Is it me or is this a RIDICULOUSLY hard maths problem for a 7 year old?

[Spidermama]

New school. He's in year three and this is the homework problem he has been given.

Keiron has three cats. Each is a different weight.
The 1st and 2nd weigh 7kg together.
The 2nd and 3rd weigh 8kg together.
The 1st and 3rd weigh 11kg together.

What is the weight of each cat?

Ok, I've just asked my 7 year old to do this. She wouldn't write it down, so did it in her head and got the correct answer. But she can't explain how she did it, so i'm no help at all! It would have taken me longer to get the answer...

Smalltown is right

Or to write that in a maths styley way...

Cat 1 = a
Cat 2 = b
Cat 3 = c

So we can express the problem as:

First statement a+b = 7
Second statement b+c = 8
Third statement a+c = 11

Which means (using first and third statements) if

a+b = 7 and a+c = 11

then c = b+4

We can then use that in second statement:

b+c = 8

becomes

b + (c+4) = 8

becomes

b + b = 8 - 4

becomes

b+b = 4

therefore b = 4 / 2

b = 2

Cat 2 is 2kg

Then you use that back in the original problem:

First statement

a+b = 7
which means

a+2 = 7
a = 7 - 2
a = 5

Cat 1 is 5kg

Second statement

b+c = 8
2+c = 8
c = 8 - 2
c = 6

Cat 3 is 6kg

No idea if this is typical for a 7 year old or not but at least he is being stretched.

Doing it properly this time..

a+b=7
b+c=8
a+c=11

Sum up all three statements.

(a+b)+(b+c)+(a+c)= 7+8+11

2a+2b+2c= 26

2(a+b+c)=26

Divide by 2.

a+b+c= 13

(a+b+c) - (a+b) = 13 - 7 (using statement 1)

So c = 6.

b+6= 8 (using statement 2) so b= 2

a+ 2 = 7 (using statement 1) so a = 5.

As Rattitude has shown, the logical method is by using simultaneous equations, which I don't think is taught until secondary school.

It is probable that a Y3 would be expected to work it out by trial and error - otherwise the teacher would have already taught simultaneous equations, which obviously isn't the case here.

This would have really upset DD1 at age 7 (Y1 or Y2).

I'll try it out on DD2 (age 7) when she gets home and let you know.

I should add DD1 really struggled with maths in KS1. So this would be a very challenging concept for her.

I'm hoping this was selected because the teacher felt this was an appropriate level for your DS spidermama and not given to every child. However - having said that I think as I kid I'd be intrigued to know how to do it - even if I did find it tricky myself.

I'll have to experiment on DD2. Interesting to see what happens.

Its a piece of cake using algebra.

However, I am not 7! There must be a child-friendly way of working it out.

I used:
a+b=7
b+c=8
a+c=11

So a+b+1 =b+c
Cancel the b as they're on both sides, leaves you with a+1=c.

Then using substitution:
a+a+1=11
So a+a=10
So a=5.

Then 5+c=11, c+6
And b+6=8, b=2.

Not too tricky, but I'm 30 years older than their target audience... :o

A simple way is:

If a & b weigh 7 kg, and b & c weigh 8 kg, then cat c must weigh 1 kg more than cat a.

If a and c weigh 11 kg, then they must weigh 5 kg and 6 kg respectively.

So if a & b weigh 7 kg, cat b must weigh 2 kg.

I don't think it is necessarily ridiculously hard for a 7 year old, it teaches problem solving and the teacher would want to see some sort of trial and error working out, showing logical problem solving.

Quite fun homework really, if that was the only question!

if you see my post above you will see how easy it is to do by trial and error without needing to use explicitt algebra.
problems such as 10 + ? = 14 are common in KS1 btw

my ten year old just did this using trial and error and she's top maths set in year 6

my 8yo just looked at it lol

I would get your son to make a chart. In the 1st column write down the possibilities, in order, for cat 1, i.e. 1 to 6. Then in column 2 write down the value that cat 2 would have to be in order to make 7 kg in total. Then in column 3 write down the value that cat 3 would have to be to make cat 2 and 3 weigh 8 kg in total. Finally look at the totals for cat 1 and 3 and find the combination that will make 11.
Hope that makes sense- can't make charts on the iPad!

For Year 3, I would expect a child to list all possibilities for the first statement:
cat 1 + cat 2
1 + 6
2 + 5
3 + 4
4 + 3
5 + 2
6 + 1

Then to list the options for cat 3 for each of the above, give statement 2 (or to separately list all the 2+ 3 combinations, but to cross-reference them to the first column at the same time or after the event IYSWIM)
1 + 6 + 2
2 + 5 + 3
3 + 4 + 4
4 + 3 + 5
5 + 2 + 6
6 + 1 + 7

Then to cross out those that don't match condition 3 (1 + 3 = 11)

Which leaves only
5 + 2 + 6

It's a fairly classic 'finding all solutions' type problem, though it is slightly more complex because you have to 'find all options' for 2 of the statements and them eliminate most of the options based on the final statement. If they have been doing 'finding all solutions' problems in school then I would expect a Year 3 child to be able to do it using the above method, though perhaps at this stage in the year it is more a 'higher end of the class' type activity. The logic is trickier than the maths IYSWIM.

Actually I better just point out that my DD is nearly 8 and just gone into Y3 so one of the 'older' ones, I doubt she would have been able to do this if she was a summer baby.

DS yr3 loves homework like this and uses basic algebra to do it.

Not saying he would get the right answer though!!

This is from a book called Mathematical Puzzles for the more Able -and your cats problem is from the Years 3 and 4 section.

I used to use them a lot when I taught year 4 for my top group extension homework.

For this one, the teacher will be looking for a trial and error method with a sensible guess to start with and a systematic recording system. She will also be seeing whether your son gives up when an answer is not immediately obvious, or whether he plugs away and tries different strategies.

It's an excellent book - it really gives the children confidence in their own problem solving strategies.

There are six different ways of making 7 - I wrote them out in two columns.
Then there are six different ways of making 8 using the 6 values in the second column.
Only one of the resulting lines adds up to 11.
Isn't it just a complicated way of doing number bonds to 7 and 8, and addition of 3 figures? My 7 yo could probably do it with guidance, but probably not alone.

y6DC pondered without writing anything down, & made educated guesses & got it right; pure trial & error stuff.

She got furious with me when I tried to explain the algebra way.

y8DC declared it "impossible!!" and ran away.

Bloomin' heck!! Hope we don't get anything like this too soon. I would be stumped without sitting down and reading through a few times.

DD is in the same year and only been doing vertical additions in the 100's carrying over and working on 100's, 10's and Units. She wouldn't stand a chance!

Marking place as I think I'm going to inflict this problem on DCs this evening.

I predict that the one in year 4 will either be totally stumped or try the scaevola/teacherwith2kids method, but am hoping the older ones will go for Rattitude's Thu 20-Sep-12 16:54:55 method.

I just googled Mathematical Puzzles for the more Able Kingscote and couldn't find it on Amazon - is it a book that parents can use or just one for teachers?

I would say that this is advanced maths for a 7 yr old.

There is often a SATs question in a similar style to this on the year 6 paper. But it is a level 5 question. I am very surprised they sent a question like that home for a 7 year old!!