Solution:

It can be solved as shown.

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:

It can be done in 8 moves as shown in the animated image below.

Solution:

Sam : 2
Dean : 5
Castiel : 7

Age diff
7 - 2 = '5' is prime
7 - 5 = '2' is prime
5 - 2 = '3' is prime

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple

Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.

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:

The guard and the prisoner cross the river
The guard comes back
The guard and the girl cross the river.
The guard and the prisoner comes back.
The mother and the girl cross the river.
Mother comes back
The mother and the father cross the river
Father comes back
The guard and the prisoner cross the river
Mother comes back
The father and the mother cross the river
Father comes back
The father and the boy cross the river
The guard and the prisoner comes back
The guard and the boy cross the river
The guard comes back
The guard and the prisoner cross the river.

Solution:

Hhe hung herself using big Ice slabs , which are now now melted and hence puddle of water.

Solution:

30,450 miles.

Explanation:
For each 5 miles driven using this rotation, each tire will lose (2/29000)+(2/21000) of its tread.So, suppose we drive 5*N miles like this. Then, each tire will lose 2*N*((1/29000)+(1/21000)) of its tread. We wish to determine N so that the latter expression gets to be 1, representing all of each tire's tread. That happens when N=6090, which is when 5*N (the total miles driven) is 30,450.

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.