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# Weighing I.

### #11

Posted 23 November 2007 - 09:55 AM

'You have one opportunity to use a balance that will return the accurate weight (in number of g).'

I was confused and thought you meant scale (as balance in English conveys a sense of being on a scale that is even...). This has confused me a minor bit in the past as when you state scale, I envision something returning numbers (rather than what you refer to as a balance)...

For amusement, I spent some time trying to figure out how you could put x bags (or coins) on one side vs another side to determine which single bag was the heavier in one go!

Thanks!

### #12

Posted 24 November 2007 - 12:20 PM

Rookie, I love your puzzles. But a note to help me read these better - when you say balance for this one (or similar ones in the future), can you state it as:

'You have one opportunity to use a balance that will return the accurate weight (in number of g).'

I was confused and thought you meant scale (as balance in English conveys a sense of being on a scale that is even...). This has confused me a minor bit in the past as when you state scale, I envision something returning numbers (rather than what you refer to as a balance)...

For amusement, I spent some time trying to figure out how you could put x bags (or coins) on one side vs another side to determine which single bag was the heavier in one go!

Thanks!

for this particular puzzle, I used words "accurate weighing-machine" and thought that it would be clear that it returns numbers ... for some other puzzles I used words "pair of scales" where I meant two panes that can be brought to balance

I see your point ... I have replaced "balancing" with 'weighing"

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### #13

Posted 27 November 2007 - 05:13 PM

If you stick with Balancing, simply place one bag on the left side of the balance, then compare it to each of the remaining bags on the right side of the balance.

If the bag on the left is is heavier than any of the one compared to then it is fake and any that balanced evenly are also fake.

If the bag on the left is lighter than any of the ones compared to then is is real and any that balanced evenly are also real.

This solution answers both having only one fake bag as well as having multiple fake bags.

### #14

Posted 28 November 2007 - 03:34 PM

"

Your solution doesn't properly handle this as if you don't know how many bags there are, you can't possibly know which solution to use. And since you can only balance/weigh only once you can't try multiple solutions.

EDIT to my above post, change balancing to balancing each bag/coin only once.

Take 9 coins from the first bag, balance the first coin to one coin from the first of the remaining 9 bags.

balance the second coin to one coin from the second of the remaining 9 bags. etc..

This is the only solution that works.

### #15

Posted 28 November 2007 - 03:39 PM

I think your solution is overly complicated.

If you stick with Balancing, simply place one bag on the left side of the balance, then compare it to each of the remaining bags on the right side of the balance.

If the bag on the left is is heavier than any of the one compared to then it is fake and any that balanced evenly are also fake.

If the bag on the left is lighter than any of the ones compared to then is is real and any that balanced evenly are also real.

This solution answers both having only one fake bag as well as having multiple fake bags.

the requested solution allows only one weighing (which is not the case of your procedure)

btw, you may use weighing machine (eg. a digital one with numbers) - so no pair of scales (that resolves your second query as well)

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### #16

Posted 30 November 2007 - 10:38 PM

I must be missing something as I don't see how if you don't know how many bags are fake, the proposed solution works, can someone explain?

### #17

Posted 30 November 2007 - 11:15 PM

try it ...

for 1 bag - imagine which bag is fake (eg. the third one) and you will see difference of 3 grams on the weighing machine (so the total weight won't be 1+2+3+4+5+6+7+8+9+10=

**55**... it will be 1+2+3.3+4+5+6+7+8+9=

**55.3**... the .3 indicates that the third bag contains forgeries)

for more than 1 bag - every combination of differences must be unique (so you can not achieve difference of X grams by more than 1 combination - eg. first and third bag and no other 2 or Y bags)

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### #18

Posted 10 January 2008 - 07:23 PM

You are given one opportunity to use the balance. Your method requires 1 or more tries.

No, taking things off the scale doesn't require more tries, you have to remove the bags anyhow so why not use it to help with the answer

### #19

Posted 08 February 2008 - 03:12 AM

Weighing I.- Back to the Water and Weighing Puzzles

Imagine you have 10 bags full of coins, in each bag 1000 coins. In one bag, there are all coins forgeries. The original coin is 1 gram light, forgery is 1.1 gram. Balancing (Edit: Weighing) just once on an accurate weighing-machine, how can you identify the bag with forgeries? And what if you didn't know how many bags contain forgeries?Spoiler for Solution

Wouldn't the bag of forgeries weigh 100 coins more than the bag originals (1000 x 0.1 = 100)

### #20

Posted 08 February 2008 - 10:55 AM

yes it would ... but the question is how would you find that 1 bag if you could use an accurate weighing machine just onceWouldn't the bag of forgeries weigh 100 coins more than the bag originals (1000 x 0.1 = 100)

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