Tricky maths question
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bromleymum1
Thank you for you those two brilliant minds who worked this out. I don't know the answer you see - this was a question on my son's SATs revision math homework. I spotted it and boastfully said it was all a matter of logic and I would show him how to do it....and failed dismally! I can see how you worked it out. His teacher hasn't marked his homework yet so I don't know I don't know how she would work it out - but I'm sure you've cracked it. Thanks again!
Yes, indeed you are right. The interesting thing here is to also explain to a kid why this is incorrect:Guest55 wrote:I suspect the question is miscopied - I will see if I can find the original
The numbers that can be calulated to reach having A and B only as unknowns are:
18 15 A
17 12 B
16 24 11
This means that from the first row and the right leaning diagonal we have:
18+15+A = 16+12+A, i.e.
33+A=30+A
which clearly is impossible.
I think it both helps and is good practice to show this to an 11 year old as an example of 2 simultnaeous equations which clearly can not hold at the same time as A plus 30 can not be equal to that same A plus 33 (or [to make it more interesting] since the A's cancel out 33 is clearly not equal to 30!) Teaching a child to translate back into plain English their results allows them to clearly see whether what they have derived makes sense or not; it also plants the first seeds for proof derivation. Here we have just prooved that this is not a magic block (is that what these are called?)
Regards,
INEX
sj355
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