Let A be the number of athletes at the soccer club who play ONLY defense
Let B be the number of athletes at the soccer club who play BOTH defense and midfield
Let C be the number of athletes at the soccer club who play ONLY midfield
Let D be the number of athletes at the soccer club who neither defense nor midfield.

A+B+C+D = 100
A+B+B+C = 110

Given D>20, Since A+B+C+D = 100, A+B+C < 80.
Given A+B+B+C = 110, and A+B+C < 80, then B>30
Given A+B = 40, So B < 40
So 30<B<40

IMO C

What is the OA?

D=40, M=70, Not D&M=20+
Total=100, find D&M?
Since we are given at least condition for Not D&M, we count this can be 30 at most and decide D&M can be D&M=D+M-(100-30), and D&M=70+40-70=40. The difference between range 20 (Not D&M) and 30 (at most Not D&M) is 10. Hence the required range D&M is 30-40

c

ranvijay87 wrote:Of the 100 athletes at a soccer club , 40 play defense and 70 play midfield.if at least 20 of the athletes play neither defense nor midfield , the no of athletes that play
both midfield and defense could be any no between
1.10 to 20
2.20 to 40
3.30 to 40
4.30 to 70
5.40 to 70.

[Moderator Edit: Moving to a relevant section]

ranvijay87 wrote:Of the 100 athletes at a soccer club , 40 play defense and 70 play midfield.if atleast
20 of the athletes play neither defense nor midfield , the no of athletes that play
both midfield and defense could be any no between
1.10 to 20
2.20 to 40
3.30 to 40
4.30 to 70
5.40 to 70.

[Moderator Edit: Moving to a relevant section]

Total = Defense + Midfield - Both + Neither.

The big idea is to SUBTRACT the overlap.
When we count the total number who play defense and the total number who play midfield, the OVERLAP -- everyone who plays BOTH positions -- is counted TWICE.
Thus, the athletes who play BOTH positions must be subtracted from the total so that they are not double-counted.

In the equation above:
Total = 100.
Defense = 40.
Midfield = 70.
Both = B.
Neither = N.

Plugging these values into the equation:
100 = 40 + 70 - B + N
-10 = -B + N
B = N + 10.

To minimize B, we need to minimize N.
It is given that the minimum value of N is 20:
B = N + 10 = 20 + 10 = 30.
Since the minimum value of B is 30, eliminate A, B and E.

In C, the maximum value of B is 40.
In D, the maximum value of B is 70.
Since only 40 athletes play defense, it is not possible that 70 athletes play both positions.
Eliminate D.

The correct answer is C.