Today we look at another kind of puzzle - weighing with a two pan balance.

*Puzzle 1: One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of balls against another i.e. you have no weight measurements)*

First of all, do we understand what a two pan balance looks like?

Here is a picture. It has two pans which will be even if the weights in them are equal.

There are various ways in which we can solve this.

We have 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins has the lighter coin.

Split these 6 coins into further two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin. So now we know that one of these three coins has the lighter coin. Now what do we do? We have 3 coins. We cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know that the pan which is higher has the lighter coin (and we would have identified our ball).

But what if both pans are balanced? The catch is that then the leftover ball would be the lighter one! So in any case, we would be able to identify the lighter coin.

Hope you understand the logic here.

Now try the next puzzle.

*Puzzle 2: One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?*

Now we can use the pan balance only twice. We have an odd number of coins so we cannot split them evenly. Recall what we did in the question above when we had an odd number of coins. We put one coin aside. What should we do here? Can we try by putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each. We can but once we have 4 coins which contain the lighter coin, we need 2 more weighings to isolate the light coin. But we have a total of 2 weighings only.

Instead, we should split the 9 coins into 3 groups of 3 coins each. Put one group aside and put the other two groups into the two pans. This will help us identify the group which has the lighter ball. If one pan rises up, it has the lighter ball. Else the group put aside has the lighter ball.

Now the question boils down to puzzle 1. We have 3 coins out of which one is lighter and we have one weighing left. We just put one coin aside and weigh the other two against each other. If one pan rises, it has the lighter coin else the coin put aside is the lighter coin!

So here we were able to identify the lighter coin in just two weighings.

Can you use the same method to answer puzzle 1 now?

We will leave you with another puzzle now:

*Puzzle 3: On a Christmas tree there were two blue, two red, and two white balls. All seemed same. However, in each colour pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls. Using a pan balance twice, identify the lighter balls.*

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