Math Problem?

I think you mean this: A few logic puzzles

Rather than necro the old thread, I would like to point out that obviously you know what the proper weight is, because truth in labling laws require you put it on the wrapper.

CM

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Original post by Conner McCloud Rather than necro the old thread, I would like to point out that obviously you know what the proper weight is, because truth in labling laws require you put it on the wrapper. CM

Yeah, but I don't think the laws require such an accuracy. For example, a 100 gram chocolate bar canbe 98 grams or 102 grams, and still pass as valid. Right?

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Original post by Raduprv Yeah, but I don't think the laws require such an accuracy. For example, a 100 gram chocolate bar canbe 98 grams or 102 grams, and still pass as valid. Right?

Sure, but presumably you are still in charge of the factory. You know your target weight, which is the weight put onto the label, and the weight that 90% of your machines will be outputting.

You could also look at it like this: since we know we aren't using balances, to know that one bar is 5g heavier than the rest, you need to know how much the rest weigh.

CM

i cant beleive my thread would have caused such a pointless discussion as how heavy the standard chocolate bar is. Lets just say its 30 grams to keep you all quiet.

it's quite easy to determine the standard weight of the chocolate bar without knowing it ahead of time. Select one bar of each, and put them on the scale. Remove one bar, note the weight differential. Remove a second bar, and if the measurements are the same, that is your standard weight, continue until you find the heavy bar. If they are different, then not only have you found the bad machine, but they lighter of the two bars is the standard weight.

if you are limited to not being able to remove a bar and remeasure, then yes, it would be impossible to determine if you did not know the standard weight. As proof, the total weight of all the bars is:
n*w_s + m*0.5 = w_t. You know n, but you do not know w_s and m. You cannot solve a linear equation in two unknowns from one equation.

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Original post by capn_midnight it's quite easy to determine the standard weight of the chocolate bar without knowing it ahead of time. Select one bar of each, and put them on the scale. Remove one bar, note the weight differential. Remove a second bar, and if the measurements are the same, that is your standard weight, continue until you find the heavy bar. If they are different, then not only have you found the bad machine, but they lighter of the two bars is the standard weight. if you are limited to not being able to remove a bar and remeasure, then yes, it would be impossible to determine if you did not know the standard weight. As proof, the total weight of all the bars is: n*ws + m*0.5 = wt. You know n, but you do not know ws and m. You cannot solve a linear equation in two unknowns from one equation.

Well, yes, one measurement means one reading. Removing stuff from the scale is kind of cheating :)
And as for being in charge of the factory, that's not part of the premise :) You might be just a cotnractor that arrived that day and has the sole task to solve that problem :)