If n(A) = 1000, n(B) = 500, n(A ∩ B) ge 1 and n(A ∪ B) = p, then

Question

If n(A) = 1000, n(B) = 500, n(A ∩ B) ge 1 and n(A ∪ B) = p, then (A) 500 le p le 1000 (B) 1001 le p le 1498 (C) 1000 le p le 1498 (D) 1000 le p le 1499

Tardigrade Admin · Accepted Answer

We know that, n(A ∪ B) = n(A) + n(B) - n(A ∩ B) ⇒ p = 1000 + 500 - n (A ∩ B) ⇒ 1 le n(A ∩ B) le 500 Hence, p = 1000 + 500 - 1 = 1499 and p = 1000 + 500-500= 1000 ∴ 1000 le p le 1499

Q. If $n (A) = 1000$ , $n (B) = 500$ , $n (A \cap B) \geq 1$ and $n (A \cup B) = p$ , then

$500 \leq p \leq 1000$

$1001 \leq p \leq 1498$

$1000 \leq p \leq 1498$

$1000 \leq p \leq 1499$

We know that, $n (A \cup B) = n (A) + n (B) - n (A \cap B)$
$\Rightarrow p = 1000 + 500 - n (A \cap B)$ $\Rightarrow 1 \leq n (A \cap B) \leq 500$
Hence, $p = 1000 + 500 - 1 = 1499$
and $p = 1000 + 500 - 500 = 1000$
$∴ 1000 \leq p \leq 1499$

Q. If n(A)=1000, n(B)=500, n(A∩B)≥1 and n(A∪B)=p, then

500≤p≤1000

1001≤p≤1498

1000≤p≤1498

1000≤p≤1499