Is there an easy way to do this question?

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Blitz
Posts: 874
Joined: Wed Aug 25, 2010 3:58 pm

Is there an easy way to do this question?

Post by Blitz »

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
stevew61
Posts: 1786
Joined: Fri Nov 17, 2006 9:54 pm
Location: caversham

Re: Is there an easy way to do this question?

Post by stevew61 »

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. :)
yoyo123
Posts: 8099
Joined: Mon Jun 18, 2007 3:32 pm
Location: East Kent

Re: Is there an easy way to do this question?

Post by yoyo123 »

I'm in awe Steve

I just did trial and error!
Belinda
Posts: 1167
Joined: Thu Nov 08, 2007 10:57 pm

Re: Is there an easy way to do this question?

Post by Belinda »

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Last edited by Belinda on Thu Nov 01, 2012 12:16 pm, edited 1 time in total.
Sugarbeach
Posts: 7
Joined: Wed May 25, 2011 2:36 pm

Re: Is there an easy way to do this question?

Post by Sugarbeach »

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.
mystery
Posts: 8927
Joined: Tue Jul 21, 2009 10:56 pm

Re: Is there an easy way to do this question?

Post by mystery »

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?
fm

Re: Is there an easy way to do this question?

Post by fm »

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.
mystery
Posts: 8927
Joined: Tue Jul 21, 2009 10:56 pm

Re: Is there an easy way to do this question?

Post by mystery »

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.
fm

Re: Is there an easy way to do this question?

Post by fm »

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.
R3ad1ngDad
Posts: 59
Joined: Thu Feb 17, 2011 2:34 pm

Re: Is there an easy way to do this question?

Post by R3ad1ngDad »

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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