Conjoined Fruit Riddle
This fruit is made of two words conjoined
The first part of it is also a tree
The second part is a different fruit
And goes on a pizza from Hawaii
What is it?
The first part of it is also a tree
The second part is a different fruit
And goes on a pizza from Hawaii
What is it?
Hint:
Unlock Your Soul
Hint:
Full Of Holes
Hint:
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
Five Haystacks Riddle
A farmer has five haystacks in one field and four haystacks in another. How many haystacks would he have if he combined them all in one field?
Hint:
One. If he combines all his haystacks, they all become one big stack. Did you answer this riddle correctly?
YES NO
YES NO
Getting Put Into Shoes
This is a body part
That can be found in twos
It is a type of length
And gets put into shoes
They are?
That can be found in twos
It is a type of length
And gets put into shoes
They are?
Hint:
The Quietest Whimper
I talk, but I do not speak my mind
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
Hint:
A Type Of Fruit Riddle
This is a type of fruit
But it is not a lime
There are no other words
With which this word can rhyme
What fruit is it?
But it is not a lime
There are no other words
With which this word can rhyme
What fruit is it?
Hint:
Strive With Wind And Wave Riddle
Off I must strive with wind and wave, battle them both
when under the sea.
I feel out the bottom, a foreign land. In lying still, I am
Strong in the strife;
If I fail in that, they are stronger than I, and
Wrenching me loose, soon put me to rout.
They wish to capture what I must keep. I can master
Them both if my grip holds out,
If the rocks bring succor and lend support, strength
In the struggle. Ask my name
when under the sea.
I feel out the bottom, a foreign land. In lying still, I am
Strong in the strife;
If I fail in that, they are stronger than I, and
Wrenching me loose, soon put me to rout.
They wish to capture what I must keep. I can master
Them both if my grip holds out,
If the rocks bring succor and lend support, strength
In the struggle. Ask my name
Hint:
Trapdoor, Water, Wolf
Hint:
Three Gods Riddle
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
What three questions can you ask?
What three questions can you ask?
Hint:
A possible solution is:
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
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
As A Stone Inside A Tree Riddle
What am I?
Hint:
Trace Your Steps Back Riddle
Hint:
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