There are 10 coin making machines. 9 out of them make coins weighing 1 gm each. One (unknown) machine is faulty and makes coins 100 mg heavier. You have an electronic weighing machine (single pan). Can you tell which machine is faulty by using the weighing machine only once?
2007-03-14 20:26:04 · 12 answers · asked by Ankit 2 in Science & Mathematics ➔ Mathematics
Take one coin from machine #1, 2 coins from machine #2
n coins from machine #n n=1,2,3,4,5,6,7,8,9,10
Weigh your 55 coins, and find the number of mg in which the weight of 55 gm is exceeded. Divide that number by 100 to get the number of the faulty machine.
2007-03-14 20:32:47 · answer #1 · answered by Amit Y 5 · 4⤊ 1⤋
Take 1 coin from the first machine,2 coins from the second machine,3 coins from the third machine and so on.Now weigh
them on the weighing machine.If all the coins are of the same
weight ,the total weight will be 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 gm.So you can deduct which machine the faulty coins are coming from by observing the number of extra
milligrams.For example, if the total weight is 55 gm and 100
milligrams ,you can know that the faulty coins are coming from
the first machine.If it is 55 gm and 200 milligrams , the second machine is faulty and so on.
2007-03-14 22:52:23 · answer #2 · answered by Christine$hotbabe 3 · 0⤊ 0⤋
Total weight of 10 coin from 10 machines = 10 coins × 1 gm
= 10 gm
Go on adding one one coin from each machine. Go on noting the weight from the balance every time. Keep an eye the change in weight different from no of coins × 1gm differ. From this you can find out the faulty machine.
2007-03-15 03:59:54 · answer #3 · answered by Pranil 7 · 0⤊ 0⤋
Take a coin from each machine and place them on the weighing machine one by one. The reading should keep on increasing by 1 g if all the coins are original. If the reading increases by 1g and 100mg after you have placed a coin, that coin has come from the faulty machine. It would do good to label the coins with the machine numbers so that the machine can be identified.
2007-03-14 20:48:17 · answer #4 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
We will take one coin from first machine, two from the scond machine, three from third machine, four from fourth machine and so on upto the last machine i.e., ten coins from tenth machine. There will be in all 55 coins and all the 55 coins will be put on electronic machine to weigh. If all the machines would have been accurate, then their total weight would have been 550 mgs. Since, one of the machine is not weighing correctly, therefore the total weight would not be 550 mgs. If the total weight of all the 55 coins, is more by 100mgs then first machine would be faulty. Similarly, if the total weight in more by 200 mgs then the scond machine is faulty. We may summerized as under
Total weight more by(mgs)------------ Faulty machine
100 ---------------- first
200 ---------------- Scond
300 ----------------- Third
400 ---------------- fourth
500 ---------------- fifth
600 ---------------- sixth
700 ---------------- seventh
800---------------- eigth
900 --------------- nineth
1000 --------------- tenth
2007-03-15 01:18:01 · answer #5 · answered by Anonymous · 0⤊ 0⤋
No the machines 1 to 10 and take no. of coins as per the machine no i.e from mc 1 1 coin and so on
u ll be having 55 coins and it should be of weight 55 gm. Now take the actual weight, it it exceeds 100 mg it is from 1st m/c and 200 mg it is from 2 mc etc.,
2007-03-14 20:39:04 · answer #6 · answered by tdrajagopal 6 · 0⤊ 0⤋
Take 1 coin from machine #1, 2 from #2, etc.
Weight will be 55 gram + x*100mg
x will be faulty machine
2007-03-14 20:34:01 · answer #7 · answered by blighmaster 3 · 2⤊ 0⤋
three times (inspite of the indisputable fact that i ought to be incorrect) Take 22 of the money, and then weigh 11 and 11. one in each and every of two issues can take place: a million) They weigh a similar. for that reason, only weigh the relax 2 money; the heavier coin is the ordinary coin out. 2) They weigh distinctive. for that reason, take the heavier of the 11 money, take 10 money from that 11, and then weigh 5 and 5. one in each and every of two issues can take place: a) They weigh a similar. if it is the case, the eleventh coin would be the ordinary coin out. b) They weigh distinctive. if it is the case, take any 4 of the heavier pile and weigh 2 and a pair of. i) in the event that they weigh a similar, the 5th coin is the ordinary coin out. ii) in the event that they weigh distinctive, weigh the final 2 money; the heavier coin is the ordinary coin out. In each and all the situations, the heavy coin is got here across with a maximum three times.
2016-09-30 22:58:59 · answer #8 · answered by carol 4 · 0⤊ 0⤋
Take1coin from the 1st machine, 2 from the 2nd.............and 10 from the 10th. Put the 55 coins in the pan
weight of the pan would be 55g+100*n mg
or 55+0.1*n g
from the weight n i.e the machine can be found
eg if it weighs 55.3g, 3rd machine is odd
2007-03-15 06:24:47 · answer #9 · answered by Anonymous · 0⤊ 0⤋
From the 1st machine put 1 coin on the Weighing machine, from the 2nd machine put 2 coins on the weighing machine, from the 3rd machine put 3 coins on the weighing machine.......from the 10th machine put 10 coins on the weighing balance.
If the final weight is
55.1 then its 1st machine
55.2 then its 2nd machine
55.3 then its 3rd machine
..
..
..
55.9 then its 9th machine
56gms then its the 10th machine
2007-03-14 20:59:44 · answer #10 · answered by eric_john 3 · 0⤊ 0⤋