A Singapore Math Age Problem

On this Singapore's 48th National Day (or Independence Day), let's look at another question from Lorraine Walker's Model Drawing for Challenging Word Problems: Finding Solutions the Singapore Way. The author solved the following grade 4 or 5 age problem, using the bar method.

Rolff is now 4 years older than Erik. 

Five years ago Rolff was twice as old as Erik. 

How old was Rolff 5 years ago?

Her model-drawing solution is as follows:

Method 1

No doubt, the second color makes it easier to visualize what's happening, but what if students aren't given the choice to draw their model in more than one color? In fact, it wouldn't be surprising for critics of the model method to argue that there's no need to "complexify" the solution with all these bars and arrows, when the question could easily be solved without a model drawing.

Are there other valid models?

In this before-and-after word problem, is there an alternative or better model that may depict the relationship between the ages of the two persons? Let's look at two quick-and-dirty methods of solution, which a number of local students would have used in solving this question.

Method 2

In this method, we assume that we're not aware that the age difference between Rolff and Erik is a constant.

Here, we make use of the fact that the distance of "1 unit + 5" is equal to the distance of "5 + 4."

From the model drawing, 

1 unit = 4

2 units = 2 × 4 = 8

Five years ago Rolff was 8 years old. 

Method 3

Since the difference between their ages at any point of time is a constant, in the "before" diagram, the difference is 1 unit (2 units – 1 unit), and in the "after" diagram, the difference is 4. So, 1 unit must be equal to 4. 

The choice of the smaller age numbers and the relatively small number of times one person is older than the other, makes it difficult for one to appreciate the power of the model method in this before-and-after age problem. 

A slightly modified question

A modified version of the original question could probably help convince one why the bar method might serve as a useful strategy in revealing the relationship between the ages of the two persons.

Rolff is now 12 years older than Erik. 

Five years ago Rolff was three times as old as Erik. 

How old was Rolff 5 years ago?

From the model drawing, 

2 units = 12

1 unit = 12 ÷ 2 = 6 

3 units = 3 × 6 = 18

Five years ago Rolff was 18 years old. 

Conclusion 

This age problem shows that in posing word problems that lend themselves to the bar method, we'd attempt as far as possible to constructing them in such a way that the model method is seen to offer a better problem-solving strategy than other traditional (or formal) methods in solving the questions. Otherwise, there's no incentive for one to wanting to draw a model drawing than use other strategies, which might even be shorter or easier.

Reference

Walker, L. (2010). Model drawing for challenging word problems: Finding solutions the Singapore way. Peterborough, NH: Crystal Springs Books.

© Yan Kow Cheong, August 9, 2013.

The Chickens-and-Cows Problem

At the end of an earlier post entitled "The Chickens-and-Rabbits Problem," I tickled readers whether the following question could be solved using the Singapore model method, or the Sakamoto method.

Mr. Yan has almost twice as many chickens as cows. 

The total number of legs and heads is 184. How many cows are there?

On his "Johnny and Mary Do Maths" blog, Chris Patterson followed up with an algebraic solution to the word problem in a post entitled Visualisation versus Algebra?. He asked me whether I have a visual proof to the question, particularly one using the Sakamoto method.

I'm not sure whether I've succeeded in answering Chris's question, but here are three quick-and-dirty non-algebraic attempts to the chickens-and-cows problem.


Method 1 (Singapore model method)

From the model,

11 units = 184 + 3 ▌= 16 × 11 + (8 + 3 ▌)

If ▌= 1, then 11 units = 16 × 11 + 11

          1 unit = 17

2 units – 1 = 2 × 17 – 1 = 33

A quick check shows that for other integral values of ▌other than 1, 11 units cannot be an integer.

Hence, Mr. Yan has 33 chickens and 17 cows.


As a student of the Singapore model method, still trying to learn how to make wise use of this visualization strategy, especially when it lends itself well to a particular word problem, I would like to hear from members of the Singapore mathematical brethren about their alternative models in solving this chickens-and-cows problem, assuming that we're discussing this question with a group of grade 5 or 6 students—with no knowledge of Diophantine equations or advanced algebraic techniques.


Method 2 (Sakamoto method)

1. Grasp the relation

Let ① represent the number of cows, and △ be a variable quantity that is much less than ①.

                                           Chickens            Cows

                                              ② – △                ①

Number of legs                 (② – △) × 2         ① × 4

                                            = ④ – 2△            = ④

                                   _________________________

                                                           184

2. Diagram

3. Number sentences

② – △ + ④ – 2△ + ① + ④ = 184
⑪ – 3△ = 184
⑪ = 184 + 3△ = 16 × 11 + (8 + 3△)

If △ = 1, then ⑪ = 16 × 11 + 11 = 17 × 11
                       ① = 17
② – △ = 2 × 17 – 1 = 33

So, there are 17 cows and 33 chickens.


Unless there's a more elegant Sakamoto solution, I find that it doesn't differ much conceptually from its algebraic cousin.


