Symbiosis National Aptitude Test SNAP 2016 Paper

Let A = set of persons who got medals in dance.
B = set of persons who got medals in dramatics.
C = set of persons who got medals in music.
Given,
n(A) = 36
n(B) = 12
n(C) = 18
n(A U B U C) = 45
n(A ∩ B ∩ C) = 4
We know that number of elements belonging to exactly two of the three sets A, B, C
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) – 3n(A ∩ B∩ C)
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) – 3 × 4 ……..(i)
n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) – n(A U B U C)
From (i) required number
= n(A) + n(B) + n(C) + n(A ∩ B ∩ C) – n(A U B U C) – 12
= 36 + 12 + 18 + 4 – 45 – 12
= 70 – 67
= 3