Monday, October 20, 2008

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.

28 comments:

Lilian said...

This is beautiful!! I'm speechless. You should write your own math assessment books. Those books I have don't present the answers like you do dammit. They don't cut it a different direction like you do here. I'm sooooo bloody impressed, pardon my french, but truly, I'm feeling extremely inadequate now.

Thank you thank you thank you, I'll be posting up more when I can. Can I send Brian over to you in December for intensive model tuition? *kowtow*

monlim said...

Maybe I shd write an installment for Murderous Maths on models, hehe...

No lah, like I said, I'm no math whiz (not trying to be modest, I was the despair of my Maths C teacher cos I just couldn't grasp the abstract concepts). It's just from constantly trying to help Lesley-Anne with the problems that I honed the skills. You haven't done enough of it that's why you can't do it. I'm sure with your math aptitude, you'll get it with some practice.

More worrying is that Lesley-Anne still hasn't grasped the more complex and trickier ones. How? I can't take PSLE for her leh...

Anonymous said...

Slightly unrelated but useful for kids at lower Pri:--

Just saw this on my kid's DVD (don't know if it's already familiar with you'all?)

9x table at your fingertips:--

Qn1: What is 9x3?

Put up your 10 fingers.

||||| |||||

Count to 3 fr left & hold down 3rd finger.

||_|| |||||

2......7

Voila! Answer is 27.



Qn2: What is 9x6?

Count to 6 fr left & hold down 6th finger.

||||| _||||

5.........4

Voila! Answer is 54.


Cool trick?

YY.

Last nite quite xiong you know, did model method until got insomnia... Brain too revved up!

YY.

monlim said...

Yep, Andre learnt that 9 times table trick from his school teacher :)

Anonymous said...

Anymore tricks for rest of times-table? I'm really interested to find out! Did L-A learnt it during lower Pri too?

First time hubby knew of this trick as well. I wonder if my girl learnt it from 'mental-math' when she was young... must get boy to 'test' her when he gets back from school...

Trying to get kid to memorize times-table at the moment.

YY.

Lilian said...

Aiyoh YY, I learnt that when I was a kid lah hahahahah! That's er 15 years ago? hee...

In Changi Airport now, just had wantan mee at T3. Mon, on my plane ride back to Sg, I tried some questions on my own (Brian's not with me, I'm back with Sean first) and still can't do it. Those that I could took me lots of time. Hope you can post more examples, even those you're doing with Andre, cos I think our basics may have been screwed up by some of the answers given in assessment books. I think we need to be "Unschooled" in our models approach.

Anonymous said...

You don't have to use models for every question. Most students, including my son and your daughter and Brian, use the ratio method for the museum question, as it was obviously a question on ratio. Faster and less time consuming than drawing the models, and both parts of the question are related.

MD

monlim said...

MD: thanks for your comment! Yes, I know you don't have to use models for every question. I was just trying to demonstrate that it could be done using models, in response to Lilian's question. Incidentally for Q1, it took me a shorter time to draw the model than for my daughter to use ratio :)

monlim said...

YY: Yes! There is a wonderful trick for the 7,8,9 times-tables - (the toughies!)

Face your palms towards you, fingers pointing towards each other. Ring finger is 7, middle finger is 8, index finger is 9. Say you want to find out 7x7. Make the 2 ring fingers touch each other. Count those 2 fingers and all others below (4) - this is the first digit. Take all fingers above ring finger on one hand (3) multiplied by fingers above ring finger on other hand (3), you get 9 which is 2nd digit. Answer: 49!

Anonymous said...

I saw your age-related question, and it gives me great delight to present my very own - which often stumps students and adults alike (usually for a few minutes...:))

Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?

MD

monlim said...

MD: That's not maths question lah, it's English! 24?

Anonymous said...

It's Maths. and your answer is wrong (Pat is already 24 years old). Hint, hint - this one got to use algebra.

Seriously this was a P6 question. Below is a P4 question.

Postman: How are your 3 daughters, Mr Lim?
Mr Lim: They are fine. Thank you.
Postman: How old are they now?
Mr Lim: The product of their 3 ages is 36. The sum of their ages is the same as my house number.
Postman: Err.. I can’t figure it out.
Mr LIm: Well, my eldest daughter is waiting for her letter now.
Postman: Oh, I got it. Thanks

Can you get it?

