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Using Venn Diagrams to Solve Word Problems

So how do we use these diagrams to solve word problems? Let's work through a few example problems that are similar to ones you may encounter.

Example 1

There are $1{,}000$ students in a high school. $800$ students enjoy learning about English, $700$ students enjoy learning about History, and $500$ students enjoy learning about both History and English. How many students enjoy learning about English but not History? How many students do not like English nor History?

Let's draw a diagram to figure it out:

Here, we let the variables $E,B,$ and $H$ represent amounts of students:

• $E=$ number of students who only enjoy English
• $B=$ number of students who only enjoy both English and History
• $H=$ number of students who only enjoy History

First let's figure out how many students only enjoy English. In other words, let's calculate $E$.

We know that $800$ students in total enjoy English. That means: $$E+B=800$$

We also know that $500$ students enjoy both English and History, i.e., $B=500$. Now, we can substitute $B=500$ into our $E+B=800$ equation: $$\begin{align} E+B&=800\\ E+\color{#8925AE}{500}&=800\\ E+500\,\color{#8925AE}{-\,500}&=800\,\color{#8925AE}{-\,500}\\ E&=300 \end{align}$$ So, $E=$ number students who only enjoy English $=300$.

Next, let's figure out how many students only enjoy History, or $H.$

We can calculate $H$ the same way we calculated $E.$ We know that $700$ students in total enjoy History. That means: $$H+B=700$$

We also know that $500$ students enjoy both English and History, i.e., $B=500.$ Now, we can substitute $B=500$ into our $H+B=700$ equation: $$\begin{align} H+B&=700\\ H+\color{#8925AE}{500}&=700\\ H+500\,\color{#8925AE}{-\,500}&=700\,\color{#8925AE}{-\,500}\\ H&=200 \end{align}$$ So, $H=$ number of students who only enjoy History $=200$.

Let's fill in the diagram with this new information, just for kicks!

Example 2

Out of the $200$ seniors at your local high school, $180$ of them take Opera Singing and $30$ of them take Speedcubing. Students must take either Opera Singing, Speedcubing, or both. How many students take both classes?

Again, let's make a Venn diagram to model the situation.

Notice that if we add the number of students taking Opera Singing and the number of students taking Speedcubing, we get $$180+30=210$$ But $210$ is $10$ greater than the total number of seniors, which is $200.$ This means that $10$ students must take Opera Singing and Speedcubing.

Example 3

Challenge Problem: Sets $A$ and $B$ have the same number of elements. If their union has $4{,}000$ elements and their intersection has $1{,}000$ elements, how many elements are in the set $A$?

Time to draw a Venn diagram!

Here, we let the variables $a,b,$ and $c$ represent how many elements are in their respective sections of the Venn diagram:

• $a=$ number of elements in only $A$
• $b=$ number of elements only in $B$
• $c=$ number of elements in both $A$ and $B$

The number of elements in the union of the sets $(A \cup B)$ is represented by $a+b+c.$

The number of elements in the intersection of the sets $(A \cap B)$ is represented by $c.$

In the problem, we are given: $$\begin{align} a+b+c&=4{,}000\\ c&=1{,}000 \end{align}$$

Let's plug in $c=1{,}000$ into our $a+b+c=4{,}000$ equation! $$\begin{align} a+b+c&=4{,}000\\ a+b+\color{#8925AE}{1{,}000}&=4{,}000\\ a+b+1{,}000\,\color{#8925AE}{-\,1{,}000}&=4{,}000\,\color{#8925AE}{-\,1{,}000}\\ a+b&=3{,}000 \end{align}$$ So, now we know that $a+b=3{,}000.$

The problem also tells us that sets $A$ and $B$ have the same number of elements. This means that $a+c=b+c.$ We can simplify this equation by subtracting $c$ from both sides: $$\begin{align} a+c&=b+c\\ a+c\,\color{#8925AE}{-\,c}&=b+c\,\color{#8925AE}{-\,c}\\ a&=b \end{align}$$

