i opened the sack and took one coin from each… now since we know the weight… i divided it into 5-5 then 3-2.. and so on… i also thought what if we put all 10 coins together and start removing one coin at a time.. and when weight reduces by 9 we will get our coin

1 weighing. We take one coin from the first one, two coins from the second one and so on…we take 10 coins from the tenth bag. Now had all the coins had been of 10 grams, we would have had the total weight of 10*(1+2+3+4+5+6+7+8+9+10)= 550. Now, if the total is coming to be 549, the first bag has faulty coins. If the total is 558, then the second bag has faulty coins and so on.

4 weighings..

Think again, Girish!

regards

J

i opened the sack and took one coin from each… now since we know the weight… i divided it into 5-5 then 3-2.. and so on… i also thought what if we put all 10 coins together and start removing one coin at a time.. and when weight reduces by 9 we will get our coin

No, the latter would count as 10 weighings, since you would be recording the weight ten times.

regards

J

3 weighing

Nope, 3 is not the answer I am looking for either!

regards

J

Is 5 the answer ?

I’m afraid not, Tanvi. All will be revealed on Thursday, though ðŸ™‚

regards

J

1 weighing. We take one coin from the first one, two coins from the second one and so on…we take 10 coins from the tenth bag. Now had all the coins had been of 10 grams, we would have had the total weight of 10*(1+2+3+4+5+6+7+8+9+10)= 550. Now, if the total is coming to be 549, the first bag has faulty coins. If the total is 558, then the second bag has faulty coins and so on.

Haan jee!

J