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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 232

We have 5 balls of identical shape and volume, but with different weights, 10, 20, 30 40 and 50 grams.

How can we determine the weight of each ball by only 5 weighings? We must use a balance scale!

*Last edited by anna_gg (2012-01-11 00:05:49)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,933

hi anna_gg

Welcome to the forum.

Five balls and five weighings. Hhmmmmmm. That doesn't sound difficult.

Just weigh each one independently. Is there some other restiction?

Bob

*Last edited by bob bundy (2012-01-10 23:38:10)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 232

Sorry Bob, I just clarified that we are using a balance scale!!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,933

Arrgg! I had a feeling that might be the case.

Working on it.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 232

I have gone through a similar problem "Again the puzzle of balls and scales", which is listed here under the same category, but can't follow the same analysis for the 5x5 problem...

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 232

I started by weighing 2 vs 2, but we have 15 different possibilities:

1st plate 2nd plate

10+20=30, 30+40=70

10+20=30, 30+50=80

10+20=30, 40+50=90

10+30=40, 20+40=60

10+30=40, 20+50=70

10+30=40, 40+50=90

10+40=50, 20+30=50

10+40=50, 20+50=70

10+40=50, 30+50=80

20+30=50, 10+50=60

20+30=50, 40+50=90

10+50=60, 20+40=60

10+50=60, 30+40=70

20+40=60, 30+50=80

30+40=70, 20+50=70

In cases 7, 12 and 15 the scale balances,

but then what?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,481

Hi anna_gg;

You are correct about the 15 possibilities but on another forum that I am a member of there seems to be a solution. Unfortunately, I do not understand it. But at least it is possible.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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