IPMAT 2020 - Quantitative Aptitude Question 8
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IPMAT 2020 - Quantitative Aptitude
Total Student = 80
n(m) = 50
n(c) = 40
n(p) = 30
n(m ∩ p) = At most 20
n(p ∩ c) = At most 20
n(m ∩ c) = At most 20
As we need to maximize the students who passed in all 3 exams so the number of students who failed in all 3 = h = 0
a + b + c + 2(d + e + f) + 3g = 50 + 40 + 30 = 120 _________ (i)
a + b + c + d + e + f + g = 80 __________ (ii)
(i) - (ii)
(d + e + f) 2g = 40
Since we need to maximize g we should minimize d + e + f
By taking d + e + f = 0
2g = 40
g = 20
By taking g = 20 no condition for at most 20 students passed in pairs of any of 3 subjects is violated
So g = 20
Maximum value for students who passed in all 3 subjects is 20