## Saturday, April 20, 2013

### Visualisation versus Algebra?

Here is a Primary 5 (Grade 5) word problem from Yan Kow Cheong in Singapore:
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? (http://www.singaporemathplus.net/1/post/2013/03/the-chickens-and-rabbits-problem.html ).    In ‘the West’ this sort of problem would not be attempted until the students have been taught algebra (i.e., y = mx + c, etc) so it would probably be a Grade 9 problem!   In a comment on another problem involving chickens and ducks, Kow-Cheong states that “the model heuristic is nothing but algebra in disguise. Instead of using the variable x, we use a unit, part or line to represent some unknown quantity ..[which] … allows us to use visualization to solve higher-order word problems.

The same author asks if we can use the bar method, or the Sakamoto method, to solve the following problem:  Mr. Yan has almost twice as many chickens as cows. The total number of legs and heads is 184. How many cows are there?

I must admit that my preferred technique for this sort of problem is algebraic.
Let the number of cows be k.
Let the number of chickens be h.
Then h = 2k – x, where x is a ‘slack’ variable such that 0 < x << k.
(4k + 2h) + (k + h) = 184
Substitute for h.
(4k + 2(2k – x)) + (k + (2k – x)) = 184
4k + 4k – 2x + k + 2k – 2x = 184
11k – 3x = 184
11k = 184 + 3x.
Apparently, k < 20 and x is much less than k, so x might be 1, 2, 3 or 4.
Assuming x = 1
11k = 184 + 3 = 187
k = 187/11 = 17
Then h = 2k – x = 2(17) – 1 = 33.
Note that if x = 2, 3 or 4, k cannot be an integer but we obviously expect there to be a whole number of cows so the problem is solved.  Mr Yan has 33 chickens and 17 cows.  I do look forward to seeing the Sakamoto solution.