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?