I am one question away from finishing my maths homework. But I am stuck please can anyone help. It will probably turn out to be really easy.
Here goes,

If you had nine identically sized objects(one of which had a greater mass than each of the other 8) and a balance, how could you determine which was the odd one out if you could only use the balance twice?

Thanks folks
Daisy
Year 5

Interesting question Daisy!

I have the answer, but to start you off, begin by dividing the 9 objects into three groups of three. Two to weigh and one as a spare...

Divide the objects into 3 groups of 3. Put one group of 3 on one side of the balance and another group of 3 on the other side of the balance. If the arms are unequal, then select the group that is dragging the balance down. Take 2 of the weights from that group and put one on each arm of the balance. If one arm is heavier than the other, the weight on it is the odd one out. If both are equal, then the third weight is the odd one out.

If after doing the first weighing, both arms are equal then remove the weights from the balance and weigh two of the weights from the third group. If one is heavier than the other, that is the odd one out. If both are equal then the third one is the odd one out.

(sorry for all this drivel, hope it makes sense to you - I hated questions like that when I was in junior school!)

Oh Rainbow, I resisted posting the solution in the hope that the original poster could work it out with an initial hint

tHANK YOU. I HAVE TRIED ALL WEEK. i COULD NOT SEE IT. I HAVE HAD 20 OF THESE QUESTIONS TO DO. I WAS JUST SO BORED WITH THEM. THERE IS ONE MORE THAT IAM NOT SURE ABOUT BUT I WILL SEE WHEN THEY ARE MARKED.
THANK YOU DAISY

Sorry, I really did spoil that! Somebody shoot me now!

Nah, you did fab... it wasn't that easy!

Yes but I see your point - it might have been a good idea to let Pudding work it out for herself.

I got stumped Was thinking of the top pan balances you have in chemistry and thinking..HUH??

I can think of a solution that means you wouldn't have to even use the balance you were given once...

First, you need some electronic scales. Then you measure every single one of the weights. Take out the heaviest. Tada! - you haven't even used the balance once, and you found the heaviest.

An alternative solution, if this is thought to be cheating, is to line up all nine weights, and push them all across the table with your arm. The heaviest will have suffered the greateset inertia (not sure if that is the correct terminology, but oh well), and thus will travel the shortest distance!

~Wobby~

I would have thought, that in picking up the weights a normal person would be able to tell which was heaviest....

I couldn't work it out, I'm in trouble!

Lol, yep, that would be the third solution!

Don't worry Noodelz, neither could I - it's not really Maths, more logic...

~Wobby~

Ah well, during the hoildays I usually grow a bit fatter and more stupid!

Same. I am year 7, and am supposed to be good at maths! Oh well...

Only if the heavier was >10% heavier than the others (It was an experiment we did during my degree)