Revisiting maths models

I haven't written a maths post in ages, since Lesley-Anne finished her PSLE, actually!

However, this year, I found that Andre frequently had difficulty with problem sums that featured two different variables in numbers, eg. number of coins vs value of coins. Somehow, he couldn't grasp the method that is commonly taught in school and used in assessment books. I guess it doesn't help that these sorts of sums tend to look very complicated and that gives him a mental block.

After some experimentation, I found a way to teach him using the model method and I'm happy to say, it really works! A similar question came out in his mid-year exam and he could solve it. Ironically, the teacher didn't understand the method and put a question mark next to it, which annoyed me a little. As long as he could arrive at the answer, I thought it shouldn't matter that he didn't use her method.

Anyway, I thought I'd share it here, for the benefit of those kids who might face the same difficulty. This probably works with kids who are visual learners.

1. In a coin pouch, there are 12 fifty-cent coins more than twenty-cent coins. If the total value of the fifty-cent coins is $21 more than the total value of the twenty-cent coins, find

a) the number of fifty-cent coins in the pouch
b) the total value of coins in the pouch

As a starting point for the model, assume the number of the coins for each denomination are the SAME and draw your model based on the VALUE of the coins, This is easy cos in terms of value, 50cts will always be 5 parts and 20cts will always be 2 parts.

Next, you draw in the value of the extra coins, in this case the 12 fifty-cent coins (always, always remind them that they're looking at VALUE, not number. This is critical!)

So you find the value of 12 fifty-cent coins, ie 12 x 0.50ct = $6 and add that to the model.

Now, you're told the value of the fifty-cent coins is $21 more than the twenty-cent coins. This is represented by the portion as drawn here.

Clearly, $21 - $6 = $15 -> 3 parts, so 1 part -> $15 ÷ 3 = $5

To find the number of fifty-cent coins simply take the total value and divide it by 0.50

$5 x 2 + $21 =$31
$31 ÷ 0.50 = 62

Answer: a) There are 62 fifty-cent coins in the pouch.

Finding total value of coins in the pouch is also a cinch, just find the value of the twenty-cent coins and add it to $31.

2 x $5 = $10
$10 + $31 = $41

Answer: b) The total value of coins in the pouch is $41.


This type of question can also be in forms other than money, eg. number of animals vs number of legs, or in this next example, number of vehicles vs number of wheels.

2. In a carpark, there are motorcycles and cars. 5/7 of the wheels are the wheels of the cars. There are 12 more cars than motorcycles. How many wheels are there in the carpark?

Similar to Question 1, first assume the number of both types of vehicles are the same and draw the model based on the number of WHEELS (4 wheels per car, vs 2 wheels per motorcycle).

Next, add in the additional wheels for 12 cars, which is 12 x 4 = 48.

Now, the question states that 5/7 of the wheels are the wheels of cars, meaning 2/7 of the wheels are the wheels of the motorcycles.

Looking at the model, there are already 2 parts to the motorcycles vs 4 parts to the cars, therefore 48 has to be equivalent to 1 part.

So total number of wheels is 48 x 7 = 336.

Answer: There are 336 wheels in the carpark.

For those of you new to my blog, I'm a firm believer of using models for maths as they've helped my kids, who are both visual learners (and algebraically-challenged), tremendously. If you're interested, you can visit some of my old posts on how to use math models (click on the 'Mathematics' label under the Blog Contents column on the right).