Thursday, October 23, 2008

Using models to solve maths problems part 3

More maths problems using models, at the request of Lilian!

These two are found in the P4 CASCO Challenging Maths assessment book. Neither Lesley-Anne nor I could solve them last year (when she was in p4). Only when I re-visited them this year could I solve them, probably due to more practice. Lesley-Anne could only get Q1 right with some guidance.

Question 1:

Jack and Ali are given a certain number of maths problems to solve. If Jack solves 3 problems and Ali solves 1 problem every minute, Jack will have 12 problems unsolved when Ali has finished solving all his problems. If Jack solves 1 problem and Ali solves 2 problems every minute, Jack will have 42 problems unsolved when Ali has finished solving all his problems.

a) How many problems were given to Ali?
b) To finish solving the problems at the same time as Ali, how many problems must Jack solve every minute if Ali solves 4 problems every minute?

I'm guessing that some of us have a mental block with this problem because it deals with time. Since time is linear and not an object, we don't know how to graphically capture it in a model. It also appears to add another variable to the problem, making it harder to pin down. But notice that the question never asks how long Jack or Ali takes to solve the problems. In other words, time is not an issue here.

In cases where you have two different models, it's often important to find the constant that is valid for both models. In this question, the constant for both scenarios is 1 minute.

So I started by drawing the basic model for the first part of the question (right). The unknown shows the number of problems solved by Jack and Ali per minute. 12 is the number of problems unsolved by Jack. Easy so far?

Now we need to figure out the model for the second part of the question. If Jack solves 1 problem and Ali solves 2 problems every minute, Jack will have 42 problems unsolved when Ali has finished solving all his problems. Some people might draw the model this way (right). Add an unknown part to Ali and the portion after Jack's one unknown part is 42. Sounds logical, right? WRONG. (I marked a big X in case it escaped you, DON'T FOLLOW THIS!!)

Here's why - if you add another part to Ali, you will need to add 3 more parts to Jack, otherwise the first model doesn't hold true anymore (ie If Jack solves 3 problems and Ali solves 1 problem every minute, Jack will have 12 problems unsolved when Ali has finished solving all his problems).

Instead of having to redraw a model by adding parts, it's easier to just cut Ali's unknown part in the original model into 2 (below). Remember one of my basic assumptions of models (mentioned in my previous post), ie every unknown part should be equal. So I also cut each of the other unknown parts into 2. Since Jack solves 1 problem when Ali solves 2 problems, the entire part after Jack's one part is 42.

5 parts + 12 = 42
5 parts = 42-12 = 30
Therefore, 1 part = 30÷5 = 6

Ali has 2 parts, so 6 x 2 = 12

Answer for a): Ali was given 12 problem sums.

Next part should be quite straightforward. If Ali solves 4 problem sums per minute, he would take (12÷4) 3 minutes to solve all his sums. Jack has 48 (42 + 1 part = 42 + 6) problem sums to solve. To finish solving all of them in 3 minutes, he needs to solve (48÷3) 16 per minute.

Answer for b): Jack has to solve 16 problem sums every minute, ie Jack's mother is a sadist.

Question 2:

There are 500 male and 200 female employees in Company A. There are 400 male and 600 female employees in Company B. Some employees are transferred from Company A to Company B. After the transfer, the number of male employees is the same as the number of female employees in Company B and the number of male employees is twice the number of female employees in Company A. How many female employees were transferred from Company A to Company B?

This is what I call a "transference" question, ie there is a change in value from one scenario to another. (By the way, all the lingo is coined by me, it's by no means the standard in the maths community! All the tips and assumptions are also mine, based on my own experience. If they don't help you, by all means chuck them.)

First, I need to draw the model depicting the number of employees in both companies before the transfer. When you need to compare models, it's usually easier to draw in comparable parts than in one block with a value. Looking at the values given, it's clear that it would be easiest to draw in 100 employee parts. So this is the initial model (right). Each part represents 100.

Now we need to do the transfer. This is an interesting problem because it's an example of a case where the PROCESS of drawing the model (and not the final model itself) helps you arrive at the answer. Sometimes, we tend to attempt to find the answer first then try to draw the model to fit the answer (come on, admit it! I know I do!) but this really defeats the purpose of the model.

We know that after the transfer, Company B had the same number of male and female employees. This means that at least 200 (2 parts) male employees were transfered from Company A. So first we move 2 parts male from A to B (right, indicated by shading). Looking at the remaining parts in Company A, we can see that the number of male employees (3 parts) is not twice that of female employees (2 parts), in other words, we need to move more people.

