(i) Here X = {1, 3, 5} and Y = {1, 2, 3}
∴ X ∩ Y = {1, 3}
(ii) Here A = {a, e, i, o, u} and B = {a, b, c}
∴ A ∩ B = {a}
(iii) Here A = {x: x is a natural number and multiple of 3} = {3, 6, 9, 12, ....} and B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5} ∴ A ∩ B = {3}
(iv) Here A = {x: x is a natural number and 1 < x ≤ 6} ={2, 3, 4, 5, 6} and B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9} ∴ A ∩ B = ϕ
(v) Here A = {1, 2, 3} and B = ϕ ∴ A ∩ B = ϕ.