Cranberry Sauce And Pumpkin Pie
Thirty days of autumn bliss
I hold the holiday you won't want to miss
The home of cranberry sauce and pumpkin pie
Which month of the year am I?
I hold the holiday you won't want to miss
The home of cranberry sauce and pumpkin pie
Which month of the year am I?
Hint:
Made Of 12 Riddle
What is a phrase thats formed of 4 letters, but consists of 3, sometimes is written with 9, never with 5, and occasionally is made up of 12.
Hint:
Eggy Vacation Riddle
Hint:
Breaking In Half
Cindy is holding something in her hands. She is able to break it into thirds by only breaking it in half. What is Cindy holding?
Hint:
Cindy is holding a fortune cookie. By breaking it in half, she has received not only two halves of the cookie, she now also has the fortune, making the cookie into thirds. Did you answer this riddle correctly?
YES NO
YES NO
My Spreading Wings
I fly to any foreign parts,
Assisted by my spreading wings:
My body holds an hundred hearts,
Nay, I will tell you stranger things:
When I am not in haste I ride,
And then I mend my pace anon;
I issue fire out from my side
Ye witty youths, this riddle con.
I'm a?
Assisted by my spreading wings:
My body holds an hundred hearts,
Nay, I will tell you stranger things:
When I am not in haste I ride,
And then I mend my pace anon;
I issue fire out from my side
Ye witty youths, this riddle con.
I'm a?
Hint:
Keeping Things Cold
Meat, milk and yogurt it does hold
The reason you put them in this
Is because it helps keep them cold
Hint:
Meeting In The Office
If you have a meeting in the office
Youll need to know the time and place
Something that can help with one of these things
Has two or three hands over its face
What is this?
Youll need to know the time and place
Something that can help with one of these things
Has two or three hands over its face
What is this?
Hint:
Under The Cup Riddle
You decide to play a game with your friend where your friend places a coin under one of three cups. Your friend would then switch the positions of two of the cups several times so that the coin under one of the cups moves with the cup it is under. You would then select the cup that you think the coin is under. If you won, you would receive the coin, but if you lost, you would have to pay.
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
Hint: Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...
The rightmost cup.
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
The Secret Santa Exchange
A group of ten friends decide to exchange gifts as secret Santas. Each person writes his or her name on a piece of paper and puts it in a hat. Then each person randomly draws a name from the hat to determine who has him as his or her secret Santa. The secret Santa then makes a gift for the person whose name he drew.
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
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. Did you answer this riddle correctly?
YES NO
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. Did you answer this riddle correctly?
YES NO
An Old Relative Riddle
Hint:
1 Rabbit Saw 6 Elephants Riddle
Every elephant saw 2 monkeys going towards the river.
Every monkey holds 1 parrot in their hands.
How many Animals are going towards the river?
Hint:
5 Animals.
Lets go through the question again.
1 rabbit saw 6 elephants while going to the river. Hence, 1 animal (rabbit) is going towards the river.
Every elephant saw 2 monkeys going towards the river. This is the tricky part, from the sentence it seems to imply each of the 6 elephants saw 2 monkeys going towards the river, hence logically will be 6 x 2 = 12 animals (monkeys) going towards the river.
However, the statement does not explicitly mention that Every elephant saw 2 DIFFERENT monkeys, hence implicit rules apply and infer that the 2 monkeys are the same.
Hence, correct answer is that every elephant saw 2 monkeys, and by inference, the 2 monkeys are the same, hence there exists only 2 monkeys which are going towards the river !!
Finally, every monkey holds 1 parrot in their hands. Hence, 2 parrots are going towards the river.
So in total, 1 rabbit, 2 monkeys and 2 parrots (5 animals) are going towards the river. Did you answer this riddle correctly?
YES NO
Lets go through the question again.
1 rabbit saw 6 elephants while going to the river. Hence, 1 animal (rabbit) is going towards the river.
Every elephant saw 2 monkeys going towards the river. This is the tricky part, from the sentence it seems to imply each of the 6 elephants saw 2 monkeys going towards the river, hence logically will be 6 x 2 = 12 animals (monkeys) going towards the river.
However, the statement does not explicitly mention that Every elephant saw 2 DIFFERENT monkeys, hence implicit rules apply and infer that the 2 monkeys are the same.
Hence, correct answer is that every elephant saw 2 monkeys, and by inference, the 2 monkeys are the same, hence there exists only 2 monkeys which are going towards the river !!
Finally, every monkey holds 1 parrot in their hands. Hence, 2 parrots are going towards the river.
So in total, 1 rabbit, 2 monkeys and 2 parrots (5 animals) are going towards the river. Did you answer this riddle correctly?
YES NO
What Brightens Mom's Day?
Here's a common riddle that'll get your brain working. It's not tough to solve but can be tricky if you don't think outside of the box a little. In regards to the riddle... for most of us this has indeed brightened up our moms day and is very big and always yellow. You have 30 seconds to solve.
Hint: Mom's children aren't always too happy about it.
Two Planes
There are two planes. One is going from New York to London at a speed of 600 MPH. The other is traveling from London to New York at a speed of 500 MPH.
When the planes meet which one will be closer to London?
When the planes meet which one will be closer to London?
Hint:
How Many Times A Day?
Hint:
22 times: 12:00:00, 1:05:27, 2:10:55, 3:16:22, 4:21:49, 5:27:16, 6:32:44, 7:38:11, 8:43:38, 9:49:05, 10:54:33. Each twice a day. Did you answer this riddle correctly?
YES NO
YES NO
How Old Could He Be?
In 1940, a correspondent proposed the following question:
A man's age at death was one twenty-ninth of the year of his birth. How old was he in 1900?
A man's age at death was one twenty-ninth of the year of his birth. How old was he in 1900?
Hint:
He was 44 years old.
From the question you know the man died between 1900 and 1940. We also know his age at death (x) is one twenty-ninth of the year of his birth (29x). If you add his age at death to the year he was born you get the year he died (30x). Only one year between 1900 and 1940 is divisible by 30, 1920 (the year he died). The year he was born can now be found: 1920 * (29/30) = 1856. So in 1900 he was (1900 - 1856) = 44 years old. Did you answer this riddle correctly?
YES NO
From the question you know the man died between 1900 and 1940. We also know his age at death (x) is one twenty-ninth of the year of his birth (29x). If you add his age at death to the year he was born you get the year he died (30x). Only one year between 1900 and 1940 is divisible by 30, 1920 (the year he died). The year he was born can now be found: 1920 * (29/30) = 1856. So in 1900 he was (1900 - 1856) = 44 years old. Did you answer this riddle correctly?
YES NO
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