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judocallin02

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Jun 20, 2009
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Problem:
At a convention of 375 butchers(B), bakers (A), and candlestick makers (C), there were
50 who were both B and A but not C
70 who were B but neither A nor C
60 who were A but neither B nor C
40 who were both A and C but not B
50 who were both B and C but not A
80 who were C but neither A nor B
How many at the conevtion were A,B, and C?

My Work:
|A U B U C| = 60 + 70 + 80 - 50 - 40 - 50 +|A ? B ? C|
|A U B U C| = 70 +|A ? B ? C|

then I've no idea what to do next. Would somebody help me?
 
Hello, judocallin02!

At a convention of 375 butchers \(\displaystyle (B)\), bakers \(\displaystyle (A)\), and candlestick makers \(\displaystyle (C)\), there were:

50 who were both B and A but not C
70 who were B but neither A nor C
60 who were A but neither B nor C
40 who were both A and C but not B
50 who were both B and C but not A
80 who were C but neither A nor B

How many at the convention were A, B, and C?
Click to expand...

Translate the given information:

. . \(\displaystyle \begin{array}{ccccc} \text{60 were A, but not B or C} && n(A \cap B' \cap C') \:=\:60 \\ \text{70 were B, but not A or C} && n(A' \cap B \cap C') \:=\:70 \\ \text{80 were C, but not A or B} && n(A' \cap B' \cap C) \:=\:80 \\ \text{50 were A and B, but not C} && n(A \cap B \cap C') \:=\:50 \\ \text{40 were A and C, but not B} && n(A\cap B' \cap C) \:=\:40 \\ \text{50 were B and C, but not A} && n(A' \cap B \cap C) \:=\:50 \end{array}\)


Insert these numbers into the Venn diagram.

Code: 
      * - - - - - - - - - - - *
      | A                     |
      |   60          * - - - + - - - - - - - *
      |               |   50  |             B |
      |       * - - - + - - - + - - - *  70   |
      |       |   40  |   ?   |       |       |
      * - - - + - - - + - - - *       |       |
              |       |          50   |       |
              |       * - - - - - - - + - - - *
              | C   80                |
              * - - - - - - - - - - - *

We have 350 people accounted for.
. . This leave 25 who are of all three professions.


So the final Venn diagram looks like this:
Code: 
      * - - - - - - - - - - - *
      | A                     |
      |   60          * - - - + - - - - - - - *
      |               |   50  |             B |
      |       * - - - + - - - + - - - *  70   |
      |       |   40  |   25  |       |       |
      * - - - + - - - + - - - *       |       |
              |       |          50   |       |
              |       * - - - - - - - + - - - *
              | C   80                |
              * - - - - - - - - - - - *

And you can now determine how many are \(\displaystyle A\),
. . how many are \(\displaystyle B\), and how many are \(\displaystyle C\).

 
Thank You very much, soroban.

Wait, I'm still confuse.

How do I find the number of elements in A, B, and C only?Am I going to add the intersections of each then subtract it to each cardinality of A,B and C

Say for example,
If I'm going to find the number of elements in A only I will have to subtract the sum of |A?B|,|A?C| and |A?B?C| to |A|. If that is so then the number of elements in A only is 55.

Same goes for set B and C

But after completing the Venn Diagram,
when adding all of the number of elements, I'm not getting a sum of 375, to determine if its correct.I'm getting a sum of 310.

It is impossible for persons to be outside of the sets because it is stated in the problem that there are 375 of butchers,bakers and candlestick makers.

To Illustrate:

click on the image to enlarge






Please help me!










Edited a couple of times for clarity and presentation.
 
I get it now.

The number of people in the convention is only 310 because if it is 375, some people are counted more than twice


Is that it?
 
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