STEM19 American University of Sharjah, UAE , Analyzing Math Word Problems with Digital Video: The Process and Potential

*Note: When watching the videos, and taking a wide view across the ten cases and path from question to outcome, notice three reference points (i.e. units of analysis) of speech, the visual, and, the act of pointing. With the act of pointing, note the synchronicity of the pointing and speech.

Click on the link below to access and download the Power Point (the videos may not play, but the videos from the power point and more are on this page.

https://cincopa.com/~AYFACCtYYllw!AAFDz4DRr1gK

The Brick Problem: Three Cases

The Question:

Scales are balanced with a whole brick on one side and an exact half of exactly the same brick, plus a 3-pound weight on the other.

What is the weight of the whole brick?

Data Case Margo

Initial Calculations:

The Larger Visual

The Video

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Transcribed Speech from the Video

The problem is asking scales are balanced with a whole brick on one side and an exact half of exactly the same brick, plus a 3-pound weight on the other.

What is the weight of the whole brick?

Step one says interpret the word problem visually

This is my scale that I’ve drawn. On the left side I’ve drawn a whole brick. And, uh, on the other side of the scale, which is supposed to be balanced. I’ve drawn—I brought to scale on half of the brick on the left side plus the three-pound weight on the right side, and this illustration is supposed to show that balance on both sides so we have a picture. Step two is to substitute the values for each brick on the left side and the right side of the scale so it will be balanced. And I chose simple numbers, one For my whole brick on my left side, but I made a mistake because half of one is point five plus three is three point five, And one does not equal three-point five, But that’s the idea and once you figure out what number is on the left equal half of the number on the left plus three Then that is the value–Then you’re solving um The weight of the whole brick

Data Case Sandra

Initial Calculations

Initial Calculations

Case Sandra: The Larger Visual

The Video

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Transcribed Speech from the Video:

Researcher comments are in parenthesis and are in italics; other words and/or phrases in parenthesis were difficult to decipher from the recording and may not be accurate. A— “dash” indicates an abrupt change of thought or realization.

The weight of a brick

Scales are balanced with a whole brick on one side and an exact half of exactly the same brick, plus a 3-pound weight on the other.

What is the weight of the whole brick?

So first I did step one I drew a picture to look like a brick; then I read the word problem and matched the numbers on the side of the brick. Well with then the brick what I do actually, I put one up here, and one down right here cause this I want to be two whole sides So I put the one there cause this side is by one half and when I read the words on this side, it said add three  But I times three—no, uh, but it said two. Then I multiplied one half times three, and it gave me three-sixths so I divided. So here you go, one half times three Equals three-sixths, so I did three sixths divided by one half which shows right—(stick)  (there’s a word here I can’t catch when she slaps the poster here. Is the poster-paper coming unglued?) three sixths divided by one half, and that gave me point twenty-five. So I took three-sixths from what I got earlier, which is one half—no I got point two five, that’s it (I mean), so (I did) three-sixths times point two five, it gave me one twenty-five. So, um, basically, I did the weight of the whole brick was point one twenty five and that’s what I wrote the weight of the brick.

Interviewer: Do you see anything now that might be weak point? What do you think is a weak point? Anything?

Sandra: yeah I feel like right there when I multiplied three, when it said add three In the word problem I multiplied

Interviewer: It should have been an addition there?

Sandra: Yeah it should have been

Interviewer: I wonder what caused the confusion

Do you think it’s the shape of that…?

Sandra: No I feel like cause I’m not used to multiplying again so I added; So this one gonna be worse cause I’m used to the word problem when you multiply, and they say add

So um—

Interviewer: that’s where you think the error might be

Shandra: yeah

Interviewer: Ok so I’m working on this with a math professor too, and we’re breaking down some of that stuff. Good; Excellent; Super; that’s it

Data Case Isabella

Initial Calculations:

Case Isabella: The Larger Visual

Case Isabella: The Video

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Transcribed Speech from the Video

Researcher yells action

My problem is called Equal Bricks The question said scales are balanced with a whole brick on one side and an exact half of exactly the same brick, plus a 3-pound weight on the other. What is the weight of the whole brick? So the first thing I did was draw a picture of a scale that showed each side We knew that there was one brick on one side, and the other had half the same—pound of brick on the other side plus three. I put B because B equals a brick in pounds. So the first step was to draw a picture and identify variables or variable. Step two was to make an equation,so I put B equals half B plus three. So then I have to solve for B. Remember what you do to one side you have to do to the other side, and to bring all like or the same variables to the same and solve it. So what I did was subtract three on both sides so I got B minus three equals one-half B; multiply that by two Uh—on both sides to get B by itself over here. So I have two B minus six equals B. Then I subtracted B on both sides: ended up with zero. So you can solve with B. So you have two minus B minus six equals zero, which simplifies out to B minus six equals zero. So then I solve for B add six to both sides and B equals six. So the Brick equals six pounds So then I plug the answer back into the initial equation; this one for verification. So right here I have B equals one-half B plus three; six equals six over two plus three. six equals six Ta Da.

