You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?

You need to conduct 7 races.
First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.
We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).
This table shows our remaining horses:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).
We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).
Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).
Let's look at our updated table, having removed these horses that can't be in the top 3:
A1 B1 C1
A2 B2 C2
A3 B3 C3
We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:
A1 B1 C1
A2 B2
A3
We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.

Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?

John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.

General Custer is surrounded by Indians and he's the only cowboy left.
He finds an old lamp in front of him and rubs it. Out pops a genie. The genie grants Custer one wish, with a catch. He says, "Whatever you wish for, each Indian will get two of the same thing." Custer ponders a while and thinks:"If I get a bow and arrow they get two. If I get a rifle they get two!" He then rubs the bottle again and out pops the genie. "Well," the genie asks "have you made up your mind?"
What did Custer ask for to help him get away?

Two trains are traveling toward each other on the same track, each at 60 miles per hour. When they are exactly 120 miles apart, a fly takes off from the front of one of the trains, flying toward the other train at a constant rate of 100 miles per hour. When the fly reaches the other train, it instantly changes directions and starts flying toward the other train, still at 100 miles per hour. It keeps doing this back and forth until the trains finally collide.
If you add up all the distances back and forth that the fly has travelled, how much total distance has the fly travelled when the trains finally collide?

The fly has travelled exactly 100 miles. We can figure this out using some simple math. Becuase the trains are 120 miles apart when the fly takes off, and are travelling at 60 mph each, they will collide in exactly 1 hour. This gives the fly exactly 1 hour of flying time, going at a speed of 100 miles per hour. Thus, the fly will travel 100 miles in this hour.

Find words to fit the clues, all the words end in the same three letters.
_ _ _ _ _ _ Eat quickly
_ _ _ _ _ _ Unverified story
_ _ _ _ _ _ _ An outline

The Miller next took the company aside and showed them nine sacks of flour that were standing as depicted in the sketch.
"Now, hearken, all and some," said he, "while that I do set ye the riddle of the nine sacks of flour.
And mark ye, my lords and masters, that there be single sacks on the outside, pairs next unto them, and three together in the middle thereof.
By Saint Benedict, it doth so happen that if we do but multiply the pair, 28, by the single one, 7, the answer is 196, which is of a truth the number shown by the sacks in the middle.
Yet it be not true that the other pair, 34, when so multiplied by its neighbour, 5, will also make 196.
Wherefore I do beg you, gentle sirs, so to place anew the nine sacks with as little trouble as possible that each pair when thus multiplied by its single neighbour shall make the number in the middle."
As the Miller has stipulated in effect that as few bags as possible shall be moved, there is only one answer to this puzzle, which everybody should be able to solve.

The way to arrange the sacks of flour is as follows: 2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need be moved.
There are just three other ways in which they might have been arranged (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.

A women walks into a bank to cash out her check.
By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars.
On the way home she spends 5 cents, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount?

The check was for dollars 31.63.
The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.
Let x be the dolars of the check, and y be the cent.
The check was for 100x + y cent
He was given 100y + x cent
Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
Which doesn't look like enough information to solve the problem except that x and y must be whole numbers, so:
199x ≡ -5 (mod 98)
98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
This quickly leads to x = 31 and then y = 63
Alternative solution by substitution:
98y = 199x + 5
y = (199x + 5)/98 = 2x + (3x + 5)/98
Since x and y are whole numbers, so must be (3x + 5)/98.
Call it z = (3x+5)/98
so 98z = 3x + 5, or 3x = 98z - 5 or x = (98z - 5)/3
or x = 32z-1 + (2z-2)/3.
Since everything is a whole number, so must be (2z-2)/3.
Call it w = (2z-2)/3, so 3w = 2z-2 so z = (3w+2)/2 or
z = w + 1 + w/2. So w/2 must be whole, or w must be even.
So try w = 2. Then z = 4. Then x = 129. Then y = 262.
if you decrease y by 199 and x by 98, the answer is the same:
y = 63 and x = 31.

While mixing sand, gravel, and cement for the foundation of a house, a worker noticed a small bird hopping along the top of the foundation wall. The bird misjudged a hop and fell down one of the holes between the blocks. The bird was down too far for anyone to reach it and the hole was too small for it to fly out of. Someone suggested using two sticks to reach down into the hole and pull the bird out, but this idea was rejected for fear it would injure the fragile bird. What would be the easiest way to get the bird out of the hole without injuring it?

Since they had plenty of sand available, they could pour a little at a time into the hole. The bird would constantly keep shifting its position so that it stood on the rising sand.

You are standing next to three switches. You know these switches belong to three bulbs in a room behind a closed door – the door is tight closed, and heavy which means that it's absolutely impossible to see if any bulb is on or not. All three switches are now in position off.
You can do whatever you want with the switches and when you are finished you open the door and go into the room. While in there you have to tell which switch belongs to which bulb.
How will you do that?

Turn on the first switch and wait for a while.
Turn off the first one and turn on the second.
Go into the room.
One bulb is shining, the second bulb is hot and the third one nothing.

Shadow drove into the Speedy Service Station and pulled up to the pumps. "Fill it up, please," said Shadow. "
This may sound strange," said the owner, "but I'd rather fill two cars from out of town than one car from this town."
Shadow looked across the small town and replied, "I know just what you mean."
Why would the owner feel this way?

The owner would rather fill two cars from anywhere than one car from town because he would make twice the amount of money.