2n (A / B) = n (B / A) and 5n (A cap B) | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$2(\mathrm{n}(\mathrm{A})-\mathrm{n}(\mathrm{AB}))=\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{AB})$

$2 n(A)-n(B)=n(A B)$

$\mathrm{n}(\mathrm{A})+

Vedclass · Accepted Answer

$63$

Vedclass · Answer

$2(\mathrm{n}(\mathrm{A})-\mathrm{n}(\mathrm{AB}))=\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{AB})$

$2 n(A)-n(B)=n(A B)$

$\mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=5 \mathrm{n}(\mathrm{AB})$

$\Rightarrow \mathrm{n}(\mathrm{A})+3 \mathrm{n}(\mathrm{B})=10 \mathrm{n}(\mathrm{A})-5 \mathrm{n}(\mathrm{B})$

$8 n(\mathrm{B})=9 \mathrm{n}(\mathrm{A})$

$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{B})}=\frac{8}{9}$

$\mathrm{n}(\mathrm{A})=8 \mathrm{k} ; \mathrm{n}(\mathrm{B})=9 \mathrm{k}$

$\Rightarrow \mathrm{n}(\mathrm{AB})=7 \mathrm{k}$

$\mathrm{n}(\mathrm{A} \cup \mathrm{B})=10 \mathrm{k} \leq 10$

$\mathrm{k}=1$

$2n (A / B) = n (B / A)$ and $5n (A \cap B) = n (A) + 3n (B) $, where $P/Q = P \cap Q^C$ . If $n (A \cup B) \leq 10$ , then the value of $\frac{{n\ (A).n\ (B).n\ (A\ \cap\ B)}}{8}$ is

Similar Questions

Consider the two sets : $A=\{m \in R:$ both the roots of $x^{2}-(m+1) x+m+4=0$ are real $\}$ and $B=[-3,5)$ Which of the following is not true?

Let $a>0, a \neq 1$. Then, the set $S$ of all positive real numbers $b$ satisfying $\left(1+a^2\right)\left(1+b^2\right)=4 a b$ is

Let $S=\{1,2,3, \ldots \ldots, n\}$ and $A=\{(a, b) \mid 1 \leq$ $a, b \leq n\}=S \times S$. A subset $B$ of $A$ is said to be a good subset if $(x, x) \in B$ for every $x \in S$. Then, the number of good subsets of $A$ is

Let $\mathrm{A}=\{\mathrm{n} \in[100,700] \cap \mathrm{N}: \mathrm{n}$ is neither a multiple of $3$ nor a multiple of 4$\}$. Then the number of elements in $\mathrm{A}$ is

Let $S$ be the set of all ordered pairs $(x, y)$ of positive integers satisfying the condition $x^2-y^2=12345678$. Then,