The Goose Problem: Three Cases

(An adaptation from an old problem from India). A goose meets a flock of geese. The goose says ‘Hello, a hundred of geese’. The leader of the geese answers, ‘No, we are not a hundred. If there were also a half of our number and you, goose, then together there will be a hundred’.

How many geese were in the flock?

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Case Dan

Case Dan, Initial Calculations

Case Dan, Visual

Case Dan Video

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Case Dan, Transcript:

Researcher: Action

Dan: Alright The question was a goose meets a flock of geese the goose says hello a hundred of geese the leader of the geese answers no we are not a hundred if there were also half of our number and you. Goose then together there will be a hundred how many geese were in the flock?

Alright so first of all I took a hundred. The number we were looking for to get minus the one goose that came up and said hello we have ninety-nine geese. Then I took muh divided that by two to get a new a new half right here but and I came with forty-nine and a half geese in the flock, which is not possible because you can’t have half a geese So after thinking about it and reading the question a few more times I came up with this problem still trying to get the hundred geese er what we’re coming to and minus your one goose that came up you have ninety-nine now when I divide that by three which gives me thirty-three so thirty-three would be the the half of the the half of their number So if there’s sixty-six there’s ah well ninety-nine minus thirty three gives you sixty-six So if there’s sixty-six and you take half of that  and add it to the flock then you got—er you got ninety nine, then you add the one goose that came up and said hello which makes a hundred which answers the question an  therefore this one has to be right and this one has to be wrong.

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Case Lana

Case Lana, Initial Calculations

Case Lana, Visual

Case Lana, Video

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Case Lana, Transcript

Researcher: Action

Lana: Ok this is called fun with Geese The question says A goose meets a flock of geese The geese says Hello a hundred geese Leader the—the leader of the geese answers No we’re not a hundred It whirl–It we’re . also sorry

Researcher: It’s Ok Keep going

If we were also a half of our number any you goose together ruled a hundred How many geese are to a flock so right here, that’s the simplest number right here, it’s forty-nine, because it says right here if you have half of half of a flock plus you will equal a hundred.Therefore, you say there is forty-nine in the flock plus the one geese, so if you get half of thatand it’s pictured on here so we want to say that x is half so a hundred times one half equals fifty,
so we have the forty-nine, which is this flock of geese and the one goose when you add that together it is fifty, so let’s just say we added—well yeah it’s going to be ninety-nine geesewe add these two so together it is going to be a hundred you know half of a hundred is fifty so yeah

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Case Diana

Case Diana Initial Calculations

Case Diana Visual

Case Diana Video

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Case Diana Question

Researcher: Action

Diana: A Flock of Geese; an adaptation from an old problem from India. A goose meets a flock of Geese The goose says hello to a hundred of Geese The leader of the Geese says that no we are not a hundred And there are also half our number And yur—and you goose then there will be a hundred. How many geese are in the flock. Step one I draw—I drew a picture up here for a better understanding, which was one flock of geese and one flock of goose. Step two, I identify the number of geese and goose; then add it up one flock of geese and one flock of goose, which is two, and I plug in the variable for it; so 2x the number geese flock. Step three I looked back at the problem and I took the one hundred and looked at, and I looked back at step and I took 2x divided by one. 2x equals one hundred divided by two 2x use—uh fifty so there was fifty number of Geese in the flock, and that’s what I got; that’s the answer.

Researcher: That’s good—that’s good.

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Tolstoy’s Hat Problem: Four Cases

 This is how the problem was presented to participants

A merchant is selling hats, whose unit price is 10 rubles.

After a few hours into the shop bursts outraged neighbor, yelling that the 25 rubles are counterfeit, and demands the merchant to return his money, otherwise the neighbor will call the police. The merchant does not want a scandal, and with sore heart gives the man genuine 25 rubles.
What loss was suffered by the merchant?

