Solution:

2hr 11min

Explanation:
Now this question cannot be solved unless you know a fact i.e. the hour hand is exactly on the minute mark five times every hour - on the hour, 12 minutes past hour, 24 minutes past hour, 36 minutes past hour and 48 minutes past hour.

Denoting the number of hours with X and the number of minutes as Y, let us begin with the calculation.

When we have the hour hand on the minute mark, the position of hour hand:
5X + Y/12
And the position of the minute hand = Y

When the first observation was taken,
Y = Y = 5X + Y/12 + 6
Or, 60X = 11Y - 72.

With the given facts, we know for sure that Y can only be 0, 12, 24, 36 or 48.
According to which, the possible values for X and Y can be 1 and 12 respectively. Thus the time must be 1:12 when the first observation was taken.

Similarly, in the second observation, let's make the equation:
60X = 11Y - 84
Here, the possible values for X and Y here are 3 and 24 respectively. According to this, the time is 3:24 for sure.

From 1:12 to 3:24, two hours and eleven minutes have passed.

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

Solution:

40 balls

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:

The young lady was presented with a bill which belonged to the elderly woman. She had asked the waiter to collect the bill from her daughter.

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Solution:

99 matches.

The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.

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:

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.

Solution:

The Password.

Explanation:
Clues 2 and 5 are correct and rest are lies.

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.