# Venn Diagram Problem

• Oct 16th 2009, 08:34 AM
fifthrapiers
Venn Diagram Problem
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?

My solution:

I let x = number of students in both the club and fencing team and A = the number of students on the fencing team

Then, the number of students in the club is 70 - x

And then we know:

A + (70-x) + 20 - x = 100

This means A = 10

So now I used the same formula (Total # = A + B - (A^B) + neither)

100 = 10 + (70-x) - x + 20

But this didn't work... why isn't this formula working?

For 3 sets, Total # = A + B + C - A^B - A^C - B^C + (AUBUC) + "others"

So I think my formula for 2 sets is right..

The answer is meant to be 61 but plugging into that formula doesn't work:

10 + 9 -61 + 20 =! 100

Help?
• Oct 16th 2009, 09:08 AM
Plato
Quote:

Originally Posted by fifthrapiers
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?

Think of it this way. There are 80 students in French or fencing.
Of those 70 are in French. So there must be 10 in fencing that are not in French.
But the number in fencing must be more than 70.
So what in the minimum in both?
• Oct 16th 2009, 10:49 AM
Soroban
Hello, fifthrapiers!

Quote:

In a group of 100 students, more students are in Fencing than in French.
If 70 are in French and 20 are neither in Fencing nor French,
what is the minimum number of students who could in both actitvities?

I placed the information into a chart . . .

. . $\begin{array}{c||c|c||c|}
& \text{Fencing} & \sim\text{Fencing} & \text{Total} \\ \hline \hline
\text{French} & & & \\ \hline
\sim\text{French} & & & \\ \hline \hline
\text{Total} & & & \\ \hline\end{array}$

The total number of students is 100.
70 are in French.
20 are in Neither.

. . $\begin{array}{c||c|c||c|}
& \text{Fencing} & \sim\text{Fencing} & \text{Total} \\ \hline \hline
\text{French} & & & 70 \\ \hline
\sim\text{French} & &20 & \\ \hline \hline
\text{Total} & & & 100 \\ \hline\end{array}$

We can fill in two more boxes:

. . $\begin{array}{c||c|c||c|}
& \text{Fencing} & \sim\text{Fencing} & \text{Total} \\ \hline \hline
\text{French} & & & 70 \\ \hline
\sim\text{French} & {\color{blue}10} & 20 & {\color{blue}30} \\ \hline \hline
\text{Total} & & & 100 \\ \hline\end{array}$

Let ${\color{red}x} \:=\:n(\text{Fencing}),\;\;{\color{red}y} \:=\:n(\text{French} \wedge \text{Fencing})$

. . $\begin{array}{c||c|c||c|}
& \text{Fencing} & \sim\text{Fencing} & \text{Total} \\ \hline \hline
\text{French} & {\color{red}y} & & 70 \\ \hline
\sim\text{French} & 10 & 20 & 30 \\ \hline \hline
\text{Total} & {\color{red}x} & & 100 \\ \hline\end{array}$

"More students in Fencing than in French": . $x \:>\:70$

Since $y+10 \:=\:x$, we have: . $y + 10 \:>\:70 \quad\Rightarrow\quad y \:>\:60$

Therefore, at least 61 students are in both activities.

• Oct 16th 2009, 09:44 PM
fifthrapiers
Thank you, this makes sense.

My question now, though, is why did my method not work? And why did that formula not work? I've used it many times in the past and it has, but not on this one.
• Oct 17th 2009, 05:33 AM
Plato
Quote:

Originally Posted by fifthrapiers
My question now, though, is why did my method not work? And why did that formula not work? I've used it many times in the past and it has, but not on this one.

This is incorrect (Total # = A + B - (A^B) + neither)
$\left| \text{Total} \right| = \left| {A \cup B} \right| + \left| {A^c \cap B^c } \right| = \left[ {\left| A \right| + \left| B \right| - \left| {A \cap B} \right|} \right] + \left| {A^c \cap B^c } \right|$