Solution:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

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:

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:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.

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:

Put one red marble in one box and all other marbles in the 2nd box.

Probability :
50 + 24/49

Solution:

146900

Explanation:
Whenever you face such type of questions, it is wise to begin from the last thing. Here in this question the last thing will be the 9th temple. He climbed down 100 steps and thus you know, he had Rs. 100 before beginning climbing down. Thus, he must have offered Rs. 100 to the god in that temple too (he offered half of the total amount). Also, he must have dropped Rs. 100 while climbing the steps of the ninth temple. This means that he had Rs. 300 before he begand climbing the steps of the ninth temple.

Now, we will calculate in the similar manner for each of the temples backwards.
Before the devotee climbed the eight temple: (300+100)*2 + 100 = 900
Before the devotee climbed the seventh temple: (900+100)*2 + 100 = 2100
Before the devotee climbed the Sixth temple: (2100+100)*2 + 100 = 4300
Before the devotee climbed the fifth temple: (4300+100)*2 + 100 = 8900
Before the devotee climbed the fourth temple: (8900+100)*2 + 100 = 18100
Before the devotee climbed the third temple: (18100+100)*2 + 100 = 36,500
Before the devotee climbed the second temple: (36500+100)*2 + 100 = 73300
Before the devotee climbed the first temple: (73300+100)*2 + 100 = 146900

Therefore, the devotee had Rs. 146900 with him initially.

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.

Solution:

Solution can be verified from the picture below.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.