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Riddle:

Who spends the day at the window, goes to the table for meals and hides at night?

Answer:

A fly.

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Riddle:

Some say we are red, some say we are green. Some play us, some spray us.

What are we?

Answer:

Pepper.

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Riddle:

A student zips on his scooter to ride to the train station to get to college. His home is close to two stops; the first one is a mile from home, and the second is two miles from home in the opposite direction. In the morning, he always gets on at the first stop and in the afternoon, he always gets off at the second one.

Why?

Answer:

The sations and his home are on a hill, which allows him to ride down easily on his scooter.

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Riddle:

There were five men going to church and it started to rain. The four that ran got wet and the one that stood still stayed dry. How did the one stay dry?

Answer:

It was a body in a coffin with the bearers.

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Riddle:

In yon vast field of cultivated space, I there am found with members of my race; Decapitate me - if you've no objection - You then will find what brings me to perfection; Take one more cut, and then you'll plainly see What Iam destined, day by day, to be. What am I?

Answer:

WHEAT
  HEAT
   EAT

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Riddle:

0,1,2,3,4,5,6,7,8,9

Use the digits above once each only to compose two fractions which when added togeather equal 1.

Answer:

35/70 + 148/296 = 1

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Riddle:

The king dies and two men, the true heir and an impostor, both claim to be his long-lost son. Both fit the description of the rightful heir: about the right age, height, coloring and general appearance. Finally, one of the elders proposes a test to identify the true heir. One man agrees to the test while the other flatly re-fuses. The one who agreed is immediately sent on his way, and the one who re-fused is correctly identified as the rightful heir. Can you figure out why?

Answer:

The test was a blood test. The elder remembered that the true prince was a hemophiliac.

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Riddle:

A mother has three sick children. She has a 24-ounce bottle of medicine and needs to give each child eight ounces of the medicine. She is unable to get to the store and has only three clean containers, which measure 5, 11 and 13 ounces. The electricity is out and she has no way of heating water to wash the containers and doesn't want to spread germs. How can she divide the medicine to give each child an equal portion without having any two children drink from the same container?

Answer:

Fill the 5 oz. and 11 oz. Containers from the 24 oz. container. This leaves 8 oz. in the 24 oz. bottle. Next empty the 11 oz. bottle by pouring the contents into the 13 oz. bottle. Fill the 13 oz. bottle from the 5 oz. container (with 2 oz.) and put the remaining 3 oz. in the 11 oz. bottle. This leaves the 5 oz. container empty. Now pour 5 oz. from the 13 oz. bottle into the 5 oz. bottle leaving 8 oz. in the 13 oz. bottle. Finally pour the 5 oz. bottle contents into the 11 oz. bottle giving 8 oz. in this container.

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Riddle:

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weight the same or that the dish that falls lower has heavier contents than the other. The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more. Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light

Answer:

Most people seem to think that the thing to do is weight six coins against six coins, but if you think about it, this would yield you no information concerning the whereabouts of the only different coin. As we already know that one side will be heavier than the other. 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 weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different coin is either 11 or 12. Weigh coin 1 against 11. If they balance, the different coin is number 12. If they do not balance, then 11 is the different coin. If 1,2 vs 9,10 do not balance, then the different coin is either 9 or 10. Again, weigh 1 against 9. If they balance, the different coin is number 10, otherwise it is number 9. 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. 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.

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Riddle:

A boy presses a side of a blue pencil to a side of a yellow pencil, holding both pencils vertically. One inch of the pressed side of the blue pencil, measuring from its lower end, is smeared with paint. The yellow pencil is held steady while the boy slides the blue pencil down 1 inch, continueing to press it against the yellow one. He returns the blue pencil to its former position, then again it slides down 1 inch. He continues until he has lowered the blue pencil 5 times and raised it 5 times- 10 moves in all.

Supposed that during this time the paint neither dries nor diminishes in quantity. How many inches of each pencil will be sneared with paint after the tenth move?

 

Answer:

At the start, 1inch of the yellow pencil gets smeared with wet paint. As the blue pencil is moved downward, a second inch of the blue pencils smears a second inch of the yellow pencil.

Each pair of down and up movesof the blue pencil smears 1 more inch of each pencil. 5 pairs of moves will smear 5 inches. This together with the initial inch, makes 6 inches for each pencil.

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