Solution:

The Lady sitting in the chair has the detector attached to the back of her blouse.

Solution:

First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0

You have successfully split the beer into two equal parts.

Solution:

If we have to measure precisely, it is impossible. Because we are asked to measure odd number of liters whereas all the jars we have can contain only even liters of water.

* Tricky question asked in ias/infosys/infoedge/ibibo interview

Solution:

Fifteen minutes can easily be measure using these two hour glasses.

Step 1: He will start both the hourglass.

Step 2: The moment the eleven minute hourglass is empty, he will invert it.

Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)

Solution:

f the last elf is taking some time to answer the question then it shall mean that the elf's before him are all wearing distinct coloured hats. However sufficient time shall be given to the last elf to give the answer.
If he views the similar coloured hats in front of him , he shall quickly tell the answer

Solution:

The probability of two numbers being relatively prime is 60%. Thus if you keep playing the bet, you will hypothetically make 20 cents on every play.

Thus, you should accept the bet.

Solution:

Difference 1
Toy plane direction is different.

Difference 2
On the clock, number 1 & 2 is missing.

Difference 3
The colors of lamp's light is changed.

Difference 4
The ribbon on the gift is missing.

Difference 5
Duck is added in the gift section.

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:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.