UPSC NDA Math Model Paper 2

Given: A, B, C and D are four sets such that A ∩ B = C ∩ D = ϕ
Statement 1: A ∪ C and B ∪ D are always disjoint.
⇒ (A ∪ C) ∩ (B ∪ D) = [(A ∪ C) ∩ B] ∪ [(A ∪ C) ∩ D]
= [(A ∩ B) ∪ (C ∩ B)] ∪ [(A ∩ D) ∪ (C ∩ D)]
= [ϕ ∪ (C ∩ B)] ∪ [(A ∩ D) ∪ ϕ]
= (C ∩ B)] ∪ [(A ∩ D) (∵ A ∪ ϕ = A)
(A ∪ C) and (B ∪ D) is not disjoint set.
Statement 2: A ∩ C and B ∩ D are always disjoint.
⇒ (A ∩ C) ∩ (B ∩ D) = [(A ∩ C) ∩ B] ∩ [(A ∩ C) ∩ D]
= [(A ∩ B) ∩ C] ∩ [(A ∩ (C ∩ D)]
= [ϕ ∩ C] ∩ [(A ∩ ϕ] (∵ A ∩ ϕ = ϕ)
= (ϕ ∩ ϕ)
= ϕ
(A ∩ C) and (B ∩ D) are always disjoint.
Hence statement 2 is correct.