1.Set Theory
medium

Let $n(U) = 700,\,n(A) = 200,\,n(B) = 300$ and $n(A \cap B) = 100,$ then $n({A^c} \cap {B^c}) = $

A

$400$

B

$600$

C

$300$

D

$200$

Solution

(c) $n({A^c} \cap {B^c}) = n[(A \cup B)^c] $ = $n(U) – n(A \cup B)$

= $n(U) – [n(A) + n(B) – n(A \cap B)]$

 $= 700 -[200 + 300 -100] = 300.$

Standard 11
Mathematics

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