Solution:

First time = 20 ft
Second time = 10ft down + 10 ft up
Third time = 5ft up + 5ft down
It will go on like this the object stops

Total distance traveled = 20 + (10 + 10) + (5 + 5) + (2.5 + 2.5) + . . .
= 20 + (10 + 5 + 2.5 + ...) + (10 + 5 + 2.5 + ...)
Assuming the object comes to rest at infinity,

Total distance traveled = 20 + [10/ (1 - 0.5)] + [10/ (1 - 0.5)]
= 20 + 20 + 20 = 60

Thus option d is correct.

Solution:

They select one of the prisoner named Jason to do the trick and frame out a full proof plan. According to their plan, whenever a prisoner other than Jason is selected, they will follows some simple steps - If the bulb is off, they will switch it on but if the bulb is already lit, they won't do anything and sit idle.

If Jason is selected and he finds out that the bulb is lit, he will add one to his count and will switch off the bulb. If the bulb is already switched off, he will sit there idly. In such a manner when his count reaches to 99, he will tell the warden that every prisoner has now been to the room.

Let us craft out the solution in simpler terms. Whenever a prisoner goes inside the room, he simply switches on the bulb if it is off. Thus one prisoner will only light the bulb once. When Jason finds out that the bulb it on, he will know that this has been done by a new prisoner and he will add one to his count. He will keep doing this till his count reaches 99. 99 because, he already has been in the room which makes a total of 100. Thus it is a full proof plan that will definitely set them all free.

Solution:

Side A = 3 Cannibals, 3 missionaries
Side B = 0 Cannibals, 0 missionaries

1 cannibal and 1 missionary goes in the boat
Side A = 2 Cannibals, 2 missionaries
Side B = 1 Cannibals, 1 missionaries

1 missionary returns back and 2 cannibals go there
Side A = 0 Cannibals, 3 missionaries
Side B = 3 Cannibals, 0 missionaries

1 cannibal returns and 2 missionaries go
Side A = 1 Cannibals, 1 missionaries
Side B = 2 Cannibals, 2 missionaries

1 missionary and 1 cannibal return and 2 missionaries go
Side A = 2 Cannibals, 0 missionaries
Side B = 1 Cannibals, 3 missionaries

One cannibal back, two cannibals go
Side A = 1 Cannibals, 0 missionaries
Side B = 2 Cannibals, 3 missionaries

One cannibal back, two cannibals go
Side A = 0 Cannibals, 0 missionaries
Side B = 3 Cannibals, 3 missionaries

Thus all have crossed successfully

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:

64

Explanation:
Assume that the number of Woolf’s sons = n
Number of brothers of each son = n-1
Number of sons for each of Woolf’s son = n-1
Woolf’s age = n-1*n-1
Which is equal to a perfect square between fifty and seventy
Which is equal to 64