Now, we know that from this point, we have to move an equal number of male and female employees from Company A to B, in order for Company B to have an equal number of male and female employees. Again by looking at the model (below), we can see that if we move 100 (1 part) male and 100 (1 part) female employees from Company A, we will be left with 2 parts male and 1 part female, ie male employees twice the number of female employees. Voila! Let the picture do the talking.

Answer: 100 female employees were moved from Company A to Company B.


Lilian said...

Thanks Guru! Appreciate you beautiful drawings :)

But Brian claims Q2 is kind of Guess-and-Check, just a more logical kind of G&C.

Give me a million more years and I still wouldn't have been able to do lah...

monlim said...

no lah, just practise :) tell Brian not everyone is as brilliant as him leh... Auntie must see picture one then can get answer :P

eunice said...

Wah, I'm really not a maths person. Looking at the problem made me see stars and my mind just blanked. Monica, you are really something else (in a good way)!

Help needed said...

There are 600 children in Team A and 30% of them are boys. There are 400 children in Team B and 60% of them are boys. After some children are transferred from Team B to Team A , 40% of the children in Team A and 60% of the children in Team B are boys. How many children are transferred from Team B to Team A?

Tilted Earth said...

With the introduction of calculators in 2009, will PSLE Maths questions be tougher? Are past PSLE Maths questions from authorised publishers still relevant in the preparation of PSLE 2009?

I believe many parents and students are waiting anxiously to know the answer. Well, I guess you’ve to wait till Oct 09.

monlim said...

Tilted Earth: This is the first year calculators are introduced, so no one really knows but I'm pretty sure the questions will be tougher. Perhaps not in the sense of the way the questions will be phrased, but it that the answers won't be nice whole numbers anymore, so you can't be sure if your answer is correct. I think they'll also introduce more parts to a question since you supposedly can reach the answer more quickly. But your guess is as good as mine :)

Help: Will try to get around to your question some time! I'm pretty sure Lesley-Anne has done this type of sum before.

wo wen tian said...

Hi,how to solve this problem without using Algebra?

At first, two shops A & B had a total of 1040 sacks of rice. After Shop A has cleared 3/4 of its stock and Shop B has cleared 3/5 of its stock. Shop B now has 52 more sacks of rice than Shop A. How many sacks of rice does each shop have at first?

wo wen tian said...

If you seek an answer, set the question free. If it comes, enjoy the sweet moments.If it doesn’t, be patient and wait. One day your question will be answered.

Q1) The total number of stamps Ken and Ryan had was 3 times as many as Morgan's. Ken and Ryan each gave 1/3 of their stamps to Morgan. Morgan then found that he had 26 more stamps than Ken. Morgan also had 130 stamps more than Ryan.

(a) How many stamps did the three boys have altogether?

(b) Express the number of stamps Ken had at first as a fraction of the total number of stamps the 3 boys had. Give your answer in the simplest form.

monlim said...

Patience, my friend :) I've just published the workings to your first question. I probably won't do the detailed workings for your Q2 unless I have more requests to do so. But just checking, are the answers: a) 312 stamps b) 5/8?

wo wen tian said...

Hi Monica,

Thank you for your speedy response to Q1.

Your answers for Q2 are quick and correct. Never mind, if you have no time for detailed answer.I understand that it's time consuming. Wish I’ve not added to your stress.

I believe someone in some parts of our beautiful world will one day see this question in your popular blog. Hopefully, it’ll arouse his Maths mood and post the solution.

Another one to see if readers of your blog can offer some fresh perspectives, without algebra.

Q)During a warehouse book sale, Sally spent 62.5% of her money on 24 books and 18 pens. She also spent 25% of her remaining money on 18 files. Each pen costs 8/9 as much as the price of one book. The file costs $7.80 less than a book. Find the total cost of one book and one pen.

monlim said...

Wo: You're welcome, I know it wasn't that speedy :P

Just out of curiosity, are you just sharing challenging problems or do you want to see how the workings are done? Cos you seem to know the answers already, so I was wondering.

wo wen tian said...

Hi Monica
It’s fast enough, at least for me.The answers are given.

I think it’s good to see workings from different angles. I also hope that sharing of questions could lead to students having more exposure to different types of questions, thus having less surprises during examinations.
Best wishes

monlim said...

Oh I see. Thanks for sharing. I assume these are p6 questions? cos I don't think my daughter will be able to do them...

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