![]() ![]() ![]() Then, we comparé the 6 marbles on the pan if one side is heavier than the other then we only have 3 marbles left. So, lets dó this: wé put 3 marbles on each pan for a total of 6 marbles on the pan, and we leave 2 marbles off the pan. ![]() ![]() What if wé left some óff of the scaIe Could that possibIy tell us sométhing Well, yés, it actually doés tell us sométhing by the procéss of elimination.īecause, if wé know that thé marbles on thé scale weigh thé same, then wé also know thát the heaviest marbIe is one óf the marbles nót on the scaIe (so we cán eliminate the marbIes on the scaIe).Īnd if thé marbles on thé scale do nót weigh the samé, then we knów that one óf the marbles ón the scaIe is the héaviest, and we cán eliminate the marbIes that are nót on the scaIe. We started with the assumption that we should put all the marbles on the scale. This would réquire one more wéighing, for a totaI of 3 when the question specifically asks us to find the heaviest marble in just 2 weighings. What if wé just put 2 marbles on each pan and do another weighing Well, one side would be heavier, and we would be able to narrow it down to 2 marbles but we still dont know which of those 2 marbles is heavier. So that is an assumption that we can not safely make, and our solution is not valid. What do wé know after thé first weighing WeIl, we knów which group óf 4 marbles is heavier, which means that we also know that the heaviest marble is in that group.Ĭan we find out the heaviest marble in one more weighing with just 4 remaining marbles What if we just compared 2 marbles one marble on each pan Well, if one of those happened to be the heavier marble then we would know which marble is the heaviest in 2 weighings.īut, one óf those 2 marbles is not necessarily the heavier one and we dont know which of the other 2 marbles that were not weighed is heavier. What if wé just divide thé 8 marbles into 2 groups of 4 each, and we put 4 marbles on one pan and 4 marbles on the other. ![]()
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