We have 10 bags of 1 Rupee coins. One bag contains all the defective coins, the weight of each coin in that bag is 1 gram lesser than the weight of a normal 1 Rupee coin. You have a spring balance, which tells the exact weight. After how many minimum no. of weighs you can separate the bag with defective coins.
Other variant of the same problem:
There are 10 machines in a factory. Each produces coins weighing 10 grams each. One day the factory owner cones to know that one of the machine is not functioning properly and produces coins of weight 9 grams. You have to find out the faulted machine. You ONLY have a weighing machine and you can use it only ONCE.
The minimum no. of weighs required is one.
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solution :
Let the weight of a normal 1 Rupee coin is W.
Now number the bags from 1 to 10 and take out the no. of coins equal to bag’s number from each bag, for example from 1st bag, take one coin, from 5th bag take 5 coins and so on.
Let P is the no. of the bag which contains the defective coins.
Let P is the no. of the bag which contains the defective coins.
Now put all the coins on spring balance and weigh them. Read the spring balance reading, say it is S grams.
Then S = (W grams)*(1+2+3….+10) – P*(1 gram)
P will come out to be a number between 1 to 10 and thats the required bag.
Then S = (W grams)*(1+2+3….+10) – P*(1 gram)
P will come out to be a number between 1 to 10 and thats the required bag.
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It would be 3 since you dont know what W is.
1. Measure one coin from a bag – it is either W or W-1
2. Measure one coin from another bag – it is either W or W-1
If one of them is W-1, you know which bag it is.
3. If both are W, then you know W and do what your solution mentions.