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

Question

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) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 5/8

Bunuel · Accepted Answer

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.

vicksikand · Answer

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.

Bunuel · Answer

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.

mmcneilly · Answer

Why did you multiply {# of students in exactly 3 clubs} by 2?

Bunuel · Answer

This is explained in details here  ADVANCED OVERLAPPING SETS PROBLEMS

Hope it helps.

Temurkhon · Answer

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

A

ScottTargetTestPrep · Answer

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

ocelot22 · Answer

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.

BrentGMATPrepNow · Answer

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: 
 https://i.imgur.com/grrmz3k.png

What fraction of the student body is in exactly 2 clubs?  
Let a, b and c = the number of students in exactly 2 clubs
 https://i.imgur.com/Gmsn7mp.png 
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:
 https://i.imgur.com/tG04vVw.png

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

CrackverbalGMAT · Answer

In this question on overlapping sets, let the total number of students be 80, since 80 is a common multiple of the denominators of the fractions given. It’s neither too small that it will end up giving fractions nor too large that you have to spend time on calculations.

Keeping 80 as the number of students in Social High, we can draw a Venn diagram like the one below:

Attachment:

20th Nov 2019 - Reply 1.jpg [ 34.81 KiB | Viewed 7795 times ]

Note that the region outside the circles is ZERO since the question says that every student is in at least one club. This is crucial information; always be on the lookout for such statements in an Overlapping sets question because they make your life easy.

3/8th of 80 i.e. 30 are in all the three clubs. The respective number of students in the Albanian, Bardic and Checkmate sets are 40, 50 and 60 respectively.
If we add these values, we get 150. But, this 150 includes repetition of certain regions – a, b and c appear twice; 30 appears thrice. So, we will have to remove a, b and c once and 30 twice. If we do that, we should obtain 80 since that is the actual total number of students in the school.

Therefore, 150 – a – b – c – 60 = 80. Solving this, we get a+b+c = 10. a+b+c represents the number of people who belong to exactly two clubs. Hence, the required fraction = 10/80 = 1/8.
The correct answer option is A.

Hope that helps!

lakshya14 · Answer

Why we have multiplied the "3" by 2?

ScottTargetTestPrep · Answer

Response:

We are using the following formula:

Total = #Group A + #Group B + #Group C - #exactly two groups - 2 * #all three groups + #none of the groups

The reason “#all three groups” is multiplied by 2 is the following: when we add the number of elements in groups A, B, and C, the elements that are in exactly two groups (or, in other words, the elements that are in two but not three groups) are counted twice and the elements that are in all three groups are counted three times. To get the correct number of items, we need to have each element counted exactly once.

Since the elements that are in exactly two groups are counted twice, we subtract the number of elements that are in exactly two groups from A + B + C so that every element that belongs to exactly two groups is counted once. Similarly, since every element in the triple intersection is counted three times, we need to subtract twice the number of elements in the triple intersection so that every element in this region is also counted only once. That’s the reason for the coefficient of 2 in the formula.

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