A little Problem...
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If you are using the dual scale, not the single scale as the iamges above. In a single scale, if you do not know how much the balls weigh, then 1 weighing doesn't tell you anything. Its the 2nd weighing where you could first find a heavier ball.
But again, it does depend what type of scale we're talking about.
Unless of couse these are bowling balls that are clearly marked how much they wegh, an no weighing is necessary at all
Chris
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None.
You have a scale, so you counter weight using balls as weights.
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@unknownuser said:
so if you have a scale such as the first one i posted, you place one ball in each side.. if one happens to be the heavier ball, you'll know it right away hence 1 weighing.
That would be a total of 4 weighings.
As you place the balls alternating sides, you start weighing 1 vs 1, then 2 vs 2, then 3 vs 3, and then finally 4 vs 4 if you haven't found it yet. -
@ecuadorian said:
That would be a total of 4 weighings.
You start weighing 1 vs 1, then 2 vs 2, then 3 vs 3, and then finally 4 vs 4 if you haven't found it yet.you're talking about the maximum amount of weighings.. the question says minimum
but going that route with the dual scale to find the max amount.. i'd try
3vs3 then mono-y-mono..
you can always find the oddball using a maximum of 2 weighings with the dual scale.edit- oh wait, that's the same method you describe earlier.. so yeah, i agree with you on that
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remus, I think you need to see a doctor.
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I guess we need another problem now.
If I want to buy seven rabbits and a half, and each rabbit and a half costs one dollar and a half, how much do I have to pay?
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Remus never said the problem gave you:
Level 99
Magic 100
Luck 1,000,000For these problems when you say "minumum" it means the minimum without luck and magic.
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@ecuadorian said:
Remus never said the problem gave you:
Level 99
Magic 100
Luck 1,000,000For these problems when you say "minumum" it means the minimum without luck and magic.
right, i agree and hope that's how it is..
but - there's nothing worse than someone asking you a logic question only to find out it was really some weak play_on_words type of thing..
(well, i guess i can think of worse things but... )someone has to build up some cred with these types of questions with me or i have to try to eliminate the loop holes first..
so who has another one?
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@ecuadorian said:
I guess we need another problem now.
If I want to buy seven rabbits and a half, and each rabbit and a half costs one dollar and a half, how much do I have to pay?
$7.50
dividing by 1.5 then multiplying by 1.5 cancel each other out so it's a buck per rabbit.
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Yup. Most people start to do strange calculations with that one.
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7.5/1.5=5
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Well done people, its not a trick question: 2 is the correct answer as ecuadorian said (and the scales are the 1st type you posted jeff.)
edit: doh, put 3 instead of 2. Corrected.
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I'd have said $10.50, Miguel. You said 7 rabbits and a half...that's 7 whole rabbits plus a half rabbit. Given, that in Ecuador, you apparently sell rabbits in the odd amount of 1.5 rabbits, you'd need to buy 7 portions of them to get 7 whole rabbits...then have a whole lot of half-rabbits left over.
I always was something of a pedant. -
I agree with Eric - go and see a doctor, Remus. And when you've done so, also a psychiatrist (since who cares which ball weighs slightly more or less?)
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Who cares! what do you mean who cares?! we're addressing fundamental problems in comparative ball analysis and scalar logic
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Another one:I have two jars full of a liquid amoebas eat. These amoebas multiply by 2 every three minutes. Jar A starts with two amoebas and after two hours it's full of them. Jar B starts with just one. How much time will it take it to fill?
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2 hours 3 minutes, assuming the jars are the same size.
reasoning: after 3 minutes it will have multiplied, so you'll have 2 amoebas and be left with the same starting conditions you had in jar A.
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Anyone has another one? -
@remus said:
Well done people, its not a trick question: 3 is the correct answer as ecuadorian said (and the scales are the 1st type you posted jeff.)
No, Remus, two weighings is enough, when you start with the 3+3 method I think Jeff posted first:
First weighing: You select randomly 3+3 balls and weigh them. The next step depends on your result.
Second weighing: If the 6 balls were of equal weight you weigh the 2 balls left, the heavier is one of them. Otherwise, you choose randomly 2 balls out of the heavier set of 3 and weigh them. If they are equal, the heavy one is the one left out.
In real life, figuring this out would probably take more time than to weigh the balls in three steps
Anssi -
About the first problem with this and 12 same aspect objects (one is different weight)
You must say what is the different object and if it's less or more weight than the current!
In how many minimum weigh-in can you made that?
Whithout see result on the Net of course!
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