Solution:

I understand

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

Solution:

There is a simple logic to solve this question.
The size of the chocolate is 2 x 8. Thus, you need to have 2 x 8 = 16 pieces.
Every time you break the chocolate, you will get one extra piece.
Thus, to get 16 pieces, you must break it (16 - 1) = 15 times.

Solution:

Birbal drew another line that was larger than the pre-drawn line. Now the pre drawn line was smaller.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

I am word "Introvert"

Solution:

45 minutes

Explanation :
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.

Thus you have successfully calculated 30+15 = 45 minutes with the help of the two given ropes.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

Solution:

The code word is TIE.

Let us assume that the word is MOD. Now if you are told one character of this word, you would not be able to identify the number of vowels in the word. For example if you are told M, then the word can be MOD or AIM but they both have different number of vowels. If you are told about O, then there are two words MOD and OAT which again have different number of vowels. If you are told about D, there are two words again MOD and DUE which have different numbers of vowels again.

In such manner, all the words that have M, O or D in them can be excluded. Then only word that remains is TIE.
If you are told about T, there are two words OAT and TIE which both have two vowels.
If you are told about I, there are two words AIM and TIE which both have two vowels.
If you are told about E, there are two words DUE and TIE which both have two vowels.

Thus TIE is the code word.