IPMAT 2024 - Quantitative Aptitude Question 3
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IPMAT 2024 - Quantitative Aptitude
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.