ISI objective

1. (c) : Let A denote the set of students who like sports , let B denote who like music and let C denote who like studies.
Let the the total no. of students be 100.
Then n (A) = 90 , n (B) = 80 , n(C) = 70.

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (C ∩ B) – n (A ∩ C) + n (A ∩ B ∩ C)

Now
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
maximum value of n (A ∪ B) is 100 : total no. of students
n (A ∩ B) = n (A) + n (B) – n (A ∪ B) = 90 + 80 - 100 = 70
n (A ∩ B) >= 70

similarly

n (C ∩ B) >= 50

n(A ∩ C) >= 60

This process was done to minimise the value of AND terms ... to do that we had to maximise the value of OR terms
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (C ∩ B) – n (A ∩ C) + n (A ∩ B ∩ C)

n (A ∪ B ∪ C) max value is 100

100 = 90 + 80 + 70 – 70 – 50 – 60 + n (A ∩ B ∩ C)
n (A ∩ B ∩ C) >= 100 - 60

Then the % of students who have chosen all three is atleast 40
option d

objective way of doing

see least of intersection occurs when all are at their least ( trivial obs)

so draw Venn diagram for given data at their leasts

keep x as req ans and obtain restrictions on x

its same as ankits method but drawing venn dia will make it faster :)

thats what i tried to do and failed at it :(