Solution:

Time

Solution:

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

Solution:

The attached picture shows you how you can do it. If you were able to deduce any other way, feel free to comment.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

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:

The numbers that have been written over the calendar are:
12, 4, 8 and 3

A few numbers have been circled under particular months. The curious thing is that the numbers that have been written are the same number of corresponding months under which the dates have been marked.

What we can judge by the overall calendar is that the dates that have been marked may refer to the letters at that position in the month's name. In addition, the written numbers can be the position of those letters. Let's simplify below:

12th month is December
The number marked under December = 5
The fifth letter of December = M

4th month is April
The number marked under April = 4
The fourth letter of April = I

8th month is August
The number marked under August is 6
The sixth letter of August = T

3rd month is March
The numbers marked under March are 4 and 5
The fourth and fifth letter of March = C and H

When you combine the letters, you get the name as MITCH.
Therefore, Mitch is the killer.

Solution:

I will be thrown at lions. Now if Akbar threw him at lions, Birbal’s statement will stand true and he will have to bury him alive. But if he bury him, the statement will emerge as false. Thus he had no choice left

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:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

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:

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