Mine Riddles To Solve
Solving Mine Riddles
Here we've provide a compiled a list of the best mine 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.
Here's a list of related tags to browse: What Am I Riddles Rhyming Riddles School Riddles What Am I Riddles Time Riddles Logic Riddles What Am I Riddles God Riddles Unknown Riddles
The results compiled are acquired by taking your search "mine" and breaking it down to search through our database for relevant content.
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On The Earth I Am Dead Riddle
Hint:
I Come From A Mine And Get Surrounded By Wood Riddle
Hint:
I'm Gone Forever Riddle
What am I?
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
Age Of Three Daughters Riddles
I was visiting a friend one evening and remembered that he had three daughters. I asked him how old they were. The product of their ages is 72, he answered. Quizzically, I asked, Is there anything else you can tell me? Yes, he replied, the sum of their ages is equal to the number of my house. I stepped outside to see what the house number was. Upon returning inside, I said to my host, Im sorry, but I still cant figure out their ages. He responded apologetically, Im sorry, I forgot to mention that my oldest daughter likes strawberry shortcake. With this information, I was able to determine all three of their ages. How old is each daughter?
Hint:
3, 3, and 8. The only groups of 3 factors of 72 to have non-unique sums are 2 6 6 and 3 3 8 (with a sum of 14). The rest have unique sums:
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
The Cheap Mp3 Player
My MP3 player is cheap 'n' nasty and has now broken: it is stuck on 'Shuffle'. In this mode it starts with whatever track you put it on, but then plays tracks in a random order. The only restriction is it never plays a song that's already been played that day.
I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There's nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)
The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don't mind songs coming in between the chapters of my book, as long as the chapters are in order. What's the probability of that happening now?
I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There's nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)
The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don't mind songs coming in between the chapters of my book, as long as the chapters are in order. What's the probability of that happening now?
Hint:
With only 6 tracks on the player:
The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.
With the music on the player as well:
Seeing as I don't care about when the music plays, it doesn't change anything. The answer is still 1/120. Did you answer this riddle correctly?
YES NO
The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.
With the music on the player as well:
Seeing as I don't care about when the music plays, it doesn't change anything. The answer is still 1/120. 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
Mining Pianos Riddle
Hint:
People Cry At My Sight
I have a name that's not mine,
and no one cares about me in their prime.
People cry at my sight,
and lie by me all day and night.
What am I?
and no one cares about me in their prime.
People cry at my sight,
and lie by me all day and night.
What am I?
Hint:
The Parrot Doors Riddle
There are two doors. One door lead to Heaven, while the other leads to Hell. A parrot stands in front of each door. One parrot always tells a lie, while the other always tells the truth. You do not know which parrot or door is which. You are allowed to only ask one question. So, what one question must you ask to determine which door is which, so you can finally go to Heaven? (Hint: The question involves what one parrot would say about the doors.)
Hint:
It doesn't matter which parrot you ask the question to, but the question would be, "What door would the other parrot say is Heaven?". Then you would choose the other door. Did you answer this riddle correctly?
YES NO
YES NO
Spiriting Faultless Pitch
Without a partner, I sit here mutely. My grace and beauty for you to reckon. Bright head above a regal neck, soft curves. And promise of my rich voice do beckon. Im inevitably hollow, the fretful type, but with practice, I could be your soul mate: If you hold me just right, I'll resonate your spiriting faultless pitch, your song to elevate. What could I be?
Hint:
Metal Rods Riddle
How can you determine which one is magnetized and which is not?
Hint:
Solution 1: Touch the end of one bar (A) to the middle of the other bar (B) forming a 'T' shape. If the bars are attracted then bar A is magnetized and if they are not attracted then bar B is the magnet. This is because magnets have fields at the poles (the ends) but not in the middle. So the end would attract and middle would not.
Solution 2: Hang a rod from the ceiling and if it turns north than it is the magnetized rod. Did you answer this riddle correctly?
YES NO
Solution 2: Hang a rod from the ceiling and if it turns north than it is the magnetized rod. Did you answer this riddle correctly?
YES NO
Five Hundred Riddle
Hint:
First In A Family Of Nine Riddle
I came to Hogwarts and graduated 1988,
I came back at the time of Harry's almost, yet terrible fate.
I came in first in a family of nine,
I handle money but it's definitely not mine.
I've got a fang on a part of my body, long hair is my style,
A woman was once staring at me, which was caught by Harry's eye.
I took a desk job and that's where my love started,
I joined the Order of which cannot be parted.
I wear dragon hide on my feet, muggle clothes is what I've got,
My charm would outwit anyone, do you think they have not?
I respect my family with pride, unlike a dear brother that is so uncool,
One quality we share is that we were both Prefects at our dear old school.
Who am I?
I came back at the time of Harry's almost, yet terrible fate.
I came in first in a family of nine,
I handle money but it's definitely not mine.
I've got a fang on a part of my body, long hair is my style,
A woman was once staring at me, which was caught by Harry's eye.
I took a desk job and that's where my love started,
I joined the Order of which cannot be parted.
I wear dragon hide on my feet, muggle clothes is what I've got,
My charm would outwit anyone, do you think they have not?
I respect my family with pride, unlike a dear brother that is so uncool,
One quality we share is that we were both Prefects at our dear old school.
Who am I?
Hint:
Counterfeit Coins Riddle
You are given eight coins and told that one of them is counterfeit. The counterfeit one is slightly heavier than the other seven. Otherwise, the coins look identical. Using a simple balance scale, how can you determine which coin is counterfeit using the scale only twice?
Hint:
First weigh three coins against three others. If the weights