Using models to solve maths problems part 2

It appears that there are quite a few parents interested in the model method. Lilian posted a few problem sums on her blog which she says she had problems drawing the models for. Her son Brian actually got all the problems right, but not using the model method. That's ok - for PSLE, the examiners don't care what method you use, as long as it's correctly applied and you get the final answer.

I tried the sums and got all the answers within half an hour using the model method. I love the model method because it appeals to my love for problem-solving and helps me understand the concepts behind the sums. I hate learning formulas and blindly following them without understanding why, so algebra puts me off. But one boy in Lesley-Anne's class loves algebra and uses it instead of the model method, so hey, different strokes for different folks!

Here are the sums that Lilian posted. I'll explain step by step how I solved them using the model method.

Question 1:

In April, 40% of the people who went to the museum were children. The rest were adults. The number of women was 3/4 the total number of adults. The rest were men. In May, the number of children increased by 20%. The number of adults was the same as in April but the number of women became only 3/5 of the adults. Then, the number of children became 336 more than the number of women.

a) What is the ratio for the number of children to the number of men to the number of women in April?
b) How many people went to the museum in May?

First, I drew the model for April (on the right). Since 40% is the same as 2/5, 2 parts of the model represent children and 3 parts represent adults.

Then I needed to cut the 3 parts adults in men and women. I find that if you need to further divide a model, it's often easier to cut in a different direction.


In this case, instead of further dividing the columns, I cut the adult portion into four equal rows (right). You can immediately see that 1 row (1/4) is men, 3 rows (3/4) are women.

Remember each unknown part of the model has to be of equal value. So I also cut the children section into 4 equal rows. From here, you can instantly get the answer to part a) just by counting the parts.

Answer for a): Ratio of children is to men is to women in April is 8:3:9

Now we move on to May. I replicated the same model but this time, instead of cutting into 4 rows each, I cut into 5 rows each because I need to find out how many parts represent 3/5. So now, 3 rows are women (3/5) and 2 rows are men (2/5).

Same thing, since I cut the adults into 5 rows, I need to cut the children into 5 rows.

The number of children was increased by 20%. Since children are represented by 10 parts, 20%=2 parts. So I added 2 more parts to children (right).

There are 3 parts more children (12 parts) than women (9 parts). Since there are 336 more children than women, 3 parts=336.

336÷3=112 (this is the value for 1 part)

Since there are 27 parts altogether (count them!), 27 x 112=3,024

Answer for b): 3,024 people went to the museum in May.

(Lilian, like Brian, Lesley-Anne also got it right but used ratio to solve the problem.)

Question 2:

Brian invited some boys and girls, there are 20 more boys than girls. 3/4 of the boys and 2/3 of the girls managed to come. 19 children did not come. How many children did Brian invite?

First, I drew the basic model (right). Then we need to figure out how to cut the model into children who came and children who didn't come. Let's start with the boys. The tricky bit is that we don't know what fraction 20 boys is in relation to total number of boys, so you can't cut the model into 4 parts including the 20. But we know that 20 can be divided by 4, so what we can do is cut the unknown portion for boys into 4 and each of these unknown parts + 5 (20÷4) is 1/4 of the total number of boys.

We can also divide the girls into 3 parts (1 part didn't come, 2 parts came). But we need to ensure that each unknown part for boys and girls is equal, otherwise we can't compare them. I found the lowest common multiple of 4 and 3 which is 12, and divided both the boys and girls portion into 12 equal parts each. Now you can easily compare how many parts came and didn't come in the model below:

19 children who did not come to the party. This is represented by 7 parts + 5 children. Therefore, 7 parts = 19-5 = 14.
14÷7=2 (this is the number of children in 1 part.)

Total number of children is 2 x 24 parts + 20.
48+20=68

Answer: 68 children were invited to the party.

Lesley-Anne solved this problem using the model method.

Question 3:

In a school, there are 45 more students in Primary 5 than in Primary 4. In Primary 4, there are 18 more girls than boys. There are 12 more boys in Primary 5 than in Primary 4. How many more girls than boys are there in Primary 5?

Again, I first drew the basic model (left) then I cut the unknown portion vertically as such (right). Since there are 18 more girls than boys in p4, I marked the model accordingly (note: the unknown parts in the p4 row are equal in value, sorry for the improportionate drawing).









Since are 12 more boys in p5 than in p4, I extended the line for boys in p4 vertically across p5 (model below). Then I marked out an extra 12 boys. This means that that little piece in the p5 portion to the immediate left of 12 has a value of 6 (since that corresponding part in the p4 portion is 18). Since we know the portion representing number of p5 boys, the rest are girls.

Usually, the key to models is finding out the value of the unknown part. BUT in this sum, we're not required to know how many children there are altogether, only how many more girls than boys there are in p5. In the p5 portion, the girls and boys have one unknown part each so they cancel each other out.

We're left with all the known numbers, ie girls = 45+6 and boys = 12.
45+6-12=39.

Answer: There are 39 more girls than boys in p5. (Note: this answer is different from the one Lilian says was given in the assessment book which is 33, but I checked and re-checked and couldn't see how the model could be wrong. If anyone spots an error, let me know).

Lesley-Anne couldn't solve this problem as she couldn't draw the model.

Question 4:

A basket of 6 apples and 3 mangoes weighed 1kg 320g. After 4 apples and 2 mangoes were eaten, the basket with the remaining fruits weighed 760g only. If a mango weighs 20g less than 4 times the mass of an apple, find

a) the mass of the basket
b) the mass of the apple

First, I drew a model of the mango and apple (right). (Note: the 20g is an approximate, at this point, we don't know if it's more or less than one unknown part).

This question threw me off for a bit because of that darn basket which gives me 2 unknown values instead of 1. Then I realised that I could get rid of the basket by just using the fruit that were eaten, ie 1kg 320g (6 apples + 3 mangoes + basket) - 760g (2 apples + 1 mango + basket) = 560g (4 apples + 2 mangoes).

So I drew the model for 4 apples and 2 mangoes (sorry for REALLY bad pic!!).

From the model, we can see that if we want to convert the parts for the 2 mangoes into whole parts, we can just add 20g x 2 mangoes on both sides. So 12 equal parts (add up the ones for 2 mangoes and 4 apples) = 560g + 40g = 600g. Since all the unknown parts are equal, we can find 1 part, which is also equal to 1 apple.

600g ÷ 12 (parts) = 50

Answer for b): The mass of the apple is 50g.

We can also find the mass of one mango from the model - 50g x 4 (parts) - 20g = 200g-20g = 180g.

And since we know the mass of the mango and the apple, we can find the mass of the basket.

2 apples + 1 mango + basket = 760g
(2x50g) + 180g + basket = 760g
Basket = 760g - 100g - 180g = 480g

Answer for a): The mass of the basket is 480g.

I don't know if it's right that I got the answer for b) first and then a)! Lesley-Anne couldn't solve this question.

I'm no expert at models - I didn't consciously study the method, it's something I picked up along the way when helping Lesley-Anne with her maths. Which is why I believe that everyone can learn this, it just takes practice.