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 Post subject: algebra
PostPosted: Mon Apr 07, 2008 10:13 pm 
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Joined: Thu Feb 21, 2008 9:48 pm
Posts: 50
Need to give the explanation to child on how to work out the following:

A, B and C had £1.68 between them. After A paid back the 3p that he had borrowed from B and C had repaid 5p she had borrowed from B, they found thhat A had twice as much as B and N had twice as much as C.

So Afterwards A had ......
B had .......
C had.......

My method 1.68/2= 84p for A
84/3=28p for C and so 56p for B

How would you work this one out and more importantly HOW would you explain this in digestabkle terms to 10 yr old child.

Thank you


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PostPosted: Tue Apr 08, 2008 6:48 am 
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Get out some money and try it


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PostPosted: Tue Apr 08, 2008 8:15 am 
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Joined: Sun Oct 29, 2006 10:05 am
Posts: 445
Location: LONDON
If I am reading this correctly the money is divided up (after loans are paid back - which I think is a bit of a red herring/irrelevent fact in this question) in a ratio of 4:2:1 (A:B:C). The money has to be divided into 7th's (4+2+1) with A receiving 4/7, B receiving 2/7 and C receiving 1/7.

1/7 of £1.68 = 24p = C
2/7 of £1.68 = 48p = B
4/7 of £1.68 = 96p = A

At least thats my interpretation. I remember explaining this sort of question to my son and he found it easier to understand in terms of fractions rather than ratio.


(As a tip get your child to add up the answers to see if they come to £1.68 and whether answer B is twice answer C and answer A twice answer B )


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 Post subject:
PostPosted: Tue Apr 08, 2008 5:57 pm 
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Posts: 515
The algebra version.

C has x pence

B has double the amount therefore

B has 2x pence

A has double the amount again therefore

A has 4x pence

Total for A,B and C is 7x

Total amount is 168 pence, therefore

7x=168 pence, therefore

x=24 pence, therefore

A=4x=96 pence
B=2x=48 pence
C=1x=24 pence

Check 96+48+24= 168 pence


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PostPosted: Tue Apr 08, 2008 9:03 pm 
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loulou and dadofkent brilliant responses.

Thanks

I personally prefer the ratio method but both methods are simple to follow.


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PostPosted: Tue Apr 08, 2008 9:54 pm 
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These methods do work but I still say you need to do practical examples with money for it to mean anything and to make sure a similar question can be tackled another time.


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PostPosted: Wed Apr 09, 2008 8:13 am 
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Joined: Mon Jun 18, 2007 2:32 pm
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Location: East Kent
I agree with Guest 55.

if you want to explain rather than just teach a method then practical examples work brilliantly, the child needs to see the nuts and bolts in order to understand the equation. Especially at 10 years old.

I often draw diagrams for questions like this, but actual money would be far better.


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PostPosted: Wed Apr 09, 2008 11:52 am 
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Posts: 625
Hi

The ratio method and the algebraic method are techniques that need to be developed to be able to complete the question provided by mh1 and for more complex questions.

Practical demonstrations are good for explaining the concepts behind the question, but a ten year old should have advanced from counting and dividing money into appropriate proportions to learning a mathematical approach.

The original starting amounts were,

A= 99p
B= 40p
C= 29p

It would be necessary to ensure that the range of coins that produces the above values could be re-arranged to provide the new values of,

A=96p
B=48p
C=24p

Using 168 1p pieces would be impractical to demonstrate the exercise to the student.

So, what would be the minimum number of coins required to demonstrate the question? and what would be the minimum range of coins required?

A= 50p, 20p, 20p, 5p, 2p, 1p, 1p
B= 20p, 20p
C= 20p, 5p, 2p, 2p

With the exchanges complete,

A= 50p, 20p, 20p, 5p, 1p
B= 20p, 20p, 5p, 2p, 1p
C= 20p, 2p, 2p

I would prefer to use more simple ratio or algebra based questions to demonstrate how to work out the question. Starting by using two values rather than three.

Explanation, Example, Exercise, Evaluation, Extension

Regards

Mike


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 Post subject:
PostPosted: Wed Apr 09, 2008 1:04 pm 
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Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11940
Mike do you hold a PGCE in Mathematics Education?


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PostPosted: Thu Apr 10, 2008 12:07 pm 
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Posts: 625
Hi Guest55

My teaching qualification is a PGCE for teaching in Further Education.

I started working in Adult Basic Education in 1990. I graduated from the University of Liverpool in 1996. I have worked in designing and delivering Mathematics courses for government training programs for young adults who have been excluded from school. I have planned, developed and delivered courses in further education for student support in application of number key skills and for BTEC courses. I have designed mathematics assesment programs for trainers delivering training programs. I set up a private tutoring business in 1996 providing tutoring for 11+ through to higher level GCSE mathematics. I have written two CDs for preparation for the Mathematics element of the 11+ tests.

No, I do not have a PGCE in Mathematics Education.

Could you explain how to set up a practical example with money for the question set?

How would you assemble the coins required for the practical example?

Would you consider it important to apply prior learning to the task?

As this site deals primarily with bright children, would it be correct to assume that this child can;

work out that 2 is double 1 and 4 is double 2

add 4+2+1=7

Divide 168 by 7 = 24

Multiply
1x24 = 24
2x24 = 48
4x24 = 96

Add to check 24+48+96=168

independently of the question set?

A practical example with money would be used at an introductory level for ratio using two values rather than three and using values lower than 168.

I think that I should be given an honarary PGCE in Mathematics Education purely on the back of the above!!!!

Regards

Mike

_________________
Mike Edwards is a co-author of The Tutors product range.


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