How to Use Venn Diagrams to Solve Problems
Learn how to solve SAT math problems using Venn diagrams.
Jason Marshall, PhD
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How to Use Venn Diagrams to Solve Problems
Now that we know how to make and interpret the meaning of Venn diagrams, it’s time to put our knowledge to the test and use them to solve problems that you might encounter in the real world. And when I say put our knowledge to the test, I mean that literally since problems like the one we’ll look at today frequently show up on standardized tests like the SAT…and lots of other places too.
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Today’s SAT-Inspired Question
Imagine you’re at a local city council meeting at which an important vote about where to build a new dog park is taking place. The options are to build the dog park at either Washington Park or Waterfront Park, or to build a new dog park at each location. When people who support building at Washington Park are asked to raise their hands, 47 votes are registered. When people who support building at Waterfront Park are asked to raise their hands, 36 votes are registered. As you’re watching the votes come in, you notice that 24 of the voters have raised their hands in support of building parks in both locations.
So the question is: How many people voted? In other words, how can you use the knowledge that 47 people raised their hands for Washington Park and 36 raised their hands for Waterfront Park, while also knowing that 24 people raised their hands for both parks, to figure out how many people cast votes?
The Wrong Answer
Some of you might immediately jump to the conclusion that since 47 and 36 people cast votes for Washington and Waterfront parks, the total number of people who voted must be 47 + 36 = 83. If you made that leap, let me emphasize that each of the voters had the option of voting for more than one park…and we know that 24 of them did exactly that. In other words, some of the 47 people who voted for Washington Park also voted for Waterfront Park. So although there were 47 + 36 = 83 votes cast, there were actually fewer voters. How can we figure out how many?
How to Set Up the Venn Diagram
Well, fortunately we now have a great tool at our disposal that’s ready to help us tackle problems like this. Of course, the tool I’m referring to is the venerable old Venn diagram. Here’s how it works in this case. Let’s start by drawing a rectangle that represents the set of everyone who cast a vote about the dog parks. Next, let’s draw a circle within that rectangle to represent the set of people who raised their hands in support of a dog park at Washington Park. Finally, let’s draw a second circle that represents the set of voters who raised their hands in support of a dog park at Waterfront Park.
As with all Venn diagrams, these two circles must overlap to show all the possible relationships between the various sets. In this case, the overlapping region between the two circles represents the set of people who raised their hands in support of both dog parks. In other words, this overlapping region is the intersection of these two sets.
That’s all well and good, but how does it help us answer our original question: How many people voted?
How to Use the Venn Diagram to Solve the Problem
Believe it or not, to solve our problem we just need to fill in our Venn diagram with some numbers. Let’s start with the overlapping region between the two circles. Since we know that 24 people cast votes in support of building dog parks at both Washington and Waterfront Parks, the overlapping region of our Venn diagram must contain 24 votes.
Now, since we know that a total of 47 people raised their hands in support of a dog park at Washington Park, and that 24 of these 47 people also supported building a park at Waterfront Park, there must be 47 – 24 = 23 voters who only voted for Washington Park. So the part of the circle representing people who voted exclusively for Washington Park—the part that doesn’t overlap the second circle—must contain 23 votes.
Similarly, since 24 of the 36 people who voted for Waterfront Park also voted for Washington Park, the non-overlapping part of the circle representing Waterfront Park supporters must contain a total of 36 – 24 = 12 votes.
So the three regions within the circles of our Venn diagram contain 23 votes for Washington Park only, 12 votes for Waterfront Park only, and 24 votes for dog parks in both locations. When we put this all together, we get the answer to our problem. Namely, that there are a total of 23 + 12 + 24 = 59 voters.
Number of the Week
Before we finish up, it’s time for this week’s featured number selected from the various numbers of the day posted to the Math Dude’s Facebook page. This week’s number is either 50 or 62 miles. Why? Because 50 and 62 miles are two commonly accepted standard definitions of the altitude at which Earth and its atmosphere end and outer space begins.
Why are there two different numbers? Well, it’s because there’s no precise definition of where exactly space begins since Earth’s atmosphere simply gets thinner and thinner until eventually there’s not much left…and 50 and 62 miles are two standards that the world has adopted to mark this boundary.
So the next time you’re wondering how far it is to get to outer space, just remember that you could drive there in your car in less than an hour—but you’d have to be able to drive straight up to do it!
Wrap Up
Okay, that’s all for today. Remember to become a fan of the Math Dude on Facebook where you’ll find a new number of the day and math puzzle posted each and every weekday. And if you’re on Twitter, please follow me there too. Finally, if you have math questions, feel free to send them my way via Facebook, Twitter, or by email at mathdude@quickanddirtytips.comcreate new email.
Until next time, this is Jason Marshall with The Math Dude’s Quick and Dirty Tips to Make Math Easier. Thanks for reading, math fans!
Venn Diagram image courtesy of Shutterstock
About the Author
Jason Marshall is the author of The Math Dude’s Quick and Dirty Guide to Algebra. He provides clear explanations of math terms and principles, and his simple tricks for solving basic algebra problems will have even the most math-phobic person looking forward to working out whatever math problem comes their way.