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# Chess 40 Drama 30 Math 25 The table above shows the number

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Director
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Chess 40 Drama 30 Math 25 The table above shows the number [#permalink]  22 May 2008, 14:00
Chess 40
Drama 30
Math 25

The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.

A 68
B 69
C 74
D 79
E 84
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Re: GMATPrep [#permalink]  22 May 2008, 14:15
jimmyjamesdonkey wrote:
Chess 40
Drama 30
Math 25

The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.

A 68
B 69
C 74
D 79
E 84

I am getting 73 ,which is not an option here !
Manager
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Re: GMATPrep [#permalink]  22 May 2008, 14:42
rpmodi wrote:
jimmyjamesdonkey wrote:
Chess 40
Drama 30
Math 25

The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.

A 68
B 69
C 74
D 79
E 84

I am getting 73 ,which is not an option here !

I got 73 as well.

35 students in Chess are taking only Chess, or Chess and Drama.
20 students in Math are taking only Math, or Math and Drama.

24 students in Drama are taking only Drama, or Drama and Chess.
14 students in Math are taking only Math.

Since 0 students are taking both Chess and Drama, add up 35 + 24 + 14 = 73.

If this is not the answer, then can someone explain what I did wrong? Thx
Senior Manager
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Schools: Ross Class of 2011
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Re: GMATPrep [#permalink]  22 May 2008, 14:53
Is it E?

I'm not very good at group problems - so this is probably wrong.

40 + 30 + 25 - 11 = 84

For three groups this is my normal approach.

Group1 + Group2 + Group3 - 2(All three groups) - (people in only two groups) + neither.

so

40 + 35 + 25 - 2(0) - 11 + 0 = 84 (from above formula)
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Re: GMATPrep [#permalink]  23 May 2008, 04:44
lucyqin wrote:
rpmodi wrote:
jimmyjamesdonkey wrote:
Chess 40
Drama 30
Math 25

The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.

A 68
B 69
C 74
D 79
E 84

I am getting 73 ,which is not an option here !

I got 73 as well.

35 students in Chess are taking only Chess, or Chess and Drama.
20 students in Math are taking only Math, or Math and Drama.

24 students in Drama are taking only Drama, or Drama and Chess.
14 students in Math are taking only Math.

Since 0 students are taking both Chess and Drama, add up 35 + 24 + 14 = 73.

If this is not the answer, then can someone explain what I did wrong? Thx

Hello,
I'm finding E here.
IMO you just count here the number of students in just one club.
Trying to correct what you did:
35 students in Chess are taking only Chess, or Chess and Drama => here only 35 Chess students only OK
14 students in Math are taking only Math. => OK with this
24 students in Drama are taking only Drama, or Drama and Chess => OK with this only 24 Drama students since C and D =0
Now we are left to find the students involved in several clubs.
Since C and D and M = 0, where and is the intersection operator
C and D =0 => (C and M), (D and M) give different students!!
So we add 73 to the number of these different students (5+6) => 84.
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Re: GMATPrep [#permalink]  23 May 2008, 05:08
jimmyjamesdonkey wrote:
Chess 40
Drama 30
Math 25

The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.

A 68
B 69
C 74
D 79
E 84

Nm I agree w/ E now. My mistake
Director
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Re: GMATPrep [#permalink]  23 May 2008, 05:18
OA is C...I was very confused by this problem since it asks for DIFFERENT students, but OA is C.
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Re: GMATPrep [#permalink]  23 May 2008, 06:30
jimmyjamesdonkey wrote:
OA is C...I was very confused by this problem since it asks for DIFFERENT students, but OA is C.

Buddy..there are 10 students in both Chess and Drama, not 0.
40+30+25 - 5 - 6 - 10 = 74
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Re: GMATPrep [#permalink]  24 May 2008, 06:43
i am getting 73 too...

BKK what do you meant chess and drama has 10?
Current Student
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Re: GMATPrep [#permalink]  24 May 2008, 08:05
nevermind..figured it out..the Poster forgot to write the question properly..the number of students taking both chess and drama is 10..
VP
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Re: GMATPrep [#permalink]  24 May 2008, 11:29
This problem can be solved using Venn diagram. Then it becomes more clear.Now firstly taking the case of students who are exclusively in a particular club:
Chess: 40-5=35
Maths:25-(5+6)=14
Drama: 30-6=24
so sum of such students =35+14+24=73
Now taking up the case of those students who are in more than 1 club
These are 5+6=11
total students=73+11=84
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Re: GMATPrep [#permalink]  27 May 2008, 04:39
ritula wrote:
This problem can be solved using Venn diagram. Then it becomes more clear.Now firstly taking the case of students who are exclusively in a particular club:
Chess: 40-5=35
Maths:25-(5+6)=14
Drama: 30-6=24
so sum of such students =35+14+24=73
Now taking up the case of those students who are in more than 1 club
These are 5+6=11
total students=73+11=84

No I made a mistake on Friday. You don't count the 5+6.
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Re: GMATPrep [#permalink]  27 May 2008, 07:20
(see attachment for explanation)
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Venn.jpg [ 95.46 KiB | Viewed 1274 times ]

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Factorials were someone's attempt to make math look exciting!!!

Re: GMATPrep   [#permalink] 27 May 2008, 07:20
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