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MGMAT - Venn Diagram

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MGMAT - Venn Diagram

by VirtualM » Mon Feb 26, 2007 12:08 am
Source: MGMAT

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?


The problem can be answered by making a Venn diagram for overlapping sets. However, there is one part of MGMAT's explanation that I can't understand.

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by Stacey Koprince » Mon Feb 26, 2007 11:07 pm
Hey - can you please post the part of the explanation you don't understand (and give me some detail on what's troubling you)?

Also, please post the entire question (including answer choices) whenever you post questions. Thanks!
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by VirtualM » Mon Feb 26, 2007 11:35 pm
Hello,

Attaching the problem and explanation...

This is the part of the explanation I can't understand.

Taking this equation and subtracting the 4th equation (Total students) yields the following:

a + e + f + 2b + 2c +2d + 9 = 84
–[a + e + f + b + c + d + 3 = 68]
b + c + d = 10

I believe in this part. we subtracted the total students from the sum of the first three equations. I assume that b + c + d = 10 refers to all students taking exactly 2 classes since we removed the "duplicate " b + c + d" total by subtracting the fourth equation. But what happened to the three instances of three students taking all classes 3's (3+3+3 = 9)? These were added together when we summed the first 3 equations. Were they also removed when we subtracted the 4th equation? I guess I just need to understand what we exactly did to the sets when we manipulated the equations...

Please let me know if my question is still unclear....
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by Stacey Koprince » Mon Feb 26, 2007 11:57 pm
Ok - so you understand how we wrote the first 4 equations:

1) History students: a + b + c + 3 = 25
2) Math students: e + b + d + 3 = 25
3) English students: f + c + d + 3 = 34
4) TOTAL students: a + e + f + b + c + d + 3 = 68

(For others - it helps to look at the diagram in the attached file to write the above 4 equations and to know what each variable represents.)

And you understand that adding the first 3 of the above equations gives you this new equation:
5) a + e + f + 2b +2c +2d + 9 = 84

In the next step, we are not subtracting just the total students from equation #5, but rather we are subtracting entire equation #4 (see above).
So, (entire equation #5) - (entire equation #4) which is the same thing as (history + math + English students) - (total students).

This subtraction gives us the overlap. The overlap is represented by people taking 2 classes and people taking 3 classes, but we have a real number (3) for the people taking 3 classes, so the only overlap variables left in the problem are those variables that represent people taking 2 classes.
eqn #5: a + e + f + 2b +2c +2d + 9 = 84
minus eqn #4: -[a + e + f + b + c + d + 3 = 68]
equals 0 + 0 + 0 + b + c + d + 6 = 16
equals b + c + d = 10

And since b, c, and d represent my three "attends 2 classes" categories, the answer is 10.

Does that make sense?
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by VirtualM » Wed Feb 28, 2007 2:39 am
Yes, thanks!