**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

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

*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.