Monday, May 23, 2011

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).


Lilian said...

Kowtow to my model math GURU! You are just amazing.

My model-math-challenged brain found it tough-going processing this, but I kind of get it now. Will let the boys see this, how elegant and beautiful math can be!

Write a book already!

monlim said...

Lilian: I don't think my methods are MOE-accredited, hehe. And as I always say, if you can do algebra, no need models lah! It's quicker and less cumbersome. This is just an alternative solution for kids like mine who cannot grasp the conceptual aspects of algebra.

In fact, I would love to have Brian teach Andre algebra!! Save me some headaches...

MrsCheng said...

I tried the method on a P3 problem but got stuck. Can it be used on the foll problem?
Bob drew 8 ants and spiders. There were 54 legs in total. How may spiders are there?

monlim said...

Mrs Cheng: No, can't use this method with this type of question cos you don't know how many more/less of one number there is.

For this type of question, there are 2 methods. First, trial and error, which for 8 insects, it's probably quicker. The second is to assume all have more legs (ie spiders). 8x8 = 64. That's an excess of 10 legs (64-54). The difference between the legs of spiders and legs of ants is 2 (8-6). 10 ÷ 2 = 5. Meaning there are five ants, the rest are spiders.

MT said...

Thanks Monica for sharing your method. You have enlightened me.

Previously, I only know the guess and check method, which can be quite time consuming (to solve Mrs Cheng's question).

I will share with my boys using your method. Maybe it works better with them too.

Agreed with Lilian, Maths really can be so elegant and beautiful!

MrsCheng said...

Thanks Monica.

Anonymous said...

Wow Monica you are a Math guru *clap clap*, all ready for Andre's PSLE haha... You can consider sideline as Math tutor too haha!


monlim said...

SL: Cannot lah, you never see all the tears and hair tearing on the sideline!

Jo said...

Had to stare long n hard to understand your model :P

Still can't understand why you make the assumption of same number of coins/vehicles at the start but I am only Pri 3 model standard ! However your method looks quite simple overall. I am impressed that you are able to help for upper pri when many parents give up !

Aiyoh ... Pri 3 and it is getting tough for me already.

During the exam revision I tried to make it fun with my daughter so I agreed to try problem sums with her and see who could get the solution first. Took me a long time to get the models right ... daughter was amused. The questions she couldn't do I also had to "peek" at the answers for help :)

monlim said...

Jo: Don't worry, it takes lots and lots of practice, you'll get there! Don't forget, I had a headstart with L-A.

I make the assumption of same no. of coins/vehicles because it's clearer for Andre to figure out how to draw that model first before he adds the other amounts. If not, he gets confused. Hope it works for you!

Dazziemon said...


I attempted to solve the question before looking at your solution and it turned out that I have solved it in exactly the same way which you have.
I vaguely remembered learning in primary school (18yrs ago) about using models to solve questions like this. Do they not do that in schools anymore?
Do you mind posting the way which they go about solving these kinds of questions nowadays?

Anyway it's pretty puzzling that the maths teacher put a question mark next to it. Does he/she not understand that the solution is correct?

monlim said...

The school's method is to find the difference between two values and do something with it, can't recall what. It's confusing cos it requires the child to keep converting back and forth from value to number.

In fact, the teacher deducted 3 marks from the 4-mark question cos she didn't understand the working! It was when I brought it up with her that she took a second look and agreed that it was acceptable, just that it was "not a common method" :P

bits-n-bites said...

Hi Monica, thanks for sharing your Math Models. My elder daughter is in P4 now and I was trying to use your method to solve one of her Math problem but do not know how to apply it.

The question is
Sam has a total of 25 twenty-cent coins and fifty-cent coins.
They add up to $8.30. How many twenty-cent coins and fifty-cent
coins does he have?

In this case, can I apply the models that you used? Could you kindly show me how if it's possible please?


monlim said...

Shirley: You can't use my method for this question cos you don't know how many more/less of one number there is, same as Mrs Cheng's example above.

Again, same method as the one for Mrs Cheng's: assume all 50ct. 25x50 = 1250. That's an excess of 420 (1250-830). The difference between the value of 50ct and 20ct is 30ct. 420 ÷ 30 = 14. 14 20ct coins, rest are 50ct coins.

bits-n-bites said...

OK Monica, thanks for enlightening me! I really have to spend some time to understand the use of Math models.

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