problem solving

if youve seen this before and know the answer, let it ride for a bit, then drop the answer.

You have 5 bags, each contains 5 pieces of gold. One bag contains fools gold, the other 4 bags contain real gold. Each piece of real gold weighs exactly 1 oz, each piece of fools gold weighs exactly 1.1 ozs. You have one scale. Using the scale only once how can you determine which bag contains the fools gold ?

my answer was "only weigh the bag of fools gold" and while correct, thats not the answer its looking for. ;)

Put all the bags on the scale, you should have 5,1oz, start removing bags one by one,

Lets say that bag4 has the fake gold

it should be:

TOTAL:5,1

REMOVE 1: 4,1

REMOVE 2: 3,1

PEMOVE 3: 2,1

REMOVE 4: 1 (THATS THE ONE)

REMOVE 5: O

I havent seen this one before, sorry for kilin it so early but it aint that hard:D

nope

:crazy:

why? i dont get it...i think it works fine

id like to tell you where youre incorrect, but i wouldnt want that info to help out the others. i'll explain later

okay....I have never been good at these but heres my guess.... take a piece of gold from each bag and put them all on the scale together....but remember which piece came from which bag...take them off one at a time.. when you notice the 1.1 jump in weight then.. you know you just took off thepeice of fools gold....and therefor which bag it came from? similar to tesseracts answer so i dunno bout this one...

no ma'am

Im sure it isnt the most conviniant answer but technically it would work would it not?

Aight beard, fair enough...i'll wait. In the meanwhile i'll delete the riddle i posted, i'll repost when this one's done

mathmaticsmathmaticsmathmatics.

insane construct of abstraction.

to me. only when i'm in the mood..

which is rarer than a 3$ billski.

?

HAHAHA Brown, thats the beauty...i aint a mathead either, but conceptualism can't get any better than math

you kick the guy in the nuts who owns the bags of gold, and break his scale, take off with all his shit. go home and weigh it out as many damn times as you want keeping the real gold for yourself while hustling the fake shit off at the local bodega???

qmud

First, you put the second bag on the scale next to the fourth and third bags, immediately followed by the fifth and first bags, respectively. Next, you calculate the proportionate density of each bag, in regards to its comparitive weight in relation to the search for an explanation of the curved geometry of gravity and the existence of sub-atomic particles. When this is complete, you simply synchronize the sounds of all 5 bags of gold as they sit in unison on the scale before being catapulted into a black void of subcutaneous matter from the seventh level of aphrodisiacal metamorphic calibration under the guise of faulty institutionalism perpetrated by the upper echelons of Hyperpluralism and the Power Elite.

You'll think I'm wrong, but thinking is overrated.

put the bags on the scale one by one. the first bag that makes the weight go to a ".1" or whatever is the fools gold.

that's wrong.

or you could bite all the gold. I think fools gold chips off or tastes weird or something..

id just like to say that i honestly didnt look at any replies to this. my method would be to weigh all the bags and start pullin em off until u drop 1.1 lbs. i dont know if its right or not.

Originally posted by Clever Name

First, you put the second bag on the scale next to the fourth and third bags, immediately followed by the fifth and first bags, respectively. Next, you calculate the proportionate density of each bag, in regards to its comparitive weight in relation to the search for an explanation of the curved geometry of gravity and the existence of sub-atomic particles. When this is complete, you simply synchronize the sounds of all 5 bags of gold as they sit in unison on the scale before being catapulted into a black void of subcutaneous matter from the seventh level of aphrodisiacal metamorphic calibration under the guise of faulty institutionalism perpetrated by the upper echelons of Hyperpluralism and the Power Elite.

