In Singapore, the chickens-and-rabbits question was in vogue in the late nineties, when the Ministry of Education then wanted teachers to formally teach problem-solving strategies (or heuristics, as we commonly call them here). Two common methods of solution favored by local teachers are “guess and check” (for younger students) and “make a supposition.” And in recent years, as Sakamoto math strategies gain currency in more local and regional schools, we’ve been blessed with no fewer than three other methods of solution to solve this type of problems.

A Grade 5 Contest Problem

In math contests and competitions, it’s not uncommon to witness some variations of the chickens-and-rabbits problem, which often pose much difficulty even to students, who are fluent in the Singapore model method. Let’s look at a grade 5 chickens-and-rabbits question, with a slight twist.

There are 100 chickens and rabbits altogether. The chickens have 80 more legs than the rabbits. How many chickens and how many rabbits are there?

Give it a try first, before comparing your solution(s) with the ones I’ve exemplified below.

Method 1

Since the chickens have 80 more legs than the rabbits, this represents 80 ÷ 2 = 40 chickens.

Among the remaining 100 – 40 = 60 chickens and rabbits, the number of chicken legs must be equal to the number of rabbit legs.

Since a rabbit has twice as many legs as a chicken, the number of chickens must be twice the number of rabbits for both their total number of legs to be equal.

From the model drawing,

3 units = 100 − 40 = 60
1 unit = 60 ÷ 3 = 20

Number of rabbits = 1 unit = 20
Number of chickens = 2 units + 40 = 2 × 20 + 40 = 80

A check shows that the answers do satisfy the conditions of the question.

Method 2

The equations resulting from the models for Methods 1 and 2 are the same, but conceptually this method is slightly different from the previous one.

The bar representing the number of chickens must be half the length of the bar representing the number of chicken legs. The bar representing the number of rabbits must be one quarter the length of the bar representing the number of rabbit legs.

From the model drawing,

3 units = 100 – 40 = 60
1 unit = 60 ÷ 3 = 20
2 units + 40 = 2 × 20 + 40 = 80

Therefore, the number of rabbits is 20, and the number of chickens is 80.

Let me leave you with another fertile chickens-and-legs problem, which should challenge most grade 5 or 6 students, not to say, their teachers and parents.

Mr. Yan has almost twice as many chickens as cows. The total number of legs and heads is 184. How many cows are there?

Could you use the bar method, or the Sakamoto method, to solve it?

© Yan Kow Cheong, March 3, 2013.