Steven told me that he was given this problem during a corporate training session to encourage problem solving and collaboration among managers:

*A man bought a horse for $50 and sold it for $60.
He then bought the horse back for $70 and sold it again for $80.
What’s the financial outcome of these transactions?*

He explained how the session was organized. People were first asked to think about the answer individually. After they had time to think on their own, the facilitator asked for answers and listed them. The facilitator then identified a different location in the room for each answer offered. He asked people to go to the area he designated for their answer, compare their reasoning with the others who agreed, and select a spokesperson to present a convincing argument for those in other groups. As arguments were presented, anyone who became convinced of a different answer was free to leave the group and join another. The goal was to see if everyone would wind up in the same group, convinced of the same solution. The kicker was, Steven told me, only then would they go to lunch.

Hmmm, sounds like the Common Core Mathematics Practice Standard 3 in action: Create viable arguments and critique the reasoning of others. And the organizational structure relates to the classroom instructional practice of think, pair, share.

I was intrigued by the Dealing in Horses problem and have used it many times since then. I also included it when making a video for parents: *Mathematics: What Are You Teaching My Child?* The parents in the video gave a variety of responses. Take a look at this clip.

I was also intrigued with the organizational structure used in the corporate training session, and I’ve used it with classes from grade 2 through high school, in professional learning sessions, and with parent groups. Also, I included the problem in *About Teaching Mathematics* with specific teaching suggestions.

The organizational structure reminds me of what I’ve read in a book I’ve owned and loved since the 1980s―*Thinking Mathematically* by John Mason with Leone Burton and Kaye Stacey. There’s lots to love about this book (I did an internet search and copies are available), and one thing that has stuck with me over the years is the advice for justifying your thinking: convince yourself, convince a friend, convince a skeptic. I think that think-pair-share is another version of this advice.

So, back to the horse problem: Do you have an answer that you’re convinced about? If you haven’t thought about it yet, I suggest you stop and do so.

**Why I Like the Horse Problem So Much**

Understanding the situation is not the difficulty with *Dealing in Horses*—the scenario is clear. Adding and subtracting the numbers in the problem also is not difficult—the numbers are easy to handle, even mentally without having to rely on paper and pencil. Yet deciding precisely what to do with the numbers isn’t obvious to everyone. The difficulty lies in knowing how to choose appropriate arithmetic operations for the situation at hand in order to arrive at a correct solution. The confusion that adults experience with this situation is similar to the confusion that students experience when presented with a page of mixed word problems that leads them to ask: “Do I need to add or subtract?”

**Classroom Suggestions**

Most recently, I presented the Dealing in Horses problem to a class of seventh graders. I first asked if there was more than one correct answer and they agreed that the problem has to have one right answer. Then here’s what occurred.

- The students suggested three answers―earned $10, earned $20, earned $30―and I identified a location in the room for each answer. I told the few students who couldn’t decide to stay seated and join a group when they were ready. After a good deal of back and forth arguments with students moving here and there, most students were in the group with the “earned $20” solution and half a dozen were in the group for “earned $10.”

- I had students act out the problem. I chose three students, one to be the first owner of the horse, one to be the second owner, and one to be the horse. I cut index cards in half to represent $10 bills and gave each of the two students doing the buying and selling ten cards to represent $100. Then, as I read each part of the problem, they acted it out, including exchanging the “money” and the “horse.” At the end, the two students counted the money they had―Lily had $120 and Danny had $80.

- This convinced most that $20 was the answer, but some were still holding on to the answer being $10. They reasoned that after earning $10 the first time, they spent that $10 to buy the horse back the second time.

- I asked the students to return to their seats and I presented a similar problem:
*I bought a lamp for $50 and sold it for $60.*

I bought a table for $70 and sold it for $80.

What’s the financial outcome of these transactions?There was agreement that I earned $10 for the lamp and earned $10 for the table, for a total of $20. But is this problem the same as the horse problem? Some thought yes, some thought no, and some weren’t sure.

Ah, there’s no answer book for most real world problems.

Hmmm…interesting, but isn’t the financial outcome that the horse, in the end, cost him $30?

Dear Marilyn:

Dealing in Horses – is your best problem todate.

Chuck

Oops, put reply in wrong place. You’ve figured out “bought the horse” in terms of the total price, but “sold the horse” in terms of the difference between that and the purchase price. If the horse had been sold for ten bucks (not 60) it would work that way. If you’re adding & subtracting, it’s got to be the same kind of thing for it to make sense 🙂

Wow! I was WAY off! My thinking was:

Bought the horse -$50

Sold horse +$10

Bought horse again -$70

Sold horse again +$10

Total -$100

A friend told me that life has no answer book. This problem reminds me of this. – Marilyn

Any chance you still have the animation of the problem being told? I’d love a copy.

