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A high school has 400 students 1/2 attend the airthmetic

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rxs0005
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A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   Updated on: 18 Mar 2011, 04:27
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74% (02:06) correct 26% (02:32) wrong based on 438 sessions
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A high school has 400 students 1/2 attend the airthmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 3/8 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.

A. 50
B. 75
C. 150
D. 200
E. 300
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A

Originally posted by rxs0005 on 17 Mar 2011, 15:55.
Last edited by rxs0005 on 18 Mar 2011, 04:27, edited 1 time in total.
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Re: venn diagram 3 sets problem [#permalink] New post   18 Mar 2011, 05:50
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rxs, a venn diagram is not necessary for this question, though it may help you visualize the solution.

Basically, this question is asking you to figure out how many students are being double-counted.

A-Club has 200 members (1/2 of 400)
B-Club has 250 members (5/8 of 400)
C-Club has 300 members (3/4 of 400)

If you add these numbers up, you can see that the three clubs combined have 750 students.

The difference between 750 and 400 is due to students being double or triple counted because they are in more than one club.

We can create an equation to solve this:
200+250+300 = n + x + 2y
where n is the number of students, x is the number of students in two clubs, and y is the number of students in three clubs.

The question provides y for us (150).
750 = 400 + x + 300
x = 50
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Re: venn diagram 3 sets problem [#permalink] New post   18 Mar 2011, 07:10
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A= 200, A only = 200-150 = 50
B= 250, B only = 250-150 = 100
C=300 , C only = 300- 150 = 150
Exactly two two courses = 50+100+150+150(All)-400=50 Ans.
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Re: venn diagram 3 sets problem [#permalink] New post   18 Mar 2011, 07:34
thanks i need some clarification on the below

For the set overlap i have the below formula

Total = A + B + C - AB - BC - CA - 2ABC ----------- ( 1 )

Applying this i get

400 = 200 + 250 + 300 - ( AB + BC + CA ) - 2* 150

sovling this we get

AB + BC + CA = 750 - 400 - 300 = 50

My Q is the Formula ( 1) correct

and does AB have only doubly counted values that is unique to only A and B not C right ?
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Re: venn diagram 3 sets problem [#permalink] New post   18 Mar 2011, 08:10
rxs0005 wrote:
thanks i need some clarification on the below

For the set overlap i have the below formula

Total = A + B + C - AB - BC - CA - 2ABC ----------- ( 1 )

Applying this i get

400 = 200 + 250 + 300 - ( AB + BC + CA ) - 2* 150

sovling this we get

AB + BC + CA = 750 - 400 - 300 = 50

My Q is the Formula ( 1) correct

and does AB have only doubly counted values that is unique to only A and B not C right ?


The way you used it, the formula looks fine. Please see the following link as well:
https://gmatclub.com/forum/overlapping-sets-problems-87628.html#p759888

AB have only doubly counted values that is unique to only A and B not C right; right

In the above formula;
Total = A + B + C - AB - BC - CA - 2ABC
A = Only A + only AB + only AC + only ABC
B = Only B + only AB + only BC + only ABC
C = Only C + only AC + only BC + only ABC
AB = Only AB, but not C
BC = Only BC, but not A
AC = Only AC, but not B
ABC = Only ABC

Total = Only A + Only B + Only C + Only AB + Only AC + Only BC + Only ABC + None
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Re: venn diagram 3 sets problem [#permalink] New post   19 Mar 2011, 00:21
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ABC = 150

A + AB + AC + ABC = 200

B + AB + BC + ABC = 250

C + AC + BC + ABC = 300

A + B + C + 2(AB + BC + AC) + 3ABC = 750

Also, A + B + C + AB + BC + AC + ABC = 400


=> AB + BC + AC + 2ABC = 350

=> AB + BC + AC = 350 - 300 = 50

Answer - A
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Re: venn diagram 3 sets problem [#permalink] New post   30 Oct 2011, 07:48
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Hi ppl,

I have a really fundamental doubt.

In some places, I see a formula
AuBuC = A + B + C - AnB -BnC - AnC + AnBnC

In others,
AuBuC = A + B + C - AnB -BnC - AnC - 2(AnBnC)

Pls clarify which is correct or when to apply what.

Thanks
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Re: venn diagram 3 sets problem [#permalink] New post   30 Oct 2011, 07:58
raghupara wrote:
Hi ppl,

I have a really fundamental doubt.

In some places, I see a formula
AuBuC = A + B + C - AnB -BnC - AnC + AnBnC

In others,
AuBuC = A + B + C - AnB -BnC - AnC - 2(AnBnC)

Pls clarify which is correct or when to apply what.

Thanks


overlapping-sets-problems-87628.html#p759888
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Re: venn diagram 3 sets problem [#permalink] New post   01 Nov 2011, 10:08
fluke wrote:
raghupara wrote:
Hi ppl,

I have a really fundamental doubt.

In some places, I see a formula
AuBuC = A + B + C - AnB -BnC - AnC + AnBnC

In others,
AuBuC = A + B + C - AnB -BnC - AnC - 2(AnBnC)

Pls clarify which is correct or when to apply what.

Thanks


overlapping-sets-problems-87628.html#p759888


Very clear!
Thanks Fluke.
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Re: venn diagram 3 sets problem [#permalink] New post   08 Mar 2012, 03:01
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(1/2)+(5/8)+(3/4)-x-2*(3/8)+0=1
x=1/8

1/8*400=50
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   10 Dec 2012, 20:26
The second formula described by Bunuel for 3 overlapping sets works here as well to find the last remaining "exactly" scenario.

Instead of the "in all three groups" value being the variable, the "in exactly 2 groups" is the variable and solve the equation the exact same way.
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   18 Aug 2013, 05:34
Total=A+B+C-(Exactly 2 Groups) - 2*All three

Exactly 2 Groups = X

400=200+250+300-X-2*150

X=50
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   18 Aug 2013, 13:44
rxs0005 wrote:
A high school has 400 students 1/2 attend the airthmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 3/8 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.

A. 50
B. 75
C. 150
D. 200
E. 300


can you please solve this using a 3 circle venn diagram


Exactly two = 200+250+300-300-400 = 50 (Answer)
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   19 Aug 2013, 10:40
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Hi Everyone :) ,

I have a simple formula two and three group overlaps.

Total = Group 1 + Group 2 - both + neither

Total = Group 1 + Group 2 + Group 3 - (Sum of two group overlaps) - 2(All three group overlaps) + neither

Using the above formula,

400 = 200 + 250 +300 -Sum of all two group overlaps - 2(150) + 0 (since it is given - every student attends at least one club)

So sum of two group overlaps = 50 :-D

Answer A.

Thanks,
Chandru
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   19 Aug 2013, 10:50
@Chandru yeah..its pretty easy!!!! :lol:


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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   20 Aug 2013, 02:35
Expert Reply
rxs0005 wrote:
A high school has 400 students 1/2 attend the airthmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 3/8 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.

A. 50
B. 75
C. 150
D. 200
E. 300


Theory on advanced overlapping sets problems: advanced-overlapping-sets-problems-144260.html
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Re: A high school has 400 students 1/2 attend the airthmetic [#permalink] New post   11 Nov 2020, 11:07
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rxs0005 wrote:
A high school has 400 students 1/2 attend the airthmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 3/8 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.

A. 50
B. 75
C. 150
D. 200
E. 300


A=200
B=250
C=300
ALL THREE = 150
EXACTLY TWO = X

TOTAL = A+B+C-EXACTLY 2-2(ALL THREE)
400= 200+250+300-X-2(150)
X=50

THE ANSWER IS A.
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