Santa Claus Wants To Arrive In New York Riddles To Solve
Solving Santa Claus Wants To Arrive In New York Riddles
Here we've provide a compiled a list of the best santa claus wants to arrive in new york puzzles and riddles to solve we could find.Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
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Arrived On Monday Riddle
A man wants to travel to New York from Miami on horse, this travel would normally take about seven days, but yet he left on Monday and arrived on Monday, how can this be?
Hint:
Santa's Arrived Riddle
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The Capital Of New York
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A Man In New York City
A man in New York City has $10. He spends $6.50 on flowers, and $3 on lunch (hot coffee and a hot dog). He then gets on the subway which will take him 7 stops for 50 cents. But he is forced to get off of the subway just 5 stops away from where he began.
Why is this?
Why is this?
Hint:
When he gets on the subway it is 6 stops away from the end of the line (end of the track). So when it reaches this point it begins to work backwards. So when it goes back one stop he has traveled 7 stops but is only 5 away from where he began. Did you answer this riddle correctly?
YES NO
YES NO
Going To New York Riddle
A old man was going to New York. Along the way he met a man with seven wives. Each wife had seven children. Each child had seven cats. Each cat had seven kittens. Kittens, cats, children, wives. How many people are going to New York?
Hint:
New York Plane Crash Riddle
If a plane carrying passengers from New Jersey crashes in New York, where do you bury the survivors?
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New York Multiplication Riddle
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Dracula In New York Riddle
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Cows Visit New York Riddle
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The Belle Of New York
My first wears my second;
My third might be what my first would acquire if he went to sea.
Put together my one, two, three,
And the belle of New York is the girl for me.
What one word am I?
My third might be what my first would acquire if he went to sea.
Put together my one, two, three,
And the belle of New York is the girl for me.
What one word am I?
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
Santa And A Space Ship
Hint:
Losing A New York Bet
You are hanging around in NYC when a person approaches you.
"Leaving the bald people aside, I can bet a hundred bucks that there are two people living in NYC who have same number of hairs on their heads," he says to you.
You say that you will take the bet. After talking to the man for a couple of minutes, you realize that you have lost the bet.
What did the person say to you that proved his statement ?
"Leaving the bald people aside, I can bet a hundred bucks that there are two people living in NYC who have same number of hairs on their heads," he says to you.
You say that you will take the bet. After talking to the man for a couple of minutes, you realize that you have lost the bet.
What did the person say to you that proved his statement ?
Hint:
This problem can be best solved using the pigeonhole principle.
The argument will go like this:
Assume that all the non-bald people in NYC have different number of hairs on their head. The population is about 9 million and let us assume that there are 8 million among them who are not bald.
Now, those 8 million people need to have different number of hairs. On an average, people have just 100, 000 hairs on their head. If we keep on assuming that there is someone with just one hair, someone with two, someone with three and so on, there will be 7, 900, 00 other people left who will have more than 100, 000 hairs on their head and need different number of hairs.
Now, as per this assumption, if we keep increasing one hair for each person, to make everybody hair different in numbers, we will come across someone with 8, 000, 000 hairs. But that is practically impossible (even 1, 000, 000 is impossible). Thus there must be two people who are having same number of hairs. Did you answer this riddle correctly?
YES NO
The argument will go like this:
Assume that all the non-bald people in NYC have different number of hairs on their head. The population is about 9 million and let us assume that there are 8 million among them who are not bald.
Now, those 8 million people need to have different number of hairs. On an average, people have just 100, 000 hairs on their head. If we keep on assuming that there is someone with just one hair, someone with two, someone with three and so on, there will be 7, 900, 00 other people left who will have more than 100, 000 hairs on their head and need different number of hairs.
Now, as per this assumption, if we keep increasing one hair for each person, to make everybody hair different in numbers, we will come across someone with 8, 000, 000 hairs. But that is practically impossible (even 1, 000, 000 is impossible). Thus there must be two people who are having same number of hairs. Did you answer this riddle correctly?
YES NO
Deep Fried Santa Riddle
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The Ghost And Santa Riddle
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