Given: n(C) = 50, n(F) = 20, n(C ∪ F) = 10.

Correct Option: A

Given: n(C) = 50, n(F) = 20, n(C ∪ F) = 10.
Number of students playing at least one of these two games = n (C ∪ F) = n (C) + n (F) – n (C ∩ F)
= 50 + 20 – 10 = 60.

(a) 1∈ A but 1∉ B and 6 ∈ B but 6 ∉ A.

(b) A = {x: x ∈ N and x ≤ 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(c) {4, 5} ∈ A but {4, 5} ∉ B and 4 ∈ B but 4 ∉ A.

Correct Option: B

(a) 1∈ A but 1∉ B and 6 ∈ B but 6 ∉ A.
∴ A and B are not comparable.
(b) A = {x: x ∈ N and x ≤ 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Clearly, A ⊂ B ⇒ A and B are comparable.
(c) {4, 5} ∈ A but {4, 5} ∉ B and 4 ∈ B but 4 ∉ A.
∴ A and B are not comparable.

NA

Correct Option: A

NA