MORE THAN ONE WAY TO SOLVE A MATHEMATICAL PROBLEM

Michael Harvey

Dip. Ed. student, Deakin University 

Michael has generated a good example of a Sweller Question.  As he only had the oportunity to test it with fellow Dip. Ed. teachers, we are not sure of the most appropriate year level to use it.  My feeling would be with senior maths students who had had a reasonable level of experience with the different possible solutions he suggests, not with students for whom the relevany maths was very new and certainly not students who had not yet experienced the Maths (Ed).  

Problem: A farmer has rabbits and hens. If there are 50 heads and 140 feet how many rabbits and hens are there?

Aim: To solve this problem using as many different methods as possible.

Rationale: This can be used for a problem solving class or as part of simultaneous equations. It develops an understanding of how different methods of solving simultaneous equations relate to each other and how to put word problems into mathematical forms.

Response: Of the fourth year B.Ed students I gave this problem too, only one had a mathematics background. He found the two mathematical solutions. The rest of the group found it a very challenging problem. They needed to ask many questions about it.

Answers:

1. Guess
2. Draw up a table, eg

   R	H	feet	heads
   0	50	100	50
   1	49	102	50
   2	48	104	50
   .	.	.	.

3. Put in a mathematical form, eg R + H = 50 (heads) 4R = 2H = 140 (feet) and then draw a graph from the equations and find the intersection (30 hens, 20 rabbits)
4. Put in a mathematical form, eg R + H = 50 (heads) 4R = 2H = 140 (feet) and then solve by substitution: R = 50 - H 4( 50 - H ) + 2H = 140 200 - 4H + 2H = 140 H = 30 R = 50 - 30 = 20
5. Put in a mathematical form, eg R + H = 50 (heads) 4R = 2H = 140 (feet) and then solve by elimination: 2R + 2H = 100 ( 2 * eqn 1 ) 4R + 2H = 140 ( eqn 2 - eqn 1) -(2R + 2H = 100) 2R = 40 R = 20 H = 50 - 20 = 30