Bar Model Method with Distributive Law For Chicken-Rabbit Approach

Ho Soo Thong

Abstract
This article illustrates a bar modelling  problem solving approach to a classical chicken-rabbit problem which involves distributive property situations. 

1. A Classic Chickens-Rabbits Problem

The following classic word problem , chickens and rabbits in a cage, posed in the ancient Chinese book Sunzi Suanjing :

There are chickens and rabbits in a cage.
Look at the top of the cage – there are 35 heads.
Look at the bottom of the cage – there are 94 legs.
How many chickens and how many rabbits are there in the cage?

 The problem involves the Actual situation:

There are 35 chickens and rabbits and there is a total of 94 legs

where each rabbit has 4 legs and each chicken has 2 legs. 

The strategy here is to add the Supposed situation :

There are 35 chickens and therefore a total of 35×2 = 70 legs.

Figure 1 shows the bar models for the number of chickens and rabbits and the corresponding bar model for the number of legs. 

 

 

 

 

 

 

 

Therefore we have 23 chickens and 12 rabbits.

  

2. Variants of Chicken-Rabbit Problems

Now, we simplify the presentation in the same approach to a counting problem.

Example 1

There is a total of 33 triangles and squares, and they have a total of 113 vertices.
How many triangles and how many squares are there?

Solution

Figure 2 shows the bar model for number of vertices in the Actual situation and the Supposed situation in which we suppose that the shapes are triangles. A square has 4 vertices and a triangle has 3 vertices.

Therefore, there are 19 triangles and 14 squares.

 

Next, we apply the Chickens-Rabbits problem solving problem to a simple finance problem.

 Example 2

Jennifer spent $ 24.80 for a total of 30 apples and oranges. 

An apple costs 70₡ and an orange costs 90₡.

How many apples and how many oranges did Jennifer buy?

Solution

Figure 3 shows the bar model for the amounts spent in the Actual situation and the Supposed situation in which we suppose that only apples are bought. An orange costs 20¢ more than an apple.

Therefore, Jennifer bought 11 apples and 19 oranges

 

3. A Challenging Problem

Finally, we post a complex problem which requires the chicken-rabbit problem solving approach to be applied twice.

A total of 31 teachers and students plan for a two-nights excursion.
Teachers prepare 96 kg of food and 43 kg of camping equipment.
Each teacher will carry 4 kg of food. Among the students, each boy will carry 3 kg of food together with 2 kg of equipment and each girl will carry 3 kg of food together with 1 kg of equipment.
How many boys and how many girls are there in the excursion?

References
[1] Ho Soo Thong, Ho Shuyuan, Bar Model Method for PSLE and Beyond – AceMath, 2011
[2] Ho Soo Thong, Ho Shuyuan and Leong Yu Kiang Problem Solving methods for Primary Olympiad Mathematics, AceMath, 2012.