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 Post subject: Is there an easy way to do this question?Posted: Tue May 24, 2011 6:04 pm

Joined: Wed Aug 25, 2010 2:58 pm
Posts: 491
I have been pondering on this question for a while (trying to help my DD). I finally cracked it though it took me about 30 mins! I used the method of trial and error and a little bit of logical thinking.

Here is the question: Hopefully one of you can suss out an easier way to work it out.

TYPE OF CROSSING COST (£)
Zebra Crossing 1800
Pelican Crossing 16800
Puffin Crossing 22000

The council spends £184,200 on 10 crossings. How many of each type do they buy?

Thanks

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue May 24, 2011 7:06 pm

Joined: Fri Nov 17, 2006 8:54 pm
Posts: 1768
Location: caversham
Blitz wrote:
I have been pondering on this question for a while (trying to help my DD). I finally cracked it though it took me about 30 mins! I used the method of trial and error and a little bit of logical thinking.

Here is the question: Hopefully one of you can suss out an easier way to work it out.

TYPE OF CROSSING COST (£)
A 1800
B 16800
C 22000

The council spends £184,200 on 10 crossings. How many of each type do they buy?

Thanks

I think this is asking for a graphical solution, but I played with the numbers.

A1800 + B16800 + C22000 = 184200

A + B + C = 10 ( the question should state at least one of each)

So two equations and three variables, need to find another relationship!

Looking at the numbers the total 184200 ends in 200 so must be a multiple of A and B.
800
1600
2400
3200 * A+B=4 see below
4000
4800
5600
6400
7200 * A+B=9, leaves one lot of C to make ten looks unlikely

assume A + B = 4, if A+B+C=10 then C=6

6*22000= 132000

184200 - 132000 = 52 200

This 52 200 is made up of four lots of A and B, which has to be 16800+16800+16800+1800=52000

So 6*22000 + 3*16800 + 1*1800 = 184 200

Even I'm not convinced.

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue May 24, 2011 7:22 pm

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6937
Location: East Kent
I'm in awe Steve

I just did trial and error!

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue May 24, 2011 9:34 pm

Joined: Thu Nov 08, 2007 9:57 pm
Posts: 1167
.

Last edited by Belinda on Thu Nov 01, 2012 11:16 am, edited 1 time in total.

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 Post subject: Re: Is there an easy way to do this question?Posted: Sat Jun 11, 2011 6:06 pm

Joined: Wed May 25, 2011 1:36 pm
Posts: 7
It's a very ill thought out maths question.

Even for 5/6 year old, I see ill thought out questions in the form of..

11 + ? = 19
12 - ? = 9

It's not like they are taught the method to work out the answer
(i.e. ? = 19 - 11, or ? = 12 - 9), they are (somehow) expected to KNOW. I can see why many youngsters struggle.

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 Post subject: Re: Is there an easy way to do this question?Posted: Mon Jun 13, 2011 12:38 pm

Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8210
Yep too many reasonable possibilities to work it out trial and error in the given time, particularly calculator free. I wonder if there was a line missed out of the question?

Where did this question come from?

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue Jun 14, 2011 8:15 am
Quote:
TYPE OF CROSSING COST (£)
Zebra Crossing 1800
Pelican Crossing 16800
Puffin Crossing 22000

The council spends £184,200 on 10 crossings. How many of each type do they buy?

There is a way of doing it relatively quickly.

If you observed that two of the choices have 8 in the hundreds place while the other has 0, and the total has 2 in the hundreds place, then can work out there must either be 4 or 9 distributed between Zebra and Pelican. It is obviously not 9 because it wouldn't come to such a large total if you only had 1 of the Puffins.

That is 4 between Zebra and Pelican, thus 6 of the Puffin.

184,200 - (6 x Puffins) = 52,200

Then it is a short step to realising that this must be mostly Pelicans as you wouldn't reach this number with Zebras so it is 3 Pelicans and 1 Zebra, ie. £16,800 x 3 + £1800

This took me 2 minutes, although obviously if I hadn't spotted the hundreds thing, it would have been a lot longer.

I always work on the premise that an 11 year old child would have only 5 minutes, tops, to do these and therefore there must be a relatively simply way to do it that does not involve algebra, simultaneous equations etc.. It is more a case of having a 'feel' for numbers and how they work.

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue Jun 14, 2011 9:16 am

Joined: Tue Jul 21, 2009 9:56 pm
Posts: 8210
That's the way I did it, but I still thought it took too long. If 2 minutes is OK on whatever paper that was then that's fine, but I thought maybe it came from a paper requiring greater speed than that.

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 Post subject: Re: Is there an easy way to do this question?Posted: Tue Jun 14, 2011 11:23 am
Sorry, I didn't read yours properly. I just glanced at all the multiples and thought you were doing it by trial and error rather than just illustrating your method. But you are quite right, it is the same reasoning.

With most types of 11 plus/entrance exams, you would allow different amounts of time depending on the question. Often on a test, the first half of a paper can be done fairly quickly (if the child is able) leaving more time at the end for the more complex ones.

Some of the independent papers have what I'd term scholarship questions at the end. The really able will manage the bulk of the paper in quite a quick time, allowing them the luxury of thought for the harder ones.

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 Post subject: Re: Is there an easy way to do this question?Posted: Wed Jun 15, 2011 4:24 pm

Joined: Thu Feb 17, 2011 1:34 pm
Posts: 59
I think I have seen that question before - is it from the primary maths challenge ?

I agree with the first poster that the obvious solution is to do a quick graph but that might be hard in 2 mins

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