Repost Puzzle [SOLUTION ADDED]
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Well, if he the scale is off by X, then one side (A) will weigh:
A + X == CounterWeightA
The other (B) will weigh:
B - X == CounterWeightB
So:
CounterWeightA + CounterWeightB = (A + X) + (B - X)
Therefore:
CounterWeightA + CounterWeightB = A + B
Fair, no? But he'll have a heck of a time balancing things if A or B is < X. Of course, he could fix it all by using dead weights to tare the balance first.
Does this math still hold if the error is
A*Xinstead ofA+X? Continuing farther does it hold forA+B*(C+X). You really need to quantify how it's erroring, otherwise there're a potentially infinite number of failure modes that would need tested.-- Rules of thumb should not be taken for the whole hand.
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Customer orders X quantity of something. Let Y = X/2 Now he measures out Y amount by putting the stuff on the Left and a weight on the Right. Assume there's an error of Z% where the balance is biased towards the Right one. So he actually weighs out (Y - Z% of Y) Now he measures out Y amount by putting the stuff on the Right and a weight on the Left. Since the error works in the customer's favor now, what's weighed out is (Y + Z% of Y) Putting the two quantities together, he gets :- (Y - Z% of Y) + (Y + Z% of Y) = 2Y = X So, the customer gets exactly the amount he wanted. No cheating at all.
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:
Let Y = X/2
how is that division accomplished?
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Nishant Sivakumar wrote:
Let Y = X/2
how is that division accomplished?
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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*) -
working. I didn't see an obvious answer and can't take half an hour to scribble on a scratch pad.
-- Rules of thumb should not be taken for the whole hand.
dan neely wrote:
working. I didn't see an obvious answer and can't take half an hour to scribble on a scratch pad. --
I believe I've answered it correctly - and if my answer is right, it's a pretty simple puzzle.
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*) -
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[^]
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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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Well, let's work through it...
- Grocer divides items, sets half aside.
- Grocer puts half items on one side of scale, weights on the other, 'till balanced. Scale indicates 10lbs.
- Grocer switches sides, adjusts weights 'till balanced, scale now indicates 11lbs.
- Grocer tells customer, "The scale is obviously biased by .5lbs, this half weights 10.5lbs"
- Grocer then proceeds to weight the rest of the items by placing them on the same side of the scale used to initially weigh the first set of items, and adding .5lbs to the result (which is 6lbs).
- Customer is charged for 17lbs.
Verdict:
- The method is unfair to the customer, as the grocer was using the items not being weighed to hide his fat thumb on the balance. Customer was then charged for 17lbs instead of the actual weight (1lbs, 11oz).
- The method is also unfair to the grocer, as it allowed his greed to become obvious to the customer, who will never again shop in his store and will report his obviously inaccurate scales to the Authorities, causing Trouble (a fair scale would have forced him to resort to a tiered pricing scheme, scamming the customer out of much more money, while letting him think the incredible effort required to calculate prices was ensuring him a better deal).
- Of course, it's even more unfair to the poor scale repair company, which was deprived of a customer and forced to lay off a young repairman.
- It's most unfair of all to the repairman's young child, who, his father deprived of money for food, was forced to milk squirrels to put milk on his cereal, which tasted just awful as a result.
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It appears that everybody is under the impression that I approve of the documentation. You probably also blame Ken Burns for supporting slavery.
--Raymond Chen on MSDN
Shog9 wrote:
It's most unfair of all to the repairman's young child, who, his father deprived of money for food, was forced to milk squirrels to put milk on his cereal, which tasted just awful as a result.
And how unfair is that to the squirrel ? poor squirrel
Omit Needless Words - Strunk, William, Jr.
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Customer orders X quantity of something. Let Y = X/2 Now he measures out Y amount by putting the stuff on the Left and a weight on the Right. Assume there's an error of Z% where the balance is biased towards the Right one. So he actually weighs out (Y - Z% of Y) Now he measures out Y amount by putting the stuff on the Right and a weight on the Left. Since the error works in the customer's favor now, what's weighed out is (Y + Z% of Y) Putting the two quantities together, he gets :- (Y - Z% of Y) + (Y + Z% of Y) = 2Y = X So, the customer gets exactly the amount he wanted. No cheating at all.
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*) -
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
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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Does this math still hold if the error is
A*Xinstead ofA+X? Continuing farther does it hold forA+B*(C+X). You really need to quantify how it's erroring, otherwise there're a potentially infinite number of failure modes that would need tested.-- Rules of thumb should not be taken for the whole hand.
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Quartz... wrote:
the question asked is Whether the method applied by the grocer is fair ?
No, since you've previously indicated that "he cannot weigh any of the halfs accurately." That constraint alone nullifies any attempt at weighing the order by 1/2, 1/4, etc.
"Approved Workmen Are Not Ashamed" - 2 Timothy 2:15
"Judge not by the eye but by the heart." - Native American Proverb
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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
Omit Needless Words - Strunk, William, Jr.
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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.
Vista? Soapbox Videogadget here
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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