maths help !!
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maths help !!
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!!
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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- Posts: 35
- Joined: Fri Oct 24, 2014 12:49 pm
Re: maths help !!
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.
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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Re: maths help !!
deleted my burblings - didn't read the question properly!!!
Re: maths help !!
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 ...
Herman's method is the way to go ...
Re: maths help !!
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!!
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!!
Re: maths help !!
Agree with Guest55, at KS2 I would be looking for it to be solved without algebra
Re: maths help !!
Algebra is totally inappropriate at whatever level for problems such as this ...
Re: maths help !!
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.
Why would you not use algebra?
I've spent ages trying to get dd to use algebra where possible.
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Re: maths help !!
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
20 / 3 = 6 remainder 2
20 / 4 = 5
So there are 20 dogs and 6 kennels
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Re: maths help !!
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).
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).