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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
We know, n(T ∪ J ∪ C) = x + y + z …(i)
And also, n(T) + n(J) + n(C) = x + 2y + 3z …(ii)
Where, x, y and z represent number of students who drink exactly, one, exactly two and all three drinks.
We have to find out the maximum number of students that like more than one drink.
It means those students who like exactly two or exactly three drinks i.e. the maximum possible value of y + z.
Given n(T ∪ J ∪ C) = 150; n(T) = 52; n(J) = 48; n(C) = 62
Subtracting (i) from (ii), we get
n(T) + n(J) + n(C) – n(T ∪ J ∪ C) = y + 2z
162 – 150 = (y + z) + z
12 = (y + z) + z
The maximum value of (y + z) can be 12, when is z is 0 & y is 12.