 

You'll think I'm wrong, but thinking is overrated.

heh! mannnnn your cadence if off, all wrong... all wrong...

i thought that was pretty good, iceburg. but im far from a poet.

all your calculations have been wrong in that you can only use the scale ONCE. putting them all on and taking away one at a time is using it more than once.

the only thing i can think of that's wrong with tesseracts solution is that it doesn't account for the bag's weight. so here is my revision to tesseract's solution, and it's only major assumption is that the bags weigh the same:

put one bag on the scale. the probability this bag is the fools gold is 1/5, and it's weight will be 5.5oz + x, where x is any positive real number. the probability that it isn't fools gold is 4/5, and its weight will be 5oz + x. now, you don't know at this point what is in the first bag, since you don't know how much the bag weighs.

put another bag on the scale. there are three possibilities now. (1) neither bag was the fools gold, (2) the first bag wasn't fools gold and the second bag was, (3) the first bag was fools gold and the second wasn't.

after the second bag is on the scale, divide the new weight by 2.

if (1) is the case, then you have (5oz + x) + (5oz + x) = 2(5oz + x) = [2(5oz + x)] / 2 = 5oz + x. so when you divide by two you should have the same number as the weight of the first bag. repeat this process, incrementing the denominator by one every time (divide by 2, then 3 after adding another bag, so on). since this scenario is more probable -- P(first bag isn't fools gold) = 4/5 and P(second bag isn't either) = 3/4; P(first)*P(second) = 4/5 * 3/4 = 3/5 = 60% -- you will have to keep going until you find the fools gold.

if (2) is the case, then you have (5oz + x) + (5.5 + x) = 10.5 + 2x = (10.5 + 2x) / 2 = 5.25oz + x. so the result is larger than the initial weight [(5.25 + x > 5 + x) ==> 5.25 > 5, which, of course, is true.], and you know the second bag is fools gold. probability of this scenario is 1/5 = 20%

if (3) is the case, then you have (5.5 + x) + (5 + x) = 10.5 + 2x = (10.5 + 2x) / 2 = 5.25oz + x, so the result is smaller then the initial weight, and you know the first bag is the fools gold. the probability of this scenario is 1/5 = 20%

so in summary: if the average of the bags equals the initial reading, then all bags on the scale are gold. if the average is smaller than the initila reading, then the first bag was the fools gold. if the average is larger, then the last bag put on the scale is the fools gold.

and after typing all that up, i will say that this method is fool proof and guaranteed to work, and since the scale never tares back to 0, technically it is only used once. if "only used once" means only one reading is taken from the scale, then of course this doesn't work. but having said all that i will go ahead and guess that this isnt the answer you're looking for either.

how many pieces of fools gold does the bag have?

nevermind.

Originally posted by beardo

if youve seen this before and know the answer, let it ride for a bit, then drop the answer.

 

 

You have 5 bags, each contains 5 pieces of gold. One bag contains fools gold, the other 4 bags contain real gold. Each piece of real gold weighs exactly 1 oz, each piece of fools gold weighs exactly 1.1 ozs. You have one scale. Using the scale only once how can you determine which bag contains the fools gold ?

 

 

 

my answer was "only weigh the bag of fools gold" and while correct, thats not the answer its looking for. ;)

thats not correct because u dont know which bag has the fools gold

but if you correctly guessed the bag of fools gold, it is.

that would make the bag of fools gold half an ounce heavier than the others. u should be able to feelt that one out probably.

Originally posted by beardo

all your calculations have been wrong in that you can only use the scale ONCE. putting them all on and taking away one at a time is using it more than once.

ah, fuck. i saw that after i posted.

and feeling out a half an ounce isn't as easy as it sounds.

maybe not

glue, youre solution is so full of jargon i dont know what the hell it is, reguardless, its incorrect because you used the scale more than once.

Seems that i'm SOOOOOO correct and on point from post 1...release the knowledge Beard!!!

what if you put the bags on the scale one at a time... the one that increases the scale by 1.1 is the answer

or is that using the scale more then once too?

i do believe that is the definition of using the scale more than once, bra

the answer

It would be answered by taking

1 gold piece from bag 1

2 from bag 2

3 from bag 3

4 from bag 4

5 from bag 5

when you weigh the 20 coins together, if they weigh 20.1 oz, the fool's gold is in bag 1. if they weigh 20.2 the fools gold is in bag 2.. follow the same logic up to bag 5, Which if the fools gold was in bag 5 the total weight would be 20.5 oz.

wonk saggin