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 Post subject: maths help !!Posted: Sat Jul 04, 2015 7:17 am

Joined: Thu Mar 17, 2011 12:58 pm
Posts: 92
A lady has some dogs and some kennels for them to sleep in.
If she puts 3 dogs in each kennel, there are 2 dogs left over. If she puts 4 dogs in each kennel, 1 kennel is left empty.
(a) How many dogs are there ?
(b) How many kennels are there ?

Many Thanks!!

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 8:04 am

Joined: Fri Oct 24, 2014 11:49 am
Posts: 35
Hi,

This is a tough question for 11 plus.

There are 2 approaches to this. Algebra and non-Algebra. The non-algebra approaches can involve some trial-and-improvement. I'll cover the algebra approach below and would welcome others' views on the non-algebra approach(es).

The algebra approach involves 2 difficult skills; firstly changing English sentences into equations; secondly solving pairs of simultaneous equations.

Here we go...

Lets say there are "d" dogs and "k" kennels.

From the sentence "If she puts 3 dogs in each kennel, there are 2 dogs left over." we can say that the number of dogs is 3 times the number of kennels plus 2. or d=3k+2

From the sentence "If she puts 4 dogs in each kennel, 1 kennel is left empty." we can say that the number of dogs is 4 times one-less-than-the-number-of-kennels. or d = 4(k-1)

So we know d=3k+2 and we know d=4(k-1). So:
3k+2=4(k-1)
expanding the brackets 3k+2=4k-4
adding 4 to both sides 3k+6=4k
subtracting 3k from both sides 6=k. The number of kennels is 6
substuting k=6 into d=3k+2 d=3*6+2
d=20. The number of dogs is 20.

The answer is 20 dogs and 6 kennels.

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 9:04 am

Joined: Fri Sep 15, 2006 8:51 am
Posts: 8113
deleted my burblings - didn't read the question properly!!!

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 9:09 am

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11940
Martin's method is totally inappropriate - the question is testing multiples. I'm not quite sure why he posted it

Herman's method is the way to go ...

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 9:18 am

Joined: Thu Mar 17, 2011 12:58 pm
Posts: 92
Thanks Martin_ Procter for the clear explanation.
Yes indeed for 11 plus it is quite hard. Algebra method is clear though.

Yes Hermanmuster, we can form two sequence,

3n+2 : 5, 8, 11,14,17,20,23
4n-4 : 0, 4, 8, 12,16,20,24

Look for the common number which is 20 and this is the 6th term in both set of sequence

so, logic guess- 6 kennels and 20 dogs.

But the problem was the rule for second sequence to figure out , 4n-4, since one cage was empty.

Many Thanks for the replies!!

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 12:27 pm

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6966
Location: East Kent
Agree with Guest55, at KS2 I would be looking for it to be solved without algebra

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 Post subject: Re: maths help !!Posted: Sat Jul 04, 2015 5:44 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11940
Algebra is totally inappropriate at whatever level for problems such as this ...

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 Post subject: Re: maths help !!Posted: Sun Jul 05, 2015 8:01 am

Joined: Sat Aug 10, 2013 11:46 am
Posts: 188
Showed the problem to dd (aged 11) and she began by writing 3k+2=4k-4.
Why would you not use algebra?
I've spent ages trying to get dd to use algebra where possible.

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 Post subject: Re: maths help !!Posted: Sun Jul 05, 2015 8:07 am

Joined: Sat Dec 03, 2011 3:14 pm
Posts: 625
this is an multiple question - what are the multiples of 4? One of them divided by 3 will give a remainder of 2 and have 1 empty kennel

20 / 3 = 6 remainder 2
20 / 4 = 5

So there are 20 dogs and 6 kennels

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 Post subject: Re: maths help !!Posted: Sun Jul 05, 2015 8:25 am

Joined: Fri Oct 24, 2014 11:49 am
Posts: 35
Deciding how much algebra to teach for entry to super-selectives and independent schools is not easy. Some 10-year-olds will lap it up; others simply won't get it. However, many of the most difficult questions in secondary entry exams can be solved easily and quickly with algebra. Some independent school exams are time-critical (30 or 40 questions in an hour or thereabouts) therefore speed is very important. If you can invest a few hours of your time in teaching a child a bit of more-difficult algebra, they can have the tools to complete tough questions quickly.

You must make a judgment based upon the individual child's area of skill - very possible if you are the parent of the child. Teaching tougher algebra is much less easy in a mixed group in a primary school (but some schools do achieve this).

In general, if a student uses an algebraic method to find the answer they can use another method to check it, time permitting (usually just substituting numbers into the original problem).

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