I HAVE A RING BUT NO HANDS I USED TO BE PLUGGED INTO THE WALL BUT NOW I FOL RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

Ship Thief Riddle

Hint:
The thief is the Sri Lankan seaman. They are on a Japanese ship, so it will bear a Japanese flag. The Japanese flag will look the same upside down.
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Clean Up Here Riddle

Hint:
The bathroom.
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Cut Into Chunks

Hint:
Pineapple
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Topping On A Hawaiian Pizza

Hint:
I am a pineapple
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White And Soft

Hint:
Coconut
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Cowboy Cows

Hint:
We’ll use A to represent the first cowboy and B for the second cowboy.

A + 1 = 2A, so A = 1.
A + 1 = B – 1, so B = 3.
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Hold Me By Neck

Hint:
A guitar
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Meeting In The Office

Hint:
A clock
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The Murder Of Mr Brown

Hint:
The chef killed Mr. Brown because he said he was cooking breakfast but it was a Sunday afternoon.
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Sherlock Holmes And The Case Of Ganpat

Hint:
Jason Kumar

The number on the calendar was written in a hurry.Sherlock matched the written number with the months of the year.
So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November.

Use the first letter of each month and it spells J-A-S-O-N.
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Play Me With A Ball

Hint:
Basketball
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Shiny And Metallic

Hint:
A bell
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Found Beneath A Chimney

Hint:
A fireplace
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The Loaded Revolver Riddle

Hint:
Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
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The Secret Santa Exchange

Hint: It's not as difficult as it seems. It's the number of ways the friends can form a circle divided by the number of ways the names can be drawn out of the hat.
1/10

For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is

(n-1)! / n!

Since n! = (n-1)! * n (for n > 1), this can be rewritten as

(n-1)! / (n*(n-1)!)

Factoring out the (n-1)! from the numerator and denominator leaves

1/n

as the probability.
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