Solution:

The words are:

1. LIMP
2. IDEA
3. MEAT
4. PATE

Solution:

Yes, you should switch the envelope. In the beginning when you picked up the envelope, you had a 1/3 probability of finding an offer letter in the envelope. There was 2/3 chance that the letter was there in the two envelopes on the table.

If you keep your selected envelope, you still have a 1/3 chance of finding an offer letter in that. However, since the interviewer has removed one empty envelope from the table, if you switch, you have a probability of 2/3 that the offer letter is inside that.

Solution:

The woman told him to take out one nut each from the rest of the wheels and use them.

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

I am Spice

Explanation:
Spice is hot if you remove first two alphabets then it will become "ice", which is really cold.

Solution:

Since every bacterium divides into two in two minutes and we know the fact that the dish was half filled with bacteria at 3 pm, after exactly 2 minutes, all will divide into two and will fill the petri dish. Thus, the petri dish will become full at 3:02 pm.

Solution:

Slumdog Millionare = Slum + Dog + Million + air

Solution:

Numbers of the first cube:
1, 2, 4, 0, 5 and 6.

Numbers on the second cube:
3, 1, 2, 7, 8 and 0.

Now you can pick any date and you will be able to show it using these cubes. Remember that you can always use 6 as 9 by inverting it.

This problem has another solution and we leave it for you to deduce.

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

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.