Solution:

24 strikes - One Striker every hour

156 Strikes - Strike depends on the number of hours
(3rd hour - 3 strikes, 11th hour - 11 strikes)
Total Strikes = 2(1+2+3+4+5+6+7+8+9+10+11+12) = 156.

180 Strikes - 156 from previous + 24 (one strike every hour)

228 Strikes - 156 Strikes(Strike depends on the number of hours) + 72(24*3, one every quarter except for the 0th hour)
= 156 + 72 = 228

Hence all clocks are working fine.
24 strikes - One Striker every hour

156 Strikes - Strike depends on the number of hours
(3rd hour - 3 strikes, 11th hour - 11 strikes)
Total Strikes = 2(1+2+3+4+5+6+7+8+9+10+11+12) = 156.

180 Strikes - 156 from previous + 24 (one strike every hour)

228 Strikes - 156 Strikes(Strike depends on the number of hours) + 72(24*3, one every quarter except for the 0th hour)
= 156 + 72 = 228

Hence all clocks are working fine.

Solution:

2 degree below zero

Solution:

When you were inside the shed, the murderer replaced the lock with his own one that was identical to your original lock. When you locked the store room, you did not require a key and there was nothing abnormal for you. When you left, the murderer opened the lock with his key, planted the dead body inside and replaced the lock again putting the original lock in the place. He then closed it without any problem.

Solution:

Never

Explanation:
In the question, the distance of x miles is given at a particular moment. The middle car is running at that time at 50 MPH. The distance will keep increasing with time as three are running at different speeds. It will increase by x miles every hour till a certain point of time and then it will start decreasing till the fastest car meets the slowest again.

As per question, all the cars stop then. Thus, the distance will never be x miles again.

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:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.

Solution:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years

Solution:

7, 4, 6, 5, 8

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.

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:

1/2

one of two is the possibility
1. If any of the first 99 people sit in Abhishek seat, Abhishek will not get to sit in his own seat.
2. If any of the first 99 people sit in Aishwarya's seat, Abhishek will get to sit in his seat.