If U be the universal set and A, B are subsets of U such that n(U) = 25, n(A) = 15, n(A ∪ B)' = 8, n(A ∩ B) = 6 then n(B) and n(B -A) are respectively given by the ordered pair

Question

If U be the universal set and A, B are subsets of U such that n(U) = 25, n(A) = 15, n(A ∪ B)' = 8, n(A ∩ B) = 6 then n(B) and n(B -A) are respectively given by the ordered pair (A) (6,2) (B) (8,2) (C) (8,8) (D) (16,8)

Tardigrade Admin · Accepted Answer

n(B - A) = n(U) - [n(A ∪ B)' + n(A)] = 25 - (8 + 15) = 2 ∴ n(B) = n(B - A) + n(A ∩ B) = 2 + 6 = 8 ∴ Required ordered pair = (8,2)

Q. If $U$ be the universal set and $A$ , $B$ are subsets of $U$ such that $n (U) = 25$ , $n (A) = 15$ , $n (A \cup B)^{'} = 8$ , $n (A \cap B) = 6$ then $n (B)$ and $n (B - A)$ are respectively given by the ordered pair

$(6, 2)$

$(8, 2)$

$(8, 8)$

$(16, 8)$

$n (B - A) = n (U) - [n (A \cup B)^{'} + n (A)]$
$= 25 - (8 + 15) = 2$
$∴ n (B) = n (B - A) + n (A \cap B) = 2 + 6 = 8$
$∴$ Required ordered pair $= (8, 2)$

Q. If U be the universal set and A, B are subsets of U such that n(U)=25, n(A)=15, n(A∪B)′=8, n(A∩B)=6 then n(B) and n(B−A) are respectively given by the ordered pair

(6,2)

(8,2)

(8,8)

(16,8)