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Club # of Students Chess 40 Drama 30 Math 25 The table

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Manager
Joined: 26 Dec 2007
Posts: 70
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Club # of Students Chess 40 Drama 30 Math 25 The table [#permalink]  07 Jan 2008, 18:35
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Question Stats:

50% (04:31) correct 50% (01:15) wrong based on 8 sessions
Club # of Students
Chess 40
Drama 30
Math 25

The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both Chess and Drama, 5 students are in both Chess and Math, and 6 students are in both Drama and Math. How Many different students are in the three clubs?

68
69
74
79
84

This problem can be solved by setting up a venn diagram. However, I was wondering how you could solve by setting up an equation (if that's even possible).
Manager
Joined: 04 Jan 2008
Posts: 85
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Re: OG PS #179 - Students question [#permalink]  07 Jan 2008, 18:52
Use the equation,
n(atleast one)= N(A)+N(B)+N(C)-N(exactly two)- 2N(ALL THREE)
total diff students= 40+30+25-10-5-6= 74 ( we are given no tudent attens all three)
Manager
Joined: 26 Dec 2007
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Re: OG PS #179 - Students question [#permalink]  07 Jan 2008, 18:55
n(atleast one)

What does "atleast one" mean?
Manager
Joined: 04 Jan 2008
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Re: OG PS #179 - Students question [#permalink]  08 Jan 2008, 10:21
it means, no of students in atleast one club= no of different students
Intern
Joined: 07 Jan 2008
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Re: OG PS #179 - Students question [#permalink]  08 Jan 2008, 10:35
n(C) = 40
n(D) = 30
n(M) = 25
n(C & D) = 10
n(C & M) = 5
n(D & M) = 6
n(C & D & M) = 0

Total = n(C) + n(D) + n(M) - n(C & D) - n(C & M) - n(D & M) + n(C & D & M) = 74
Re: OG PS #179 - Students question   [#permalink] 08 Jan 2008, 10:35
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