Anonymous said...

monlim:

Wow! Cool but dunno if my boy will get it all confused... Oh I get it, it only works if both numbers are 7 or above, right? For e.g. it's not meant for 7 x 3.

So he'd better learn all the multiples for numbers 6 & below.

Anywhere on the internet where I can learn about these tricks?

Lil:"That's er 15 years ago? hee..."

Oh shuurrree.... Makes me feel so old like dat. :)

YY.

monlim said...

Chris is 18?

Postman question seems to have many possible answers? Not sure what the house number has got anything to do with it (since can be anything), unless I'm missing something. 2x3x6? 3x3x4? 1x6x6? 2x2x9? etc

monlim said...

YY: yes, only for 7,8,9. But these are usually the ones the kiddos have problems remembering :)

Think I may post a quick one on this. Guess where my kids learnt this trick... from our domestic helper!!!! Maybe it's a Philippino trick, haha!

Anonymous said...

Clap! Clap!

Let me show the "model" algebra method:

WAS IS
Pat X 24
Chris 12 X

We can then clearly see that:
24 - X = X - 12 ⇒ X = 18 ;)

For the postman question, perhaps by listing down the various combinations, you can see the solution.

Anonymous said...

My P3 boy was also taught this trick by his maths teacher. :) Then he went on to teach his P6 brother, who wanted to know out of curiosity. Of course, he didn't know his older brother knew how to multiply two digit figures by two digit figures mentally when he was only 5. :)

monlim said...

err... actually, I just did trial and error, I suck at algebra.

Anonymous said...

""Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?""

Ans is 18.

Can use model method also!

YY.

Lilian said...

MD your second question is more like a brain bender leh...not math! I know the answer cos I google lah haha. No one who hasn't come across this question & its solution will ever be able to solve it!

Anonymous said...

aiyah, this was indeed a maths question given to my eldest when he was in P4. Anyway, you can check out Nanyang Maths Prelim 2007, Q40. Seems like a brain bender, but it's in it as a maths question.

This year PSLE maths' brain bender was the millipede question, the very last question of the paper (but strangely only 4 marks; usually the last few questions carry 5 marks)

monlim said...

yah lah, but you have to read into each statement, like the postman couldn't figure out the answer because there were several combinations... maybe he's just a dumb postman?????

Yve said...

Interesting post. I had to learn how to use models as well (and cheat at the back by using algebra!) to help with the kids' homework. Here's my post about a question that we got from one of the top schools' paper. Primary 3.
http://kembangankids.blogspot.com/2008/10/what-we-did-last-night.html

tianzhu said...

Hi
I think 39 is correct.
http://farm4.static.flickr.com/3300/3224896115_d22ea64109_o.jpg

tianzhu said...

Hi
The earlier link was incomplete after posting.The revised link is http://farm4.static.flickr.com/
3300/
3224896115_d22ea64109_o.jpg

Anonymous said...

Hi Mon, sorry to disturb you. I need to retrieve this question,

{"Pat is twice as old as Chris was when Pat was as old as Chris is now. If Pat is 24 now, how old is Chris now?""

Ans is 18.}

I was trying to do this sum yesterday but the english got into me, I could not figure out. My older gal managed to understand the question on her own and got the ans 18. She tried to explain to me, but still, I catch no ball....anyway, as long as she understand what she is doing, I no need to squeeze my brain juice for this. But, prior to letting her try out the sum, I email one of my galfriend, her husband did this sum and got 16 as the answer, which after working out the model, Chris could possibly be 16.

Eg,

Was
Pat [][]=16 (twice the age)
Chris [] =8

Now
Pat [][][] = 24
Chris [][] = 16

Is this correct too?

Many tks.

Chris

monlim said...

Chris: 16 cannot be right because of the first bit - Pat is 24 now and Pat IS twice as old as Chris was, so Chris was 12. At the point when Chris was 12, the age that Pat was and Chris is now is the same. This number is exactly between 12 & 24 (since Pat is always the same number of years older than Chris), hence 18.

Hope that makes sense! I used this logic to work it out, didn't use algebra or models :P

Anonymous said...

Mon, thanks so much for your reply. After reading & reading...& reading your reply...I sought of getting it *so embarrassing :p*. Now can show off to my friend's hb.....heehee

Chris

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