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Riddle:

My first is a creature whose breeding is unclear. My second, a price you must pay. My whole can be found in the river of Time and refers to events of today. What am I?

Answer:

Current.

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Riddle:

Marking mortal privation, when firmly in place. An enduring summation, inscribed in my face.

What am I?

Answer:

A Tombstone.

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Riddle:

When I'm used, I'm useless, once offered, soon rejected. In desperation oft expressed, the intended not protected.

What am I?

Answer:

A poor alibi or excuse.

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Riddle:

If you like pretty gems that sparkle and shine, I invite you to dig in my virtual mine. My first is purple, fit for a king, My second is green where Dorothy did her thing. My third is red, July's birthstone as well, My fourth is seen in strings and is found inside a shell. My fifth is hard, pure Carbon and expensive to buy, My sixth is Crocodolite, striped like the big cat's eye. Seventh is two words, a man-made fake of April's stone, Eighth is very dark and found at Lightning Ridge alone. Now take from each gem, one letter in its turn, And you will find the stuff for which even the god's yearn.

Answer:

Answer: Ambrosia The gems are: Amethyst Emerald ( Emerald City in the Wizard of Oz ) Ruby Pearl Diamond Tiger's eye Cubic Zirconium (Fake diamond) Black Opal Taking the first letter of Amethyst, second of Emerald, etc.. gives Ambrosia - The food of the gods.

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Riddle:

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, …). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, …). This continues until all 100 people have passed through the room. What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

Answer:

First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

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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:

In 2000, a 40-year-old doctor told his son that when a little boy he decided to be a doctor by seeing a internet web site about performing a heart transplant on a puppy with a dfective heart so that the puppy would live a normal life. I then thought that I would be a doctor so that I could help people in a similar way. What is the defect in this story?

Answer:

The internet did not exist when the doctor was a little boy.

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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:

How many cats are in a small room if in each of the four corners a cat is sitting, and opposite each cat there sit three cats, and a each cat's tail another is sitting?

Answer:

Four cats, each near a tail of a cat in an adjacent corner.

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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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