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$2n (A / B) = n (B / A)$ और $5n (A \cap B) = n (A) + 3n (B) $, जहाँ $P/Q = P \cap Q^C$ है। यदि $n (A \cup B) \leq 10$ हो, तो $\frac{{n\ (A).n\ (B).n\ (A\ \cap\ B)}}{8}$ का मान क्या है?
$63$
$72$
$90$
$70$
Solution
$2(\mathrm{n}(\mathrm{A})-\mathrm{n}(\mathrm{AB}))=\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{AB})$
$2 n(A)-n(B)=n(A B)$
$\mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=5 \mathrm{n}(\mathrm{AB})$
$\Rightarrow \mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=10 \mathrm{n}(\mathrm{A})-5 \mathrm{n}(\mathrm{B})$
$8 n(\mathrm{B})=9 \mathrm{n}(\mathrm{A})$
$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{B})}=\frac{8}{9}$
$\mathrm{n}(\mathrm{A})=8 \mathrm{k} ; \mathrm{n}(\mathrm{B})=9 \mathrm{k}$
$\Rightarrow \mathrm{n}(\mathrm{AB})=7 \mathrm{k}$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B})=10 \mathrm{k} \leq 10$
$\mathrm{k}=1$
