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BITSAT 2023Paper 2

Exam: BITSAT
Year: 2023
Paper: Paper 2
Subject: Mathematics
Chapter: Sets and Relations

BITSAT 2023 · Paper 2 · Mathematics · Sets and Relations

If n( A)=1000, n( B)=500 and if n( A B) ≥ 1 and n( A B)=p , then

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

Options

A $500 \leq p \leq 1000$
B $1001 \leq p \leq 1498$
C $1000 \leq p \leq 1498$
D $1000 \leq p \leq 1499$

Correct answer

D. $1000 \leq p \leq 1499$

Step-by-step 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

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