2 Set Venn Diagram

Let I be the percentage of students who like exactly one activity
II be the percentage of students who like exactly two activities.
III be the percentage of students who like exactly three activities.
IV be the percentage of students who like exactly four activities.
V be the percentage of students who like all the five activities.
I+II+III+IV+V=100
I+2II+3III+4IV+5V=410
We have to maximize the percentage of students who like exactly four activities.
Let us consider I, II, III be 0
IV+V=100
4IV+5V=410
On solving both the equations, we get
V=10, IV=90
90 is the correct answer.

Let the number of people who play both football and basketball be x.

We are given that at least 55 people play football and at least 55 people play basketball. Also, 100 people play at least one of the two sports.

$$n(F∪B)=100$$
$$n(F)\ge55\rightarrow1$$
$$n(B)\ge55\rightarrow2$$
Let $$n(F∩B)=x$$

Adding eq. 1 and eq. 2 - 
$$n(F)+n(B)\ge110$$
Subtracting $$n(F∩B)$$ from both sides - 
$$n(F)+n(B)-n(F∩B)\ge110-x$$
We know that $$n(F∪B)=n(F)+n(B)−n(F∩B)$$ - 
$$n(F∪B)\ge110-x$$
$$100\ge110-x$$
$$x\ge10$$

Thus, the minimum possible number of people who play both sports is 10.

Now, the number of people who play exactly one sport is given by: (Football only)+(Basketball only)

This can also be written as: $$n(F∪B)−n(F∩B)$$

Since $$n(F∪B)=100$$ and the maximum number of people playing exactly one sport will occur when the overlap is minimum, we take: $$x=n(F∩B)=10$$. Therefore,

Maximum number of people playing exactly one sport $$=100-10=90$$

Hence, the maximum possible number of people who play exactly one sport is 90.