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Riddle:

As I traveled up and down our great glorious country, I found myself in a place where the tempature goes up sharply in the day and down at night. This had an effect on my watch, I noticed it was 1/2 a minute faster at nightfall, but at dawn it had lost 1/3 minute, making it only 1/6 minute fast.

One morning- May 1- my watch showed the right time. By what date was it 5 minutes faster?

Answer:

In 24 hours the watch gained 1/2-1/3= 1/6 minute. It would seem it would be 5 minutes fast in 5 X 6= 30 days; that is,the morning of May 31. But already on the morning of May 28 the was 27/6= 4 1/2 minutes fast. At the close of the day the watch gained 1/2 minute more, so it was 5 minutes fast on may 28.

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Riddle:

(A) Use two digits to make the smallest possible positve intiger.

(B) Five 3s can express 37:  37=33+3+3/3

Find another way to do it.

(C) Use six identical dgits to make 100. (Several solutions are possible.)

(D) Use five 4s to make 55.

(E) Use four 9s to make 20.

Answer:

(A) 1 X 1; 1/1;2/2;ect....;1-0;2-1;and many others.

(B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3

(C) 99 + 99/99; 55+55- 5- 5; (666-66)/6

(D) 44 + 44/4=55.

(E) 9 + 99/9=20.

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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:

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:

While playing with a metal washer shaped like a ring, Dave accidently pushed it on his finger too far and couldn't get it off. Trying to remove it using soap and water didn't work. The hospital sent him to a service station thinking they could cut the metal. Since the ring was made with a specially hardened steel, it couldn't be cut. Just then Bob arrived on the scene and suggested an easy way to remove the washer in just a few minutes. What was his solution?

Answer:

Bob suggested that Dave hold his finger in the air while someone wound a piece of string tightly around his finger just above the metal ring. The string forced the swelling down. As they unwounded the string from the end nearest the ring, someone else slid the ring up. They continued winding and unwinding the string until the ring could be easily removed.

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Riddle:

Many-manned scud-thumper, Maker of worn wood, Shrub-ruster, Sky-mocker, Rave! Portly pusher, Wind-slave.

What am I?

Answer:

The ocean!

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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:

It's always 1 to 6,
it's always 15 to 20,
it's always 5,
but it's never 21,
unless it's flying.

Answer:

The answer is: a dice. An explanation: "It's always 1 to 6": the numbers on the faces of the dice, "it's always 15 to 20": the sum of the exposed faces when the dice comes to rest after being thrown, "it's always 5": the number of exposed faces when the dice is at rest, "but it's never 21": the sum of the exposed faces is never 21 when the dice is at rest, "unless it's flying": the sum of all exposed faces when the dice is flying is 21 (1 + 2 + 3 + 4 + 5 + 6).

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Riddle:

A hunter met two shepherds, one of whom had three loaves and the other, five loaves. All the loaves were the same size. The three men agreed to share the eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze coins as payment for his meal. How should the two shepherds fairly divide this money?

Answer:

The shepherd who had three loaves should get one coin and the shepherd who had five loaves should get seven coins. If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd gave the hunter one-third of a loaf and the second shepherd gave the hunter two and one-third loaves. The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a loaf should get seven coins.

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