Since $a=b,$ we can plug $a$ in for $b$ in our $a+b=3{,}000$ equation: $$\begin{align} a+b&=3{,}000\\ a+\color{#8925AE}{a}&=3{,}000\\ 2a&=3{,}000\\ \frac{2a}{\color{#8925AE}{2}}&=\frac{3{,}000}{\color{#8925AE}{2}}\\ a&=1{,}500 \end{align}$$ So, now we know that $a=1{,}500$ and we were given that $c=1{,}000.$ Since the number of elements in set $A$ is equal to $a+c,$ we can calculate that the total number of elements in set $A$ is: $$\begin{align} A&= a+c\\ A&=1{,}500+1{,}000\\ A&=2{,}500 \end{align}$$

Using Venn Diagrams to Show Union and Intersection

A great way to visualize sets is through the use of Venn diagrams. Also known as Euler diagrams, John Venn created these types of diagrams to help visualize set theory. These are essentially overlapping Eulerian circles to show what two subsets in common. We saw similar diagrams when we were looking at subsets and complements.

You've probably seen one of these before. Check it out:

Here is another example. We have stuff that is unique to $A, B$, and $C$, as well as things that are shared by $A$ and $B, A$ and $C$, $B$ and $C$, and all $3$ ($A, B,$ and $C$).

$2$nd graders' favorite foods are everything in the purple circle, including the part that overlaps with the pink circle. So we know that $2$nd graders' favorite foods are pudding, applesauce, yogurt, pizza, and ice cream.

The foods that are the favorites of both $2$nd and $8$th graders are where the circles overlap. There's only two things in this area: pizza and ice cream.

We can also use these diagrams to represent sets. They are especially helpful in visualizing the union and intersection of sets.

In the case of the Venn diagram above, set $A$ is in purple and set $B$ is in pink. The intersection of set $A$ and set $B$ is the part where these two sets overlap.

(Video) Venn Diagrams: Shading Regions for Two Sets

by PatrickJMT

This video by PatrickJMT demonstrates how to use Venn diagrams to represent different set interactions, including intersections and unions.

$\large{\text{Summary}}$

Definitions

When shading, there are three main cases: unions, intersections, and complements (everything but the indicated set).

Example 1

What does the Venn diagram look like for $A\cup B\,'$.

The best way is to go one step at a time. First, draw $A$.

Next, we draw $B\,'$.

Finally, we find their union, $A\cup B\,'$.

A Venn diagram shows all the possible logical relations between a finite collection of sets where a set is a collection of distinct objects.

$\cup$ is read as "union" and $\cap$ is read as "intersection"

Let $A$ and $B$ be sets. Then $A \cup B$ is the set of all the elements in $A$ and $B$. $A\cap B$ is the set of all the elements that belong to both $A$ and $B$.

Example:

$$\begin{align} P &= \{5, 17, 25, 28, 0, 2, 10\}\\ Q &= \{0, 2, 10, 7, 6, 9, 4, 11, 12, 8\} \end{align}$$

This is the union of $P$ and $Q$. $$P \cup Q = \{5, 17, 25, 28, 0, 2, 10, 7, 6, 9, 4, 11, 12, 8\}$$ It includes all the elements in both sets, but only one of each (notice that there aren't two $20$'s or $0$'s in the set).

This is the intersection of $P$ and $Q$. $$P\cap Q = \{0, 2, 10\}$$ It includes only the elements that are in both of $P$ and $Q$, and nothing else.

The Venn diagram for the two sets is below:

Try it yourself!

Look at the following information, then draw a Venn diagram. Notice that we're working with three sets, rather than just two, so you'll need three circles.

$$\begin{align} A &= \{1, 4, 5, 6, 8\}\\ B &= \{6, 7, 8, 9, 10, 13, 25\}\\ C &= \{4, 5, 8, 13, 22, 25, 31, 42, 55\}\\\ \\ A \cup B &= \{1, 4, 5, 6, 7, 8, 9, 10, 13, 25\}\\ A \cup C &= \{1, 4, 5, 6, 8, 13, 22, 25, 31, 42, 55\}\\ B \cup C &= \{6, 7, 8, 9, 10, 13, 25, 4, 5, 22, 31, 42, 55\}\\\ \\ A \cap B &= \{6, 8\}\\ A \cap C &= \{4, 5, 8\}\\ B \cap C &= \{8, 13, 25\} \end{align}$$