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:

Divide the 9 gems into a group of three (randomly)

Now select any two group of gems and put them in a balance. If they are equal then, the heavier gem must be in the group you have left. If one of them is heavy, then that group contains the heavier gem.

In either case, take the group that contains the heavier gem and select any two gems among them (randomly). Weigh the two gems in the balance. If one is heavier, you know which one is the heavier gem of them all. If the weights are same, then the third gem is the heavier one in all.

In this manner, it will take you just two tries before you can determine the heavy gem out of the nine gems.

Solution:

Dice 1: 0 1 2 3 5 7
Dice 2: 0 1 2 4 5 8

Please keep in mind that you will have to show 11 and 12 as well, thus 1 and 2 must be present in both the dices. In the same manner, to show 01, 09, 0 must be present in both the dices. Now if you are wondering how you will show 9 as we have not written it in any of the faces of the dices, the trick is to reverse 6 to show it as 9.

You may now try any day of the month and you will easily be able to represent it through the dices.

Solution:

W02

F21: Friday the 21st.
S23: Sunday the 23rd.
T25: Tuesday the 25th.
T27: Thursday the 27th.
S29: Saturday the 29th.
M31: Monday the 31st.

Avoiding every other day, the next term must be 'Wednesday the 2nd' i.e. 'W02'

Solution:

Let the simple interest be x
According to the question
1120 - 4x = Amount ----- eq (1)
1200 - 5x = Amount ----- eq (2)
Equating eq (1) and eq (2)
x=80
Thus the Amount is 1200 - 400=800

Option b is correct

A simpler way to solve will be

Interest for one year = Interest for five years - Interest for four years
= 1200 - 1120 = Rs.80

Interest for four years = 4*80 = Rs.320
Thus sum = 1120 - 320 = Rs. 800