Solution:

Let us first assume that the number of drops required are N.

If the egg breaks at maximum number of tries, we will have N - 1 drops till it does not break. Thus we must drop the first egg from the height N. Now if the first drop of the first egg does not break the egg, we can have N - 2 drops for the second egg if the first egg breaks in the second drop.

Let us put that into a valid example for better understanding. Suppose we need 16 drops. Now let us drop the egg from the sixteenth floor, if it breaks, we will try all the floors below sixteen. Suppose it does not breaks, then we have 15 drops left and we will drop it from 16+15+1 = 32nd floor. This is because if it breaks at 32nd floor, we can try all the way down to 17th floor in fourteen tries making the total tries to be 16.

Let us assume that 16 is the correct answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45.....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 we can easily do in the end as we have enough drops to complete the task.

Seeking the above case, we must note that we have achieved it in 14 or 14 drops but we still need to find out the optimal one. We know from above that the optimal one will require 0 linear trials in the last step.

Thus we can say that

(1+p) + (1+ (p-1)) + (1+ (p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

q=14

Thus the optimal number of drops required are 14.

Drops will be tried at floor number 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100. How? Just move up 14, 13, 12 .... Floor till it breaks or does not.

Solution:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

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:

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

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:

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

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

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:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

1. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)

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.