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Riddle:

Can you divide the watch face with 2 straight lines so that the sums of the sums of the numbers in each part are equal? 

Answer:

The sum of the numbers on the watch face is 78. If the two lines cross, there must be 4 equal parts, but 78 is not divisible by 4. Then the lines do not cross, giving three parts with the sum of 26 each. Once you see the pairs on the face that add to 13(12+1,11+2, and so far) the answer will be easy to find.

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Riddle:

You have accidently left out the plug and are attempting to fill the bath with both taps full on. The hot tap takes 6 minutes to fill the bath. The cold tap takes 2 minutes and the water empties through the plug hole in 4 minutes.

In how many minutes will the bath be filled?

Answer:

2 minutes and 24 seconds.

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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:

I am a perching barrel, filled with meat, Taking hits from leaps and dives. Look inside, but do not eat, The meat in there is still alive!

Answer:

A thimble on a finger.

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Riddle:

Peter celebrated his birthday on one day, and two days later his older twin brother, Paul, celebrated his birthday.

How could this be?

Answer:

When the mother of the twins went into labor, she was travelling by boat. The older twin, Paul, was born first, barely on March 1st. The boat then crossed a time zone, and the younger twin was born on February the 28th. In a leap year the younger twin celebrates his birthday two days before his older brother.

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Riddle:

One sunny afternoon, three men go for a ride on a hot air balloon over the Sahara desert. An hour into the trip, the balloon begins to lose altitude. A month later, someone found one of the ballooners laying on the desert sand dead, naked, and holding half a toothpick. What happened to him?

Answer:

As the balloon lost altitude, the men took of their clothes and threw them overboard to decrease the weight of the balloon. The balloon continued to drop so the men drew straws to see who would be forced to jump. The dead man in the desert drew the shortest one (the half toothpick).

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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:

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:

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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