Tolstoy’s Hat Problem

Case A

Case A Initial Calculations

Case A Visual

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Transcript Case Andy

As with all cases, Case A reads the problem first. The transcript begins after Case A reads the problem

So to solve this problem, made a table. This column, we have the merchant. This column, we have the customer. This column, we have the neighbor. So we did an accounting of sorts, what’s coming in, what’s coming out. Originally the customer gives a 25 ruble note to the merchant, so the customer’s, at the moment, down 25, and the merchant is up 25. The merchant then gives the 25 ruble note to the neighbor, the neighbor is up 25 at the moment, and he gives change: 10, 10, 5. So at this point the neighbor is even, and the merchant gets the 10, 10, and 5. At this point, the merchant then give a hat, which is where we count 10 for the value, and a 15 ruble note to the customer. So he’s up 10 plus 15, 25. Then we find out the 25 ruble note is counterfeit, so now the neighbor is down 25. The counterfeiter, the customer, is up 25. To resolve this, the merchant has to give 25 to the neighbor. When we calculate this out, the merchant, the 25 he took in, plus the 25 he gave out to break change, those are a wash. This is the 25 that he received in change from the neighbor, he gave a hat valued at ten and change of 15. So, that’s a wash. So, this washes out, all right? Then he has to give 25 back to the neighbor to resolve the counterfeit. So his total loss is 25. So then if we look at the customer, since he gave the counterfeit bill, he had no loss here. So his profit is a hat valued at 10 and a 15 ruble note. So the customer is up 25. The neighbor, he broke even because he’s resolved. He received a counterfeit bill, which made him down 25, but then he received another legitimate bill from the merchant, making him even. So we have the merchant had a 25 loss and the customer with a 25 ruble gain.

Case Barbara

Case B Initial Calculations

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Case B Transcript

The transcript begins after Case B reads the problem

So, we’re gonna solve this problem. We’re gonna use three, or four basic steps. Step one, we’re gonna break the problem apart. We know that the 10 rubles per hat is given information.  Step two, we know that one hat will be bought from a customer who has 25 rubles. That’s also given information. And step three, we’re gonna practice basic math. The given information that was given was the boy gives the merchant three ruble notes, 10 plus 10 plus 5, that’s 25. And step four, we’re gonna subtract 25, what the customer will give, minus the 15, from what the merchant will give the customer. So basic math, 25 minus 15 equals 10. And what did the merchant lose? The merchant lost a reliable customer and possibly 5 dollars.

Case Carl

Case C Initial Caculations

Case C Visual

Case C Video

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Transcript Case C

The transcript begins after Case C reads the problem.

Step one; read scenario. Step two: solve and explain. The hats sell at 25 rubles. Right here, the R stands for rubles. The merchant has a 25 dollar ruble note. He has no change so he sends a boy to get some change and the boy comes back with three notes in the amount of 10, 10, and five, which add up to 25 rubles. Okay, a hat costs 25 rubles. So subtract 25 from 10, you get 15 rubles. The merchant returns the 15 rubles and a hat to the customer. At first, the merchant gains 10 rubles, but then the neighbor comes in wanting a refund for his 25 rubles, which were counterfeit, so the merchant has to refund 25 rubles. The merchant loses a total loss of 15, which is the change he gave back, a 10, the cost of the hat, and the 25 which he has to give it back, which he had to refund, which is a total of 50 rubles, if I’m correct.

Case Derek

Case D did not have any initial calcuation sheet.

Case D Visual

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Case D Transcript

The transcript begins after Case D reads the problem

So the cost of the hat is ten Rubles and then this is showing the customer’s 25 Rubles which we now know is counterfeit. And then the seller sends a boy to go make change, so this is representative of that change, ten, ten, five is 25 Rubles. And then, the 15 Rubles goes back to the customer, plus the hat. And then step five, the 25 Ruble note was fake, and had to refund the money back to where he had gotten the change from. So then the seller suffered a loss of the 15 Rubles that he gave in change, plus the 25 Rubles which he gave back to the person he had made change with, plus ten Rubles for the hat. Assuming that the seller’s cost is the same as his selling price. So we don’t know what his actual loss is for the hat, but we can assume that it was ten Dollars because that’s what he could’ve sold it for. So that would be a grand total of 50 Rubles.

This model has evolved from the ideas of Tomasello (2003), McCafferty (2002) and many others. Specifically, The model below of a Shared Attentional Frame is an adaptation of Tomasello’s (2003) rendering of the “Structure of a linguistic symbol” (p. 29) and a “Joint attentional frame” (p. 26).

McCafferty, S. (2002). Gesture and creating zones of proximal development for second language learning. The Modern Language Journal, 86, 192-202.

Tomasello, M. (2003). Constructing a language: A usage-based theory of language acquisition. Cambridge, MA:

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