Question

Let A and B be the two sets such that $$n(A-B)=60+3x$$, $$n(B-A)=8x$$ and $$n(A\cap B)=x-4$$ then draw a Venn diagram to illustrate this information. If $$n\left(A\right)=n\left(B\right)$$ then find
(a) The value of $$x$$
(b) $$n(A\cup B)$$

Answer 1

Hint: We solve this problem by using the Venn diagrams of sets. The Venn diagrams represent the diagrammatic representation of sets inside the universal set $${}^{\prime }{\mu }^{\prime }$$
For solving the first part we use the given condition $$n\left(A\right)=n\left(B\right)$$ along with the formulas of sets that is
$$\begin{array}{rl}& n\left(A\right)=n(A-B)+n(A\cap B)\\ & n\left(B\right)=n(B-A)+n(A\cap B)\end{array}$$
For solving second part we use the general formula of sets that is
$$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$$

Complete step-by-step solution
We are given that $$n(A-B)=60+3x$$, $$n(B-A)=8x$$ and $$n(A\cap B)=x-4$$
Let us draw a Venn diagram that represents the given information then we get

(a) The value of $$x$$
We are given that
$$\Rightarrow n\left(A\right)=n\left(B\right).......equation(i)$$
We know that the formulas of sets that is
$$\begin{array}{rl}& n\left(A\right)=n(A-B)+n(A\cap B)\\ & n\left(B\right)=n(B-A)+n(A\cap B)\end{array}$$
By using the above formulas to equation (i) we get
$$\begin{array}{rl}& \Rightarrow n(A-B)+n(A\cap B)=n(B-A)+n(A\cap B)\\ & \Rightarrow n(A-B)=n(B-A)\end{array}$$
By substituting the required values in above equation we get
$$\begin{array}{rl}& \Rightarrow 60+3x=8x\\ & \Rightarrow 5x=60\\ & \Rightarrow x=12\end{array}$$
Therefore, the value of $$x$$ is 12
(b) $$n(A\cup B)$$
We know that the direct formula of union of sets that is
$$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$$
By substituting the required values from the formulas we used before in above equation we get
$$\begin{array}{rl}& \Rightarrow n(A\cup B)=(n(A-B)+n(A\cap B))+(n(B-A)+n(A\cap B))-n(A\cap B)\\ & \Rightarrow n(A\cup B)=n(A-B)+n(B-A)+n(A\cap B)\end{array}$$
Now by substituting the required values in terms of $$x$$ in above equation we get
$$\begin{array}{rl}& \Rightarrow n(A\cup B)=60+3x+8x+x-4\\ & \Rightarrow n(A\cup B)=12x+56\end{array}$$
Now, by substituting $$x=12$$ in above equation we get
$$\begin{array}{rl}& \Rightarrow n(A\cup B)=12\times 12+56\\ & \Rightarrow n(A\cup B)=200\end{array}$$
Therefore the value of $$n(A\cup B)$$ is 200.

Note: Students may make mistakes in the Venn diagram representation.
Venn diagrams are the diagrammatic representation of sets in the universal set $${}^{\prime }{\mu }^{\prime }$$
So the Venn diagram must be drawn as

But students may miss the universal set $${}^{\prime }{\mu }^{\prime }$$ and draw the Venn diagram as

This will be the wrong representation because all the sets are subsets of a universal set $${}^{\prime }{\mu }^{\prime }$$ which is very important to represent in the Venn diagram.