Method 3 (Using the "Make a supposition" strategy)

Suppose there were exactly twice as many chickens as cows.

Then each group of 2 chickens and 1 cow would have a total of [(2 + 2 × 2) + (1 + 1 × 4)] = 11 legs and heads.

Now, 184 = 11 × 16 + 8

16 groups of 2 chickens and 1 cow would have a total of 16 × 11 = 176 legs and heads.

How many chickens and cows have 8 legs and heads altogether?

Clearly, 1 chicken and 1 cow have 8 legs and heads altogether.

So, Mr. Yan has (16 × 1 + 1) = 17 cows and (16 × 2 + 1) = 33 chickens.


Conclusion

With Chris's algebraic solution, we're sharing four methods of solution to the chickens-and-cows problem. Which method do you prefer? Which one would you use with your students? Share yours with the rest of us.


© Yan Kow Cheong, April 24, 2013.

The Dogs-and-Ducks Problem

Recently, while reviewing some grade 5 olympiad questions for a local company, which specializes in conducting olympiad math programs in elementary schools, I came across the following nonroutine dogs-and-ducks question.



The number of ducks is 10 fewer than the number of dogs. 
The dogs have three times as many legs as the ducks. 
How many ducks and how many dogs are there?



When I realized that my mental calculated answer differs from the one given by the problem poser, I thought maybe my incorrect solution might serve as a "logic exercise" to tickle others' mathematical bones to point out the flawed reasoning. My faulty argument followed something along these lines:

Since there are 10 fewer ducks, and each duck has 4 legs, there are 10 × 4 = 40 fewer duck legs.



From the model drawing,



2 units = 40


1 unit = 20



Number of duck legs = 1 unit = 20


Number of ducks = 20 ÷ 2 = 10



Number of dog legs = 3 units = 3 × 20 = 60


Number of dogs = 60 ÷ 4 = 15



A quick check shows that the answers don't satisfy the given conditions in the question. Although the dogs have three times as many legs as the ducks, however, the number of ducks is only 5 fewer than the number of dogs, which is expected to be 10.




When the why is harder than the how



The lesson here is to identify the illogical step, although seasoned problem solvers would probably not commit this kind of error or blunder. 

In general, it's one thing not to make an error, and get the correct answer; it's another thing to be able to explain someone else's error.

At a deeper level, it's sometimes conceptually harder to explain someone's flawed reasoning to an incorrect answer than just know how to solve a problem. Think of mathematical fallacies and paradoxes, where in a number of cases the illegal step is anything but obvious.




A fabricated faux solution

Alternatively, if we argued that a dog has 2 more legs than a duck, we'd then have a model that looks like this:

In this case, students would easily realize that the answers are also incorrect, when they find out that the number of dogs isn't an integer.


Two quick-and-dirty solutions

Why don't you try solving the dogs-and-ducks question on your own first, before comparing your solution with the ones I've worked out? If yours is an alternative (or elegant) solution, the mathematical brethren couldn't wait to reading it!

Method 1

Since a dog has twice as many legs as a duck, and there are 3 times as many dog legs as duck legs, there are 3/2 times as many dogs as ducks.

A model depicting the above relationship may be drawn as follows:

From the model drawing,

1 unit = 10

Number of ducks = 2 units = 2 × 10 = 20

Number of dogs = 3 units = 3 × 10 = 30

Check: Ducks: 20 × 2 = 40 legs

Dogs: 30 × 4 = 120 legs = 3 × 40 legs

Method 2

Given: Number of dogs = Number of ducks + 10

Number of duck legs = 2 × Number of ducks

Number of dog legs = 4 × Number of dogs

Given: Number of dog legs = 3 × Number of duck legs

A quick-and-dirty model representing the above information may be drawn as follows:

From the model drawing,

3 × 2 units = 6 units = 4 units + 10 + 10 + 10 + 10 

2 units = 10 + 10 + 10 + 10

1 unit = 10 + 10 = 20

1 unit + 10 = 30

Therefore, there are 20 ducks and 30 dogs.

Method 3 (Sakamoto method)

Those of you who are versed with the three-step Sakamoto method in solving word problems, may proceed as follows:

1. Grasp the relation

Let ④ represent the number of duck legs.

Duck legs           Dog legs

④                     ⑫

__________________________

Number of ducks     Number of dogs

④ ÷ 2 = ②                 ⑫ ÷ 4 = ③

                                             – 10

__________________________

1              :              1

                             + 10

2. Diagram

3. Number sentences

③ – ② = ① = 10

② = 2 × 10 = 20 (ducks)

③ = 3 × 10 = 30 (dogs)

The number of ducks is 20, and the number of dogs is 30.

Two bonus problems

If you like being challenged by similar dogs-and-ducks or chickens-and-rabbits problems, may I direct you to solving two more questions at Yan's One Minute Math Blog?

Happy Mathematical Problem Solving!


© Yan Kow Cheong, March 3, 2013