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 Posted: Fri Jan 04, 2013 8:17 pm

Joined: Tue Jul 31, 2007 9:37 am
Posts: 424
Hi

Stuck on the following, and just need help on a method more than an answer:

David's savings is four times that of Peter. When he spends £15 and Peter saves £45, both of them have the same amount of money. How much does each of them have in their savings at first?

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 Posted: Fri Jan 04, 2013 8:28 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14005
Well I said that if one spends £15 and the other saves £45 and they end up the same then this changes their difference by £60.

Difference = 3 lots of what Peter has (one has 4 times as much as the other)

so Peter must have £20.

Test:
David £80, Peter £20

then David £80 - £15 = £65
Peter £20 + £45 = £65

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 Posted: Fri Jan 04, 2013 8:45 pm

Joined: Tue Jul 31, 2007 9:37 am
Posts: 424
Thanks Guest55

3 units = £45 + £15 = £60

1 unit = £60 / 3 = £20

4 units = £20 X 4 = £80

David has £80 and Peter has £20 - both at first.

I prefer your (Guest55) explanation.

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 Posted: Fri Jan 04, 2013 8:48 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14005
Thanks! You are welcome ...

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 Posted: Fri Jan 04, 2013 8:52 pm

Joined: Tue Jul 31, 2007 9:37 am
Posts: 424
Guest55 wrote:
Thanks! You are welcome ...

The worrying thing is that it is a Year 4 question!

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 Posted: Fri Jan 04, 2013 9:00 pm

Joined: Thu Oct 18, 2012 9:41 am
Posts: 444
David's savings is four times that of Peter. When he spends £15 and Peter saves £45, both of them have the same amount of money. How much does each of them have in their savings at first?

Hi.. Would have done..

4a -15 =a + 45

So 3a =60

And a = 60/3

Very surprised it is a yr 4 question, please may I ask what source it's from?

Thanks

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 Posted: Fri Jan 04, 2013 9:28 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 14005
No need to use algebra - you should really avoid it until it's abolutely necessary.

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 Posted: Sun Jan 06, 2013 7:53 am

Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1348
Location: Birmingham
Quote:
No need to use algebra - you should really avoid it until it's abolutely necessary.

Not sure I agree with that - the Durham CEM test often contains algebra questions in the Maths section so better that they get used to putting this practice

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 Post subject: Need help on the following question.Posted: Wed Jun 12, 2013 9:30 pm

Joined: Tue Jul 31, 2007 9:37 am
Posts: 424
Need help on the workings for the following question, please.

It takes 5 minutes to fry a beef burger. One side of the beef burger takes 3 minutes to fry and the other side takes only 2 minutes. Two beef burgers can be placed on the frying pan at a time. What is the shortest time to fry all five beef burgers?

The answer is not 15 mins, it is 13 minutes.

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 Post subject: Re: Need help on the following question.Posted: Wed Jun 12, 2013 9:52 pm

Joined: Mon Aug 22, 2011 8:20 pm
Posts: 1706
Location: Warwickshire
Code:
minute | left side of pan | right side of pan
1     |         A1       |        B1
2     |         A1       |        B1
3     |         A1       |        B1
4     |         C1       |        B2
5     |         C1       |        B2
6     |         C1       |        E1
7     |         C2       |        E1
8     |         C2       |        E1
9     |         D1       |        E2
10     |         D1       |        E2
11     |         D1       |        A2
12     |         D2       |        A2
13     |         D2       |
or you could slot A2 (2nd side of burger A) in between B1 and B2 if you didn't want it to cool down too much between cooking each side!

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