Thanks

Chris

I’m searching. Stay tuned.

You’ve talked about the “money spent” in terms of the whole amount, but the “money taken in” in terms of the difference between what it cost and what it sold for. If you’re adding things, you have to be adding the same things for it to make sense 😉

But you’re not just getting +$10 back. You’re getting $60 and $80 back.

Wait…!!! I am wrong (cost of $40 would be if he didn’t sell it in the end)!! You could make an argument for both gaining $10 and gaining $20, but It seems that just following the buying and selling, he makes $20.

I did this problem with my student teachers’ classroom few years ago. Acting out is the best strategy. The profit is only $20.

This question made for interesting conversation at work today. Most everyone came to the conclusion that the financial outcome was +20. Then everyone wanted to know what the answer really is.

This problem brings back such good memories. I was introduced to it at Bank Street College, in the early 1990s, where I was in the Math Leadership Program. I used the problem, as set up, with my 8th grade class in East Harlem and for 90 minutes they argued, discussed and moved from group to group. I knew it was a good lesson since no student asked to use the restroom during the class and they were still discussing it as they left the classroom at the end of the period as it didn’t get resolved. I told the students I would tell them the answer at the 8th grade graduation ceremony a couple of months later, which I did as I spoke from the lectern to the graduates and families. There was a loud cheer from those who had said the correct answer. I had the lesson videotaped (VHS) and need to get it digitized so I can use it with my preservice math teachers as an example of an engaging task where students try to convince others with justification.

I’d love to see the video when you get it digitized. I always learn from listening to students.

Mark, what was the answer you gave them at the graduation ceremony?

I treated this as if two separate horses were being bought and sold. Two purchases at $50 and $70 comes to $120. Two sales at $60 and $80 comes to $140, the difference being $20.

The simple mathematical answer appears to be $20 but from a business perspective the problem does not address the time value of money and operating costs. If the man had to borrow the money to buy the horse, then the interest on this loan must be deducted from his final profit. Not significant if he sold the horse the same day, but the problem doesn’t mention that. Maybe it took 1 year or more to sell the horse. You would also need to account for feeding, housing, insuring, and maintaining the health of the horse during the time the man owned it, and deduct these expenses from the final profit as well. He may very well be in debt after these two transactions, and get out of the horse business.

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I think the simplest representation is this:

A purchase is represented by a negative number. A sale is represented by a positive number. So, from the man’s perspective, the transaction history is:

(-50) + (60) + (-70) + (80) ,

which sums to positive 20. Therefore, he profited $20.

The abstract structure is the same whether we are talking lamps, tables, apples, or oranges. The essence is money going out (negative) and money coming in (positive).

I agree that the man (owner 2) comes out with $20 but the student’s task with the fake bills is a bit deceptive. The first owner ended up $20 short, but he gained an ostensibly $80 horse, making his equity (or cash, were he to liquidate) $60. As he started with a $50 horse he’s gained $10. Of course owner 1’s $10 and owner 2’s $20 profits sum to $30, which is the appreciation of the horse over time.

It may be informative for students to keep track of the value of the horse as well as the cash transactions.

Ah, yes, more reality. To think nothing of the cost of feeding, etc. Yes to your idea of dealing with students with the full idea. I’ve found at the dinner table the argument about the money is as far as I’ve been able to push. Thanks for your comment.

i think its $20

It is $20, and as proven with using the 20 strips of paper, each representing $10 and doing the exchanges back and forth which end up in the person who sold the goat netting $20, $20 is the right answer.

HERE’S WHY: The purchase of the goat was an investment, NOT AN EXPENSE. Most people don’t distinguish between the two, but this is an ideal example to show. There was no expense; each time the investment was paid back in full, +$10. You may be surprised how much this simple problem actually plays out in your real life. The PRIMARY reason that most people stay paycheck to paycheck is the lack of financial literacy. There’s an understanding of money and how it works that we are never taught in school, and the entity that profits from it is the big banks, and the government. I literally show people how to get the right tools to gain these basic understandings so that people aren’t squeezed so much by the banks! Apply them, and you’ll notice that you’ll start piling money into your savings account, even if you aren’t making more money. Email me @ TheCardDoc@gmail.com and I’m glad to get you the info. Society needs more financial literacy so people can finally get ahead!

He started with $50 and ended up with $80, so his profit is $30

($80-$50 =$30)

That’s not the conclusion I come to after acting it out. The $80 – $50 doesn’t account for the intermediate steps.

haha wrong

So, what is the answer? my head hurts

I’ve convinced myself that it’s $20, and I acted it out to come to this conclusion. My first thought was that it’s $10, but acting it out was persuasive. Ah, life has no answer book!