Question

Let A and B be the two sets such that $n\left( A-B \right)=60+3x$, $n\left( B-A \right)=8x$ and $n\left( A\cap B \right)=x-4$ 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) $n\left( A\cup B \right)$

Answer 1

We solve this problem by using the Venn diagrams of sets. The Venn diagrams represent the diagrammatic representation of sets inside the universal set $'\mu '$
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

& n\left( A \right)=n\left( A-B \right)+n\left( A\cap B \right) \\\ & n\left( B \right)=n\left( B-A \right)+n\left( A\cap B \right) \\\ \end{aligned}$$ For solving second part we use the general formula of sets that is $$n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$$ **Complete step-by-step solution** We are given that $$n\left( A-B \right)=60+3x$$, $$n\left( B-A \right)=8x$$ and $$n\left( A\cap B \right)=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{aligned} & n\left( A \right)=n\left( A-B \right)+n\left( A\cap B \right) \\\ & n\left( B \right)=n\left( B-A \right)+n\left( A\cap B \right) \\\ \end{aligned}$$ By using the above formulas to equation (i) we get $$\begin{aligned} & \Rightarrow n\left( A-B \right)+n\left( A\cap B \right)=n\left( B-A \right)+n\left( A\cap B \right) \\\ & \Rightarrow n\left( A-B \right)=n\left( B-A \right) \\\ \end{aligned}$$ By substituting the required values in above equation we get $$\begin{aligned} & \Rightarrow 60+3x=8x \\\ & \Rightarrow 5x=60 \\\ & \Rightarrow x=12 \\\ \end{aligned}$$ Therefore, the value of $$x$$ is 12 (b) $$n\left( A\cup B \right)$$ We know that the direct formula of union of sets that is $$n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$$ By substituting the required values from the formulas we used before in above equation we get $$\begin{aligned} & \Rightarrow n\left( A\cup B \right)=\left( n\left( A-B \right)+n\left( A\cap B \right) \right)+\left( n\left( B-A \right)+n\left( A\cap B \right) \right)-n\left( A\cap B \right) \\\ & \Rightarrow n\left( A\cup B \right)=n\left( A-B \right)+n\left( B-A \right)+n\left( A\cap B \right) \\\ \end{aligned}$$ Now by substituting the required values in terms of $$x$$ in above equation we get $$\begin{aligned} & \Rightarrow n\left( A\cup B \right)=60+3x+8x+x-4 \\\ & \Rightarrow n\left( A\cup B \right)=12x+56 \\\ \end{aligned}$$ Now, by substituting $$x=12$$ in above equation we get $$\begin{aligned} & \Rightarrow n\left( A\cup B \right)=12\times 12+56 \\\ & \Rightarrow n\left( A\cup B \right)=200 \\\ \end{aligned}$$ **Therefore the value of $$n\left( A\cup B \right)$$ is 200.** **Note:** Students may make mistakes in the Venn diagram representation. Venn diagrams are the diagrammatic representation of sets in the universal set $$'\mu '$$ So the Venn diagram must be drawn as  But students may miss the universal set $$'\mu '$$ and draw the Venn diagram as  This will be the wrong representation because all the sets are subsets of a universal set $$'\mu '$$ which is very important to represent in the Venn diagram.