I don’t see dead people but I see rectangles all the time.

The first time I saw a multiplication fact, like 3 x 5, as a rectangular array was after I’d graduated from college.

And because I was so very late to this game, I thought drawing rectangles must be how everyone else — at least math teachers — solved math problems.

But apparently not so. I started going to math workshops and often the teachers sitting next to me would look at my drawings and ask, “What are you doing?” I glanced over at their papers and saw mostly numbers and equations and thought, What are you doing?

So this is how I’ve always solved these rather mundane but classic problems.

**Problem: In a town, 3/7 of the men are married to 2/3 of the women. What fraction of the people in the town are married? **

My mind sees these two rectangles.

I notice that the blues have to match the pinks, so I make them match.

The answer is 12/23 of the people in town are married. (So fun to add fractions straight across.)

**Problem: Danielle and Jennifer can do a job in 2 hours working together. Danielle could do it in 3 hours alone. How many hours would it take Jennifer to do the job alone?**

I see a 2 x 3 rectangle as the “job,” and I choose 2 and 3 because other dimensions just get messy.

It’s easy then to see how much Jennifer can do.

The answer is it would take Jennifer 6 hours to do the job alone. I mean the answer is looking at me.

**Problem: [mathforum.com] Dad bakes some cookies. He eats one hot out of the oven and leaves the rest on the counter to cool. He goes outside to read. Dave comes into the kitchen and finds the cookies. Since he is hungry, he eats half a dozen of them. Then Kate wanders by, feeling rather hungry as well. She eats half as many as Dave did. Jim and Eileen walk through next, and each of them eats one third of the remaining cookies. Hollis comes into the kitchen and eats half of the cookies that are left on the counter. Last of all, Mom eats just one cookie. Dad comes back inside, ready to pig out. “Hey!” he exclaims. “There is only one cookie left!” How many cookies did Dad bake in all?**

Dad eats 1. Dave eats 6. Kate eats 3. So, 10 cookies are gone already.

Now, I use the rectangle to help me figure out where the rest of the cookies went. I’m reading “thirds” in the problem so I make sure I draw something that’s easy to divide into thirds. (I’m smart like that.)

Now, I read that Mom eats 1 cookie and there’s only 1 left for Dad. So the last three white squares in the rectangle represent 2 cookies.

I see 12 cookies. These 12 plus the 10 from the beginning equals 22 cookies in total that Dad baked.

**Problem: A class has 5/9 girls. If the number of boys were doubled and 12 girls were added, there would be an equal number of boys and girls. How many students were in the class at the outset?**

Naturally, I see this rectangle.

Then I do what the problem says.

The 5 pink boxes equal the 5 white boxes, so I cancel them out, leaving me with this.

At the outset there were 9 boxes, each box worth 4 students, so the answer is there were 36 students in the class.

So this is how I teach it to my students too. No equations. I remember a few years back one of my 6th graders’ tutors was not happy with me that I’d assigned a PS involving “systems of equations.” He said it was an algebra topic and how was a 6th grader supposed to solve it except for using guess-and-check. Really? Stuff people say that makes me fart.

Anyway, I’ve always looked at a square as two interlocking staircases, each a sum of consecutive integers. (Right, Chris Hunter?) I also see a non-square rectangle as two interlocking staircases, one is the sum of consecutive odds and the other sum of consecutive evens.

## 5 Comments

In the first problem, you are adding 3/7 and 2/3 and you state the answer is 12/23.

The correct answer should be 23/21.

Okay, Joie, I’ll bite…

Your 23/21 is from 9/21 and 14/21, right? What does the 9 represent in this context? The 14? Is this problematic? I know that my wife would find it so.

Also, what are the implications of 23/21 (or about 110%) of the people in the town being married to each other?

Your adding of the two fractions is, far and away, the most common approach teachers take at first. This speaks to our desire to use a standard algorithm.

Chris

Nice! It’s very interesting how a formal maths education actually makes you incapable of doing things *without* algebra. I constantly have to tell my staff *not* to “help” the nursing students with their drug calculations by teaching them algebra. They are perfectly capable of solving their problems with rectangles and/or proportional reasoning.

Hey Joie,

Try thinking about it like this: you are giving a test in 2 parts, multiple choice and short answer. The points on the multiple choice are 75 and the short answer are 25, so 100 total points possible. You score one student and she gets 70/75 on the multiple choice and 20/25 on the short answer. So her total points are 90/100. You aren’t going to triple her points on the short answer portion to get a common denominator of 75, that would give her 130/100!

In Fawn’s problem there are 23 people total. 14 men and 9 women 6 of each are married. So of the 14 men, 6 are married (3/7) and of the 9 women 6 are married (2/3). That makes 12 of the 23 people married.

The marriage problem’s pretty cool mathematically, but in 2015, there’s not enough information! For instance, you could have a town with EVERYONE married: 24 opposite-sex couples, 8 male-male couples, and 3 female-female couples.

(I’m not trying to be snarky here. Unlike when the problem was originally written, same-sex married couples are now a reality and not even uncommon among many of our students’ parents, and it would be hurtful to imply otherwise in a classroom.)

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