How to solve this question?

Nov 10, 2020
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12 landlords gift 12 lumps of gold each weighing 1kg. each to their king, one of whom used vacuum in each of his lump of gold which makes his lump of gold weigh less by 100 gms each. Now King orders PM to identify the culprit weighing once only. How?
 
Sep 30, 2020
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Well let's try to look at the problem in a logical way.
I am assuming that the minister has a standard set of weights to keep on one side of the balance.
What would the PM do if there was no constraint of weighing once ?
Of course he/she would weigh each landlord's lumps and then would identify who's the culprit. But what is the implication of this constraint ?
This constraint implies that
1.) Whatever be the technique we cannot leave any two landlord's gift lump out of the one measurement that we are allowed to make because that would lead to ambiguity.
2.) We have to devise a technique to index the measurement that we get to the culprit.
Naturally we can use different amount of lumps from each of the landlord's gifts for our measurement. This would satisfy our need of not leaving any two from the measurement and then having an index of everyone's identity in the measurement. Let me explain how we would index the measurement with the identity of landlord.
We keep 1 lump from 1st landlord's gift, 2 lumps from 2nd landlord's gift, 3 lumps from 3rd landlord's gift … 12 lumps from 12th landlord's gift. A
So if measurement weighs x*100 grams less from the expected weight of 144 lumps then xth landlord is the culprit.
Hope my answer helps !
 
Jan 1, 2020
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12 landlords gift 12 lumps of gold each weighing 1kg. each to their king, one of whom used vacuum in each of his lump of gold which makes his lump of gold weigh less by 100 gms each. Now King orders PM to identify the culprit weighing once only. How?

I sorry I have whats known as Dyscalculia which means an inability to solve maths problems -so dont know -Sorry
 
Mar 6, 2020
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well, aside from the general silliness chaman keep foisting on us. the problem is too vague. the twelve separate gifts of gold must Retain their 'identity'; or it is impossible. whichever gift weighs less than twelve kilos points to the landlord in question. the end
Yes, but if we can only weight once, how can we choose that gold piece is the question. This reminds me of a question where you must determine whether some coins are counterfeits through weight only, but I don't remember the details.
 
Nov 12, 2020
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How about using a pseudo Archimedes' solution. The total correct weight of 12 lumps of gold will displace an amount of water. Tell each of the landlords that their individual 12 lumps of gold will be weighted separately, and if the displaced water is unequal from the known amount by anyone's weightings, all of the landlords will suffer some arcane, cruel, frightful torture. On the other hand, if the short changing culprit landlord confesses, pays up the shorted amount of gold, and a hefty fine, all will be forgiven. Hence, instead of one weighing, none is needed. Just an application of modern taxation policy for the average citizen? N.B: This method won't work if the landlords are politicians or big campaign donors. In which case the average citizens will have to make up the shortfall.
 
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Mar 4, 2020
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This is a very old puzzle, and I didn't realize it needed to be explained. We can only use the scale for to measure one weight one time to solve this.

As stated, take one lump form landlord #1. Take 2 lumps from landlord #2. And so on.

This will give you 78 lumps. And if all lumps are good gold, it will be 78 kg. But if the weight is less than that, one of the landlords is cheating. The fools gold weighs .9 kg. The number of missing 100 gram units.......tells you the number of the landlord who is cheating.

Can you see and follow this reasoning?
 
Jul 2, 2020
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Put all the gold on a scale and then remove it brick by brick.

If the scale drops each time by 1kg then the gold brick removed is ok if not the gold brick removed is underweight
 
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