10 machines (numbered 1-10) producing coin of 10 gms. find by weighing once if any producing of wrong weight's

Question

product of every machine is kept separate and there is no chance to mix with other. there may be other ways also to find which machine is producing coins of less or more weight but tell the procedure which take only one chance of weighing.

Answer 1

You put 1 coin from machine 1, 2 coins from machine 2 .....10 coins from machine 10 together on the scales.

So if machine 7 produces coins which are e.g. 9,9 grams only the final weight will differ 0,7 grms from the calculated weight.
This method works only if you know the difference of weight of the coins the faulty machine produces. (you can't determine if machine 5 produces coins which are 0,2 grms out of weight or machine 10 produces coins that are 0,1 grms out of weight e.g.).
So if you don't know the difference in weight between a faulty coin and a correct one there is no chance to get a result with weighing 1 time only IMHO

Answer 2

If there is only one machine producing with a different weight, the problem is easy:

Put one coin from each machine on the scale. If the total weight is different from 100 gms, you know there is a machine not producing correctly.

To find which machine, make a note of the difference. Let w be the total weight on the scale. Then d = w - 100 gms denotes the difference in weight of the coin with the different weight.

Now empty the scale.

Put 1 coin from machine 1 on the scale.
Put 2 coins from machine 2 on the scale.
...
Put 9 coins from machine 9 on the scale.
Put 10 coins from machine 10 on the scale.

Let u denote the total weight of the coins on the scale. There are10+9+...+2+1 = 55 coins on the scale. Let v = 55 * 10 gms = 550 gms. Then D = u - v denotes the difference in weight between 55 coins of 10 gms and he 55 coins on the scale.

D/d = the number of the machine with the wrong weight.

Note that d and D may be negative.


If you suspect that there may be two or more machines with differing weights, you need more than one weighing. To see this, observe that if you have decided on a distribution of coins from each machine on the weight, you can always find a set of machines they'll give you a total weight of an average of 10gms per coin.

In general, if you suspect that there are at most n machines with differing weights, you need n weighings to find them.

I hope this answers your question.