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PostPosted: Sun Feb 03, 2013 1:58 pm 
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Joined: Fri Feb 20, 2009 2:05 pm
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Hi was just doing some of the Haberdashers sample papers with my DS. We both got stuck on one of the Questions:

The list below shows the prices at greasy Joe’s roadside snack bar. Four friends,Peter, Jon, Simon and Duncan decide to stop by for a meal on their way home from work one day.


A Portion of Chips £1.20
Burger £0.99
Meat Pie £2.20
Pasty £1.90
Sausage £0.45
Cup of Tea £0.55
Cup of Coffee £0.65
Can of Day-Glo Sparkling Drink £0.35

Simon generously orders for both himself and his friend Duncan.He orders four items in total and receives £44.40 change from a £50 note. What did he order?

This seems like a trial and error question however it would take lots of time to go through each of the possibilities of this order. Please could somebody share a strategy with us.
Thanks in advance
Queenma


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PostPosted: Sun Feb 03, 2013 2:20 pm 
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Hi... this is how I did it... working backwards

1) He must have spent £5.60 (50-44.40)
2) Took away the largest amount first being 2.20

5.60 - 2.20 = 3.40 left
3) this is still too much to be used up in 3 goes by the smaller items so
4) took away another meat pie, so 1.20 left

5) looked for two smaller items where units added up to 0, and tens to 2
6) choosing cup of tea and cup of coffee seems to work

all in

2 meat pies at 2.20 each = 4.40
1 coffee at 0.65
1 tea at 0.55
total spent = 5.60

Change from 50
50 - 5.60 = 44.40

I am not sure how logical this is, :lol: but I didn't go through too many combinations by working backwards

completely ruled out using the 0.99 item as there isn't another item ending with a 1 in the units column IFSWIM

Not a maths teacher, (just a mum who likes maths :roll: ) so I am sure someone else will come up with a quicker method.. would also be interested to know if there are alternative methods.
SH


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PostPosted: Sun Feb 03, 2013 8:32 pm 
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Thanks SleepyHead

Yes, would be very interested in knowing if there is any trick or knack involved - or is it really trial and error ?

Queenma


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PostPosted: Sun Feb 03, 2013 10:09 pm 
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Location: Warwickshire
I'd do this as:
1) He must have spent £5.60 (£50.00-£44.40)
2) For 4 items this an average of £1.40 per item
3) Only Meat Pie £2.20 and Pasty £1.90 are above £1.40 so there must be at least 1 of these.
4) There can't be more than 2 Pies/Pasties as 3 x £1.90 is £5.70 which is more than is spent in total.
5) So combinations of more expensive items are:
2 pies = £4.40 - leaves £1.20 to account for
2 pasties = ...
1 pie and 1 pastie = ...
1 pie = ...
1 pastie = ...
6) The £1.20 left with 2 pies can be easily seen as a possible combination of Cup of Tea £0.55 + Cup of Coffee £0.65 so no need to go any further.


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PostPosted: Sun Feb 03, 2013 11:07 pm 
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Okanagan that's a great method. Didn't think to look at the average cost of one item.... Thanks for the insight. :)


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PostPosted: Mon Feb 04, 2013 9:05 am 
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Yes, thanks Okanagan

It is these ideas that I was looking for !

Tell me all you enlightened ones, is there somewhere these tactics are written /mentioned,some book -or do you just have to be one smart brain to figure it out ? Lots of commonsense and wisdom ?


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PostPosted: Mon Feb 04, 2013 9:18 am 
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Location: Warwickshire
No real generic how to do explanation for this one. Just aim on simplifying it as on face value it looks very complex.

So get rid of the extra detail on how much cash was handed over and how much change there was to give a "this much was spent" figure instead.

Cross out the 99p item as that can't be part of the answer.

Then look at reducing the number of combinations to a manageable level - hence identifying that the answer had to include one or two of the pies/pasties.

That leaves the rather simpler task of finding two items which add up to x (the amount spent less the cost of those 1 or 2 items) which can be done mentally.

Worded questions can almost always be made to look simpler - maybe by converting from words to a sum first, or as in this case removing extraneous detail.


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PostPosted: Sun Feb 10, 2013 12:56 am 
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QUEENMA wrote:
Yes, thanks Okanagan

It is these ideas that I was looking for !

Tell me all you enlightened ones, is there somewhere these tactics are written /mentioned,some book -or do you just have to be one smart brain to figure it out ? Lots of commonsense and wisdom ?


These types of questions are very common, but more commonly use two variables, e.g., 'A Meat Pie is £2.20, and a Pasty is £1.90. If John spends £23 on meat pies and pasties, how many of each does he buy?'

You can solve this by writing out multiples of £2.20:

1m £2.20
2m £4.40
3m £6.60
4m £8.80
5m £11
6m £13.20
7m £15.40
8m £17.60
9m £19.80
10m £22

and then deduct £1.90s from £23:

1p £21.10
2p £19.20
3p £17.30
4p £15.40
5p £13.50
6p £11.60
7p £9.70
8p £7.80
9p £5.90
10p £4
11p £2.10
12p £0.20

And see that the solution is 4 pasties and 7 meat pies.

A slight simplification of the above would be to note that the number of pasties sold must be even, since £1.90 * an odd number, when added to 20p * any number, cannot possibly add up to an even pound total.


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PostPosted: Sun Feb 10, 2013 9:23 am 
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Location: Warwickshire
twelveminus wrote:
These types of questions are very common, but more commonly use two variables, e.g., 'A Meat Pie is £2.20, and a Pasty is £1.90. If John spends £23 on meat pies and pasties, how many of each does he buy?'
When the quantities involved are too high to make writing out all the combinations a sensible strategy you could also approach this by taking the amount spent and dividing by the minimum and maximum prices to get an approximate number of items:
£23/£1.90 = 12.1 (round this one down)
£23/£2.20 = 10.45 (round this one up)
N.B. - you don't actually have to get to 12.1 or 10.45, just to determine that its 12.something (but less than 13) and 10.something (but less than 11) and then round appropriately

So 11 or 12 items altogether.

Start by assuming all of one type of item, and then adjust to get the answer:
11 pasties would be £20.90 - so need to increase spend by £2.10
Each time you replace 1 pastie with a pie the spend increases by 30p
To get to £2.10 you'd need to do this 7 times.
So 7 pies and 11-7 (i.e. 4) pasties

Had the amount of adjustment needed (£2.10) not divided equally by the price difference (30p) you could have moved on to trying the other option of 12 items.


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