Repost Puzzle [SOLUTION ADDED]
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Chris Losinger wrote:
how is that division accomplished?
Say the customer wants 4 pounds of wheat. Y is now 4 pounds. So X becomes 2 pounds. You take a 2 pound weight and put that on the Right balance and weigh out 2 pounds (with error). Then you swap balances, putting the 2 pound weight on the Left balance. So you basically weigh twice, but 2 pounds each to get 4 pounds. The balance is faulty, but because you do it twice but from opposite balances, the errors cancel out.
Regards, Nish
Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
Currently working on C++/CLI in Action for Manning Publications. (*Sample chapter available online*)Nishant Sivakumar wrote:
Then you swap balances, putting the 2 pound weight on the Left balance
i don't see it. to balance 2lbs in one setup let's assume you have to add E to your pile of wheat; to get 2lbs in the other setup, you then have to subtract 2E from that same pile. the Es don't just cancel out, if you're always changing the amount on the wheat side of the balance - you're either balancing a 2lb weight against X + E or a 2lb weight against X - E, you can't balance X. unless... if the error is constant, there's no need to do any double measuring at all - you figure just figure out what the balance is off by, then offset all your measurements by that amount; ex. if it always weighs 2oz higher on the left side, just add 2oz to the right side, any time you measure anything. if it isn't constant, there isn't enough info given to solve the problem. -- modified at 18:22 Thursday 8th March, 2007
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OK, so if you're saying that the "fault" is of the second type, why not state that to begin with? And is it any different than if he weighed the whole order rather than half?
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PIEBALDconsult wrote:
And is it any different than if he weighed the whole order rather than half?
i didn't understand your question
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Neither do I. In what way is weighing half at a time different (more accurate?) from weighing the whole thing?
--| "Every tool is a hammer." |--
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Quartz... wrote:
The method is unique no doubt, but is the method fair also, to both his customers and himself ?
The answer is NO A "natural" fault in an otherwise fair balance is always proportional to the weight. CASE 1 : The case when there can be an addition error in the measurement is IFF A weight is added to one side intentionally or One of the pan has more weight. THIS can be fairly dealed by the method acquired by the grocer. but since we don't know that this is the case as also pointed out by Dan neely here[^] (only one who came close to the reasoning ) CASE 2 Error due to the beam which is related to the weight. Assume we have 1800 gms of weight in two lots of 900 gms each 1. 900 gms of order is put on the left pan and was found to be 1000 gms on first weighing 2. When we put 900gms on right pan , due to the error the left pan will need 900 x 900 ---------- = 810 gms 1000 So the total weight for which he charges becomes 1000 + 810 = 1810 = 1800 gms (original) + 10 gms (error) This method does reduce the bias of error but still its not accurate QED
Omit Needless Words - Strunk, William, Jr.
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Quartz... wrote:
Error due to the beam which is related to the weight.
I'm wondering how this is possible with a balance. I may need to experiment.
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Nishant Sivakumar wrote:
Then you swap balances, putting the 2 pound weight on the Left balance
i don't see it. to balance 2lbs in one setup let's assume you have to add E to your pile of wheat; to get 2lbs in the other setup, you then have to subtract 2E from that same pile. the Es don't just cancel out, if you're always changing the amount on the wheat side of the balance - you're either balancing a 2lb weight against X + E or a 2lb weight against X - E, you can't balance X. unless... if the error is constant, there's no need to do any double measuring at all - you figure just figure out what the balance is off by, then offset all your measurements by that amount; ex. if it always weighs 2oz higher on the left side, just add 2oz to the right side, any time you measure anything. if it isn't constant, there isn't enough info given to solve the problem. -- modified at 18:22 Thursday 8th March, 2007
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a picture? that's not a solution
image processing toolkits | batch image processing | blogging
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a picture? that's not a solution
image processing toolkits | batch image processing | blogging
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Quartz... wrote:
The method is unique no doubt, but is the method fair also, to both his customers and himself ?
The answer is NO A "natural" fault in an otherwise fair balance is always proportional to the weight. CASE 1 : The case when there can be an addition error in the measurement is IFF A weight is added to one side intentionally or One of the pan has more weight. THIS can be fairly dealed by the method acquired by the grocer. but since we don't know that this is the case as also pointed out by Dan neely here[^] (only one who came close to the reasoning ) CASE 2 Error due to the beam which is related to the weight. Assume we have 1800 gms of weight in two lots of 900 gms each 1. 900 gms of order is put on the left pan and was found to be 1000 gms on first weighing 2. When we put 900gms on right pan , due to the error the left pan will need 900 x 900 ---------- = 810 gms 1000 So the total weight for which he charges becomes 1000 + 810 = 1810 = 1800 gms (original) + 10 gms (error) This method does reduce the bias of error but still its not accurate QED
Omit Needless Words - Strunk, William, Jr.
Vista? Soapbox Videogadget here
hmm. i thought that's what i said here[^] maybe not. close enough for me, though.
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Quartz... wrote:
Error due to the beam which is related to the weight.
I'm wondering how this is possible with a balance. I may need to experiment.
PIEBALDconsult wrote:
I'm wondering how this is possible with a balance.
a flexible beam would probably do it
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hmm. i thought that's what i said here[^] maybe not. close enough for me, though.
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Neither do I. In what way is weighing half at a time different (more accurate?) from weighing the whole thing?
--| "Every tool is a hammer." |--
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yes :cool: you might be right there, I didnt see that because your reasoning was posted 24 minutes after i posted the solution
Omit Needless Words - Strunk, William, Jr.
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Quartz... wrote:
24 minutes after i posted the solution
:laugh:
image processing toolkits | batch image processing | blogging
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Quartz... wrote:
The method is unique no doubt, but is the method fair also, to both his customers and himself ?
The answer is NO A "natural" fault in an otherwise fair balance is always proportional to the weight. CASE 1 : The case when there can be an addition error in the measurement is IFF A weight is added to one side intentionally or One of the pan has more weight. THIS can be fairly dealed by the method acquired by the grocer. but since we don't know that this is the case as also pointed out by Dan neely here[^] (only one who came close to the reasoning ) CASE 2 Error due to the beam which is related to the weight. Assume we have 1800 gms of weight in two lots of 900 gms each 1. 900 gms of order is put on the left pan and was found to be 1000 gms on first weighing 2. When we put 900gms on right pan , due to the error the left pan will need 900 x 900 ---------- = 810 gms 1000 So the total weight for which he charges becomes 1000 + 810 = 1810 = 1800 gms (original) + 10 gms (error) This method does reduce the bias of error but still its not accurate QED
Omit Needless Words - Strunk, William, Jr.
Vista? Soapbox Videogadget here
Quartz... wrote:
QED
You demonstrated it? Did you take video of it? I'd like to see the balance that's faulty that way.
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Since nobody was able to solve it yesterday, i wanted to give a chance to those guys who did not tried Puzzle of the (YESTER)Day A grocer discovered his beam balance was faulty, So he started a new method for weighing customer's orders He divides the order into two halves, putting the first half in the left hand of the balance and weights in the right, then do the opposite. The method is unique no doubt, but is the method fair also, to both his customers and himself ? You can hide , you can run, but you cannot escape, Vote it down if you want to escape i mean if you think the puzzle is not worth a repost. HERE is a sample of PAN Balance[^] SOLUTION[^]
Omit Needless Words - Strunk, William, Jr.
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hi all , Customer is getting extra order (as much the fault) suppose the beem balance is x% faulty (+ or -) by dividing two parts i. e 50 ( let assume) from first procedure he will give 50 + or - x/2 and from second procedure he will give 50 + or - x/2 . here i am not doing any wrong .. totally getting 'x' (I.E fault ) extra
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hi all , Customer is getting extra order (as much the fault) suppose the beem balance is x% faulty (+ or -) by dividing two parts i. e 50 ( let assume) from first procedure he will give 50 + or - x/2 and from second procedure he will give 50 + or - x/2 . here i am not doing any wrong .. totally getting 'x' (I.E fault ) extra
more clearly saying.. in second procedure u may think 50 - or + x/2 yes u'r correct but due to change in direction to opposite.. 50 - or + ( - x/2) or (+ x/2) hence 50 + or - x/2 will getting
