In a survey it was found that 40 % like tea, 50 % like coffe

Say there are total 100 people. 40 like tea, 50 like coffee and 60 like milk so there are total 40 + 50 + 60 = 150 instances (Say they are 'likes' a person gives). A person could give a 'like' to only 1 thing, to 2 things or to all 3 things. So we have a total of 150 'likes'.

Each person likes at least 1 thing so 100 'likes' are given by 100 people. Now we have 50 'likes' leftover.

We want to maximise the overlap of 3 likes given by one person so we minimise 2 'likes' and and try to utilise all leftover 50 for 3 'likes'.
50 'likes' can be given by 25 people while they have already given 1 'like'. So maximum overlap of all 3 will be 25.

Now the situation is like this: 15 people like only tea, 25 people like only coffee and 35 people like only milk. Rest 25 people like all three.
So total likes = 15 + 25 + 35 + 25*3 = 150

We have minimised "2 likes" cases by making them 0.

To minimise the overlap of all 3, we can simply give the 50 leftover 'likes' to 50 people so that 50 people like exactly 2 things and nobody exactly all 3 things.

Maximum value = 25
Minimum value = 0

Answer (A)

No.of people, say 100.

Coffee: 50, Tea: 40 and Milk: 60=> 150

Excess= 150-100= 50.

This 50 gives us us the maximum: Prime factorize 50: 2*5^2.
To maximise we always take squares=> 5^2= 25. And for min, here it yields 2, but there's no option here with 2, so we can still consider 0 as there could be that 'no one' that likes all the three.

that leaves us with 25 and 0.

40 & 0 is wrong because:
"Every person likes at least one of the three items"; thus the sum of % must be 100.

25 & 0 satisfies this condition. Therefore, (A) 25 & 0 is correct