Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Of the 45 students in a certain class, 30 joined math club, [#permalink]
26 Sep 2007, 18:49

2

This post was BOOKMARKED

Of the 45 students in a certain class, 30 joined math club, 25 joined physics club, and 20 joined biology club. The students who joined physics club also joined the math club. If 12 students joined both math and biology club, and 8 students joined all of the three clubs, how many students joined none of the three clubs?

Of the 45 students in a certain class, 30 joined math club, 25 joined physics club, and 20 joined biology club. The students who joined physics club also joined the math club. If 12 students joined both math and biology club, and 8 students joined all of the three clubs, how many students joined none of the three clubs?

I am getting 7 for the answer using the Venn diagram

Of the 45 students in a certain class, 30 joined math club, 25 joined physics club, and 20 joined biology club. The students who joined physics club also joined the math club. If 12 students joined both math and biology club, and 8 students joined all of the three clubs, how many students joined none of the three clubs?

It's 7.

a. Math + Physics : 25
b. Math + Physics + Biology = 8
From a & b ==>
Only Physics = 0 & Only Physics + Math = 17 --> I

a. Math + Biology = 12
b. Math + Physics + Biology = 8 --> II

From a & b ==>
Only Math + Biology = 4 --> III

From I II & III : Only Math = 30 - ( 17 + 8 + 4 ) = 1

a. Physics + Biology : 0 ;
b. Math + Physics + Biology = 8 ;
c. Math + Biology = 4

a, b , c together ==> Only Biology = 20 - ( 12 + 8 ) = 8

Total : 45
No Clubs = k

k = 45 - ( Only Math + Only Physics + Only biology + ( Only Math & Physics) + ( Only Math & Biology ) + ( Only Physics & Biology) + ( All three)

However, I notice that you don't actually need to consider Physics students since all physics students are in Math. I wish I can draw, but basically, the physics circle is in math circle. Therefore, if you use just math and biology and m&b, then you should get the answer...
30+20 - 12 + None = 45
None = 7

However, I notice that you don't actually need to consider Physics students since all physics students are in Math. I wish I can draw, but basically, the physics circle is in math circle. Therefore, if you use just math and biology and m&b, then you should get the answer... 30+20 - 12 + None = 45 None = 7

However, I notice that you don't actually need to consider Physics students since all physics students are in Math. I wish I can draw, but basically, the physics circle is in math circle. Therefore, if you use just math and biology and m&b, then you should get the answer... 30+20 - 12 + None = 45 None = 7

how do you know that p&b=8 ?

Since all P is in M, and you know that P&M&B = 8
This means P&M&B = P&B = 8

Just want to share strategy about cracking Venn diagram problem.

Basically, I don't know any Venn formula. What I do is I keep track of how many items has been counted. The total always consist of counting everything ONCE!

Forget about None for now...

For example, if you are give two circle Venn diagram, you have total of A and B and the overlap (A&B). You know that if you add A and B, the overlap has been count twice, therefore, for formula will be:
Total = A+B - A&B

If you are given three circle Venn diagram of A, B, C, it is a bit more tricky, but same trick.
If you are given A&B, B&C, C&A, you know that A,B,C already count A&B&C three times. A&B count A&B&C one time. B&C count A&B&C one time. C&A count A&B&C one time. Therefore, to make everything count ONCE, you can come up with the following:

A+B+C - A&B - B&C - C&A + A&B&C = Total
A+B+C => count A&B&C three times
A&B => count A&B&C one time
B&C => count A&B&C one time
C&A => count A&B&C one time
From this, you know that the "counts" are all three and they cancel out and therefore, you must ADD A&B&C at the end to make everything count ONCE!

Now, some complicated problems mention of "exactly two". This means that you are given A&B, but it doesn't count A&B&C. Using the same trick, if you are given exactly A&B without A&B&C, B&C with out A&B&C, C&A without A&B&C, your formula becomes:

A+B+C - A&B - B&C - C&A - 2*A&B&C = Total

Why? Because A, B, C count A&B&C three times. You know that A&B, B&C, C&A don't count A&B&C at all. Therefore, to make everything count ONCE, you need to subtract 2*A&B&C to the equation.

If you know how to do the "count", you don't even need to remember any formula. I find a lot of problems that the formula will just make it confusing, though mostly tough problem. Using this trick, you won't make a mistake in Venn diagram type problem.