In an earlier post, I shared about the following chickens-and-rabbits problem.

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?

Other than using a guess-and-guess strategy and an algebraic method, both of which offering little pedagogical or creative insight, let me repeat below one of the two intuitive methods I discussed then.

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 in order for the 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

The Bicycles-and-Tricycles Problem

Again, if we decided to ban any trial-and-error or algebraic method, how would you apply the intuitive method discussed above to solve a similar word problem on bicycles and tricycles?

There are 60 bicycles and tricycles altogether. The bicycles have 35 more wheels than the tricycles. How many bicycles and tricycles are there?

Go ahead and give it a try. What do you discover? Do you make any headway? In solving the bicycles-and-tricycles question, I find that there are no fewer of half a dozen methods or strategies, which could be introduced to elementary school students, three of which lend themselves easily to the model, or bar, method, excluding the Sakamoto method.

© Yan Kow Cheong, March 4, 2014.