Two convicts are locked in a cell. There is an unbarred window high up in the cell. No matter if they stand on the bed or one on top of the other they can't reach the window to escape. They then decide to tunnel out. However, they give up with the tunnelling because it will take too long. Finally one of the convicts figures out how to escape from the cell. What is his plan?

His plan is to dig the tunnel and pile up the dirt to climb up to the window to escape.

A blind man walks into a hardware store to buy a hammer. There are hammers hanging behind the front desk, but obviously the blind man isn't able to see them. And yet a few minutes later, he happily walks out of the store, having just purchased a new hammer.
How did he do it?

He walks up the the front desk where the clerk is working and says "I'd like to buy a hammer."

You are standing in a house in the middle of the countryside. There is a small hole in one of the interior walls of the house, through which 100 identical wires are protruding.
From this hole, the wires run underground all the way to a small shed exactly 1 mile away from the house, and are protruding from one of the shed's walls so that they are accessible from inside the shed.
The ends of the wires coming out of the house wall each have a small tag on them, labeled with each number from 1 to 100 (so one of the wires is labeled "1", one is labeled "2", and so on, all the way through "100"). Your task is to label the ends of the wires protruding from the shed wall with the same number as the other end of the wire from the house (so, for example, the wire with its end labeled "47" in the house should have its other end in the shed labeled "47" as well).
To help you label the ends of the wires in the shed, there are an unlimited supply of batteries in the house, and a single lightbulb in the shed. The way it works is that in the house, you can take any two wires and attach them to a single battery. If you then go to the shed and touch those two wires to the lightbulb, it will light up. The lightbulb will only light up if you touch it to two wires that are attached to the same battery. You can use as many of the batteries as you want, but you cannot attach any given wire to more than one battery at a time. Also, you cannot attach more than two wires to a given battery at one time. (Basically, each battery you use will have exactly two wires attached to it). Note that you don't have to attach all of the wires to batteries if you don't want to.
Your goal, starting in the house, is to travel as little distance as possible in order to label all of the wires in the shed.
You tell a few friends about the task at hand.
"That will require you to travel 15 miles!" of of them exclaims.
"Pish posh," yells another. "You'll only have to travel 5 miles!"
"That's nonsense," a third replies. "You can do it in 3 miles!"
Which of your friends is correct? And what strategy would you use to travel that number of miles to label all of the wires in the shed?

Believe it or not, you can do it travelling only 3 miles!
The answer is rather elegant. Starting from the house, don't attach wires 1 and 2 to any batteries, but for the remaining wires, attach them in consecutive pairs to batteries (so attach wires 3 and 4 to the same battery, attach wires 5 and 6 to the same battery, and so on all the way through wires 99 and 100).
Now travel 1 mile to the shed, and using the lightbulb, find all pairs of wires that light it up. Put a rubberband around each pair or wires that light up the lightbulb. The two wires that don't light up any lightbulbs are wires 1 and 2 (though you don't know yet which one of them is wire 1 and which is wire 2). Put a rubberband around this pair of wires as well, but mark it so you remember that they are wires 1 and 2.
Now go 1 mile back to the house, and attach odd-numbered wires to batteries in the following pairs: (1 and 3), (5 and 7), (9 and 11), and so on, all the way through (97 and 99).
Similarly, attach even-numbered wires to batteries in the following pairs: (4 and 6), (8 and 10), (12 and 14), and so on, all the way through (96 and 98).
Note that in this round, we didn't attach wire 2 or wire 100 to any batteries.
Finally, travel 1 mile back to the shed. You're now in a position to label all of the wires here.
First, remember we know the pair of wires that are, collectively, wires 1 and 2. So test wires 1 and 2 with all the other wires to see what pair lights up the lightbulb. The wire from wires 1 and 2 that doesn't light up the bulb is wire 2 (which, remember, we didn't connect to a battery), and the other is wire 1, so we can label these as such. Furthermore, the wire that, with wire 1, lights up a lightbulb, is wire 3 (remember how we connected the wires this round).
Now, the other wire in the rubber band with wire 3 is wire 4 (we know this from the first round), and the wire that, with wire 4, lights up the lightbulb, is wire 6 (again, because of how we connected the wires to batteries this round). We can continue labeling batteries this way (next we'll label wire 7, which is rubber-banded to wire 6, and then we'll label wire 9, which lights up the lightbulb with wire 7, and so on). At the end, we'll label wire 97, and then wire 99 (which lights up the lightbulb with wire 97), and finally wire 100 (which isn't connected to a battery this round, but is rubber-banded to wire 99).
And we're done, having travelled only 3 miles!

There is a barrel with no lid and some wine in it.
"This barrel of wine is more than half full," said Curly.
Moe says, "No it's not. It's less than half full."
Without any measuring implements and without removing any wine from the barrel, how can they easily determine who is correct?

Tilt the barrel until the wine barely touches the lip of the barrel. If the bottom of the barrel is visible then it is less than half full. If the barrel bottom is still completely covered by the wine, then it is more than half full.

If you were to put a coin into an empty bottle and then insert a cork into the neck, how could you remove the coin without taking out the cork or breaking the bottle?

Push the cork into the bottle and shake the coin out.

A man has two ropes of varying thickness (Those two ropes are not identical, they aren’t the same density nor the same length nor the same width). Each rope burns in 60 minutes. He actually wants to measure 45 mins. How can he measure 45 mins using only these two ropes.
He can’t cut the one rope in half because the ropes are non-homogeneous and he can’t be sure how long it will burn.

He will burn one of the rope at both the ends and the second rope at one end. After half an hour, the first one burns completely and at this point of time, he will burn the other end of the second rope so now it will take 15 mins more to completely burn. so total time is 30+15 i.e. 45mins.

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.

Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.

If I were in Hawaii and dropped a bowling ball in a bucket of water which is 45 degrees F, and dropped another ball of the same weight, mass, and size in a bucket at 30 degrees F, them at the same time, which ball would hit the bottom of the bucket first?
Same question, but the location is in Canada?

Both questions, same answer: the ball in the bucket of 45 degree F water hits the bottom of the bucket last.
Did you think that the water in the 30 degree F bucket is frozen?
Think again.
The question said nothing about that bucket having anything in it.
Therefore, there is no water (or ice) to slow the ball down...