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?
Hello, fifthrapiers!
I placed the information into a chart . . .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?
. .
The total number of students is 100.
70 are in French.
20 are in Neither.
. .
We can fill in two more boxes:
. .
Let
. .
"More students in Fencing than in French": .
Since , we have: .
Therefore, at least 61 students are in both activities.