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Riddle:

A watchmaker was telephoned urgently to make a house call to replace the broken hands on a clock. He was sik so he sent his apprentice.

The apprentice was thorough. When he finished inspecting the clock it was dark. Assuming his work was done, he attached the new hands and set the clock by his pocket watch. It was sic o'clock, so he set the big hand at the 12 and the little hand at the 6.

The apprectice returned, but soon the telephone rang. He picked up to his angry client:

"You didn't do the job right. The clock shows the wrong time."

Surprised he hurried back. He found the clock showing not much past eight. He handed is watch to the client and showed her that her clock was not even one second late. The client had to agree.

Early the nect morning, the client telephoned to say the clock has apparently gone berserk, hands were moving around the clock at will. The apprentice again rushed over, the clock showed a little past seven. After checking his watch he yelled:

"You are making fun of me! Your clock shows the right time!"

Have you figured out whats going on?

Answer:

As the problem says the apprentice mixed up the hands so that the minute hand was short and the hour hand was long.

The first time the apprentice returned to the client was about 2 hours and 10 minutes after he had set the clock at six.The long had moved olny from twelve to a little past two. The little made two whole circles and an additional 10 minutes. Thus the clock showed the correct time.

The next day around 7:o5 a.m.he came a second time,13 hours and 15 minutes after he had set the clock for six. The long had, acting as the hour hand,covered 13 hours to reach 1. The short hand made 13 full circles and 5 minutes, reaching 7, So the clock showed the correct time again.

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Riddle:

A great banquet was prepared for a Roman emperor and his courtiers. 22 Dormice, 40 Larks' Tongues, 30 Flamingos and 40 Roast Parrots were served.

How many portions of Boiled Ostrich were served?

Answer:

42. Each vowel is worth 2 and each consonant 4, so Dormice gives 22, ect.

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Riddle:

When the celebrated German mathematician Karl Friedrich Gauss (1777-1855) was nine he was asked to add all the integers from 1 through 100. He quickly added 1 to 100, 2 to 99, and so on for 50 pairs of numbers each adding to 101.

Answer: 50 X 101=5,050.

Now find the sum of all the digits in integers from 1 through 1,000,000,000.

That's all the digits in all the numbers, not all the numbers themselves.

Answer:

The numbers can be grouped by pairs:

999,999,999 and 0;

999,999,998 and 1'

999,999,997 and 2;

and so on....

There are half a billion pairs, and the sum of the digits in each pair is 81. The digits in the unpaired number, 1,000,000,000, add to 1. Then:

(500,000,000 X 81) + 1= 40,500,000,001.

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Riddle:

A woman is walking down a street night at a constant pace.  As she passes the street light, she notices that her shadow becomes longer.  Does the top of her shadow move faster, slower or the same when the shadow is longer as when it is shorter?

Answer:

This point maintains a constant speed, independent of the lenght of the shadow.

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Riddle:

okay! you have 5 kids and your the mom and you gotta try to get them all into the car at once.....Tommy and Timmy are twins but they hate eachother so they cant go together and Sarah and Sally hate eachother too, so they cant go together and Max is gay so he can only sit by the guys and theres only 5 seats you have to put them in order going in the car and have them sit where they would be happy.......my friend made it up so its kinda weird

Answer:

sarah, tommy, max, timmy, and then sally

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Riddle:

Their are three gaurds and three prisoners who need to cross a river. Their boat only holds two people at a time, and the number of prisoners must NEVER be allowed to outnumber the number of gaurds on either side of the river; otherwise, the prisoners will overpower the gaurds and, well, the story will come to an abrupt end.

Determine how many trips it will take to safely transport all of the gaurds and prisoners across the river, list each of the trips that need to be made and who is in the boat and who is on each of the riverbanks during each trip.

Answer:

Inless some one can tell me a way that 2 prisoners, at some point, don't out number the guards whether they are just dropping off and still in the boat or actually on land (because even if they are just dropping off and remain in the boat they are still on the other side of the river) I conclude this to be impossible. Please let me know an alternative if you figure one out because i'm stumped.

thanks

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Riddle:

A car's odometer shows 72927 miles, a palindromic number. What are the minimum miles you would need to travel to form another?

Answer:

110 miles. (73037)

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Riddle:

Two travellers spend from 12 o'clock to 6 o'clock walking along a level road, up a hill and back again. Their pace is 4 mph on the level, 3 mph uphill, and 6 mph downhill.

How far do they walk and at what time do they reach the top of the hill?

Answer:

24 miles half past three.

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Riddle:

Robert and David were preparing to have a water balloon fight. "No Fair" cried Robert, "You have 3 times as many as I do!" David said "Fine!" and gave Robert 10 more balloons. "Still not fair!" argued Robert, "You still have twice as many as I do." How many more balloons must David give Robert for them to have the same number?

Answer:

David must give Robert another 20 water balloons, giving them each 60. Robert started with 30 water balloons and David with 90.

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Riddle:

Jack has 8 bricks 7 of them weights the same amount and one is slightly heavier. Using a balance scale, how can Jack find the heavier brick in two weighings?

Answer:

First he split them in to piles of 3, 3, and 2 bricks. Then he weighs both groups of 3 with each other. If they balance he knows the brick is one of the 2 unweighed bricks and he can weigh them to find the heaver one. If the the stacks of 3 bricks do not balance, he will weigh 2 of the 3 bricks. If they balance he will know the brick left unweighed is heavier, or if they do not balance, he will find the heavier one.

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