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.

Legs + heads = 184

(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.