Representing this number by xx, we get 9−x=89−x=8, so x=1x=1. The fraction of people in exactly 2 clubs is 1818.

Is my logic below correct or was I just lucky this time?

There are 3 students in all the groups. Therefore 1 in A or A&B or A&C; 2 in B or B&A or B&C; 3 in C or A&C or A&B
Since each student has to be in at least one group, A has to be uniquely 1. There can be no student in A&B or A&C.
By the same logic, B has to have at least 1 unique member so the overlap can only be 1 in B&C.
Answer is 1 (out of 8)
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22 May 2016, 10:14
why are we doing 2* no. of students in all three???
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22 May 2016, 10:22
vishnu440 wrote:
why are we doing 2* no. of students in all three???

Hi,

when we add # in all 3 individual groups, we add the portion overlapped in all three groups THRICE, But the TOTAL should have that ONLY once and that is why we subtract this set of people TWICE( 2 times) from the total..
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10 Jan 2017, 11:12
I think this is a high-quality question and I agree with explanation.
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25 Jul 2017, 21:47
I think this is a high-quality question.
Re S96-17   [#permalink] 25 Jul 2017, 21:47
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# S96-17

Moderators: chetan2u, Bunuel

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