YOU SAW ME WHERE I NEVER WASAND WHERE I COULD NEVER BEAND YET WITHIN THAT VERY PLACEMY FAC RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

The Queen's Hat Riddle

Hint:
It was off with her head!
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1968 Penny

Hint:
Its one more penny. $19.68 is more than $19.67.
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What You Use To See

Hint:
An eyeball
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Here To Make You Happy

Hint:
I am a cake
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A Tank Without A Military

Hint:
Toilet
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Meeting In The Office

Hint:
A clock
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Three Hunters Riddle

Hint:
They didn't really pay $9 each, remember? The bell-boy was too lazy to add up the actual sum that they would pay. They reeeally payed about a $8.66 each. So $8.66 times the three of them equals about $25, plus the $5 in the bell-boys equals $30
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Throwing A Basketball Riddle

Hint:
He threw the ball straight up in the air.
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Two Camels Riddle

Hint:
The two camels were facing each other the entire time. Hence facing in opposite directions.
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Born In Mourning

Hint:
A tombstone
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Mortal Privation Riddle

Hint:
A tombstone
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The Blind Mammals Riddle

Hint:
They are bats!
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Lazily In The Sun

Hint:
A daisy
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Roll The Dice

Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.

If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.

If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.

If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.

In each case, the dealer neither wins nor loses. Hence it's a fair game.
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The Blue And Red Dice Riddle

Hint:
Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6 - X)

X = 3

The second die must have 3 red faces and 3 blue faces.
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