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:

It can be down as shown below.

Solution:

2 degree below zero

Solution:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)

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:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

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:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.

Solution:

8 liters jar: 8; 5 liters jar: 0; 3 liters jar: 0
Fill 5 liters jar entirely.
8 liters jar: 3; 5 liters jar: 5; 3 liters jar: 0
Fill 3 liters jar with 5 liters jar.
8 liters jar: 3; 5 liters jar: 2; 3 liters jar: 3
Pour entirely from 3 liters jar to 8 liters jar.
8 liters jar: 6; 5 liters jar: 2; 3 liters jar: 0
Pour entirely from 5 liters jar to 3 liters jar.
8 liters jar: 6; 5 liters jar: 0; 3 liters jar: 2
Fill 5 liters jar with water from 8 liters jar.
8 liters jar: 1; 5 liters jar: 5; 3 liters jar: 2
Fill the 3 liters jar with water from the 5 liters jar.
8 liters jar: 1; 5 liters jar: 4; 3 liters jar: 3
Empty 3 liters jar in 8 liters jar.
8 liters jar: 4; 5 liters jar: 4; 3 liters jar: 0

Now you have 4 liters of water in 8 liters jar as well as 5 liters jar.

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.