(i) X = {1, 3, 5}, Y = {1, 2, 3} X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c} A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …} B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5} ∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6} B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9} A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ. So, A ∩ B = Φ