Hi,
Let
e0 - number of people who do not like any fruit
e1 - number of people liking only 1 fruit
e2 - number of people liking only 2 fruits
e3 - number of people liking only 3 fruits
e0+e1+e2+e3=100. let e0 be 0
e1+2.e2+3.e3=70+75+80 =225 =>e1+2e2+3e3=225
From these 2 equations (100 -e3)+e2+2e3=225 => e2+2e3=125
e3 should be minimum so we have to make e2 as large as possible.As e1+e2+e3=100, we make e1=0 to accommodate large value for e2.
So we have e2+e3=100 and e2+2e3+125 =>e3=25
So, (e3)min = 25
I have assumed that the number of people who do not love any of the 3 fruits as zero so that e2 can be maximized

Hence, C

Hi,
There is in fact another way of doing this without mathematical approach. The way of distributing those 225 fruits to the pool of 100 students. As the number of students getting all 3 fruits should be minimized we try to give 2 fruits to the maximum number of students(100). So, 200 fruits will be over and we are left with 25 fruits. Any student can get only a maximum of 3 fruits, so each of the 25 students will get 1 more fruit.

Total = A + B + C - AB - BC - AC + ABC
Max(ABC) = Total + max(AB) + max(BC) + max(CA) - min(A) - min(B)- min(C)

min(A)=70, min(B)=75, min(C)=80

Use 2 set eqn. to find max(AB) => total = A + B - AB => AB = 70+75-100 = 55
Similarly BC = 55, AC = 50

Using these values in eqn 1, Max(ABC)=25

My method (Correct me if I am wrong)
let total 100 people.
We have to minimize % of people who like all 3.
Then people who like single fruits should be minimized first (remove the atleast part from all of apples, bananas and cherries)
(because if they are more, chances that all the three fruits are liked increase
Thus 70 people like A,75 like B and 80 like C.
let 100 people be a1 to a100
and let a1-a30 don't like A.
a31-a55 don't like B
a56-a75 don't like C
Thus a76-a100 like all 3
=25
25%
IMO C