Maths Venn Diagram Problem

LostBoy101

Hey there I was trying to work this out for ages so this is what I come with so far.

Question:

In a survey of 215 students, the following data was found:
54 took Programming
72 took Chemistry
97 took Mathematics
26 took Mathematics and Programming
21 took Programming and Chemistry
18 tool Mathematics and Programming and Chemistry
59 students did not take any of the three
courses

Represent the information given above in a Venn Diagram.
How many students took Mathematics and Chemistry?

Venn Diagram I drew up:



I know I'm trying to find the variable x which is how many students took Mathematics and Chemistry but here is my attempt.

215 - 59 = 156.

My forumla:

|P u C u M| = |P| + |C| + |M| - |P n M| - |M n C| + | P n C n M| = 156

=> 54 (Prog) + 72 (Chem) + 97 (Maths) - 21 - 18 - 26 + x = 156

=> 158 + x = 156

=> x = 156 - 158

=> x = -2

I must be wrong here because I know it has been to a positive number..

Any help would be appreciated to help my sore head :P

Razzen

your diagram is wrong..

26 took Mathematics and Programming
18 tool Mathematics and Programming and Chemistry

the overlap of just M+P is therefore 26-18..hope that helps point you in the right direction.

LostBoy101

Razzen wrote: »

your diagram is wrong..

26 took Mathematics and Programming
18 tool Mathematics and Programming and Chemistry

the overlap of just M+P is therefore 26-18..hope that helps point you in the right direction.

Ah so M n P = 21 - 18 = 8?

bnt

I just tried this, based on what Razzen said, and think I have an answer

x=20

.
I basically filled in every slot in the Venn diagram with either a number or a formula based on x. The sum of all those slots must be equal to 156, so you create a formula for that, simplify and solve for x.

LostBoy101

bnt wrote: »

I just tried this, based on what Razzen said, and think I have an answer

x=20

.
I basically filled in every slot in the Venn diagram with either a number or a formula based on x. The sum of all those slots must be equal to 156, so you create a formula for that, simplify and solve for x.

I think I got exactly what you did.

Here is my solution.

|P u C u M| = |P| + |C| + |M| - |P n M| - |M n C| - |P n C| + |P n C n M| = 156

=> 54 + 72 + 97 - 8 - x - 21 - 18 = 156

=> 176 - x = 156

=> -x = 156 - 176

=> -x = - 20

=> x = 20

Mwalimu

There are some mistakes in what you have done so far, and I think you still do not understand the process of solving this type of problem.:(

In your formula, you include "+ |P n C n M| " at the end, but you finished up subtracting 18.

What does the process suggested by the formula tell you? It says that when you add up all the three sets, you have added some 'slots' more than once and you need to compensate. Thus, for this example, when you add 54+72+97 you have included the 3 (21-18), the 8 (26-18) and the x twice (assuming, as in your diagram, that x is the bit of |M n C| which does not contain the 18). You have also included the 18 three times! So, you need to compensate by subtracting the 3, the 8 and the x once each and the 18 two times.
The final total you get must equal the 156 contained within the three sets.
Then, when you get x = 20 as a result, you have one more step to do. What were you asked? Remember, |M n C| contains both the x and the 18.:)
Note, the formula as you have set it out is correct. Since each of the (complete) intersections is subtracted once, the centre 'slot' is in reality subtracted three times. Hence, the final part adds it back in once!

LostBoy101

I got the problem solved with help. I managed to get 38 for |M n C| and subtracted 18 from 38 (|M n C| to get 20. Thanks for your help guys.