Q41 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
If $n(\mathrm{~A})=1000, n(\mathrm{~B})=500$ and if $n(\mathrm{~A} \cap \mathrm{B}) \geq 1$ and $n(\mathrm{~A} \cup \mathrm{B})=p$, then
- A$500 \leq p \leq 1000$
- B$1001 \leq p \leq 1498$
- C$1000 \leq p \leq 1498$
- Correct answerD$1000 \leq p \leq 1499$
Solution
$\mathrm{n}(\mathrm{A})=1000, \mathrm{n}(\mathrm{B})=500, \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \geq 1$, $n(A \cup B)=p ; n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $\mathrm{p}=1000+500-\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \Rightarrow 1 \leq \mathrm{n}(\mathrm{A} \cap \mathrm{B}) \leq 500$
Hence $p \leq 1499$ and $p \geq 1000 \Rightarrow 1000 \leq p \leq 1499$
Hence $p \leq 1499$ and $p \geq 1000 \Rightarrow 1000 \leq p \leq 1499$
Hence, the correct answer is option: 'D'.