Question

If A and B are two sets, $n\left( A-B \right)=8+2x$, $n\left( B-A \right)=6x$ and $n\left( A\cap B \right)=x$. If $n\left( A \right)=n\left( B \right)$, then $n\left( A\cap B \right)$
(a). 26
(b). 50
(c). 24
(d). None of these

Answer 1

Hint: We had to find the value of $n\left( A\cap B \right)$, which is equal to ‘x’. So we just need to find out the value of ‘x’ which we can find out by simplifying the given info $n\left( A-B \right)=8+2x$ and $n\left( B-A \right)=6x$ through the concept of relations.

Complete step-by-step solution:
Moving ahead with the question in step wise manner, we have $n\left( A-B \right)=8+2x$ and $n\left( B-A \right)=6x$, further it given that $n\left( A\cap B \right)$ is equal to ‘x’ and we need to find out $n\left( A\cap B \right)$. So if we can find out the value of ‘x’ then we will get the value of $n\left( A\cap B \right)$ which is equal to ‘x’.
So to find out the value of ‘x’ we can use the given info i.e. $n\left( A-B \right)=8+2x$ and, and simplify them using the concept of sets. So according to sets we know that $n\left( A-B \right)$ is equal to $n\left( A \right)-n\left( A\cap B \right)$ i.e. $n\left( A-B \right)=n\left( A \right)-n\left( A\cap B \right)$ so we can write $n\left( A \right)-n\left( A\cap B \right)$ in place of $n\left( A-B \right)$, similarly we can say that $n\left( B-A \right)=n\left( B \right)-n\left( A\cap B \right)$.
So now let us simply add the given two information, so we will get;
$\begin{align}
  & n\left( A-B \right)+n\left( B-A \right)=8+2x+6x \\
 &\Rightarrow n\left( A-B \right)+n\left( B-A \right)=8+8x \\
\end{align}$
Since above we find out that $n\left( A-B \right)=n\left( A \right)-n\left( A\cap B \right)$ and $n\left( B-A \right)=n\left( B \right)-n\left( A\cap B \right)$, so replace $n\left( A-B \right)$ and $n\left( B-A \right)$ with $n\left( A \right)-n\left( A\cap B \right)$ and $n\left( B \right)-n\left( A\cap B \right)$ in the above equation respectively. So we will get;
$\begin{align}
  & n\left( A-B \right)+n\left( B-A \right)=8+8x \\
 &\Rightarrow n\left( A \right)-n\left( A\cap B \right)+n\left( B \right)-n\left( A\cap B \right)=8+8x \\
 &\Rightarrow n\left( A \right)+n\left( B \right)-2n\left( A\cap B \right)=8+8x \\
\end{align}$
Since in the question it is given that $n\left( A \right)=n\left( B \right)$ and $n\left( A\cap B \right)$ is equal to ‘x’. So by using the given info we can write above equation as;
$\begin{align}
  & n\left( A \right)+n\left( B \right)-2n\left( A\cap B \right)=8+8x \\
 &\Rightarrow 2n\left( A \right)-2x=8+8x \\
 &\Rightarrow n\left( A \right)-x=4+4x \\
\end{align}$
Since we know that $n\left( A-B \right)=n\left( A \right)-n\left( A\cap B \right)$, so from here we can say that $n\left( A \right)$ is equal to $n\left( A-B \right)+n\left( A\cap B \right)$, so replace $n\left( A \right)$ by $n\left( A-B \right)+n\left( A\cap B \right)$ in the above equation. So we will get;
$n\left( A-B \right)+n\left( A\cap B \right)-x=4+4x$
As we know that $n\left( A\cap B \right)$ is equal to ‘x’, and $n\left( A-B \right)$ is equal to $8+2x$, so by putting these value we will get;
$\begin{align}
  & 8+2x+x-x=4+4x \\
 &\Rightarrow 4=2x \\
 &\Rightarrow x=2 \\
\end{align}$
So we got ‘x’ equal to 2, which is equal to $n\left( A\cap B \right)$.
Hence the correct answer is 2, i.e. option ‘d’ since the value of $n\left( A\cap B \right)$ equal to 2 is not given in any other option, so none of these is the answer.

Note: Rather than addition of two given information which we had done $\left( n\left( A-B \right)+n\left( B-A \right) \right)$ we can subtract also, then also we will get the same answer. As there is not any specific reason that we should add or subtract them other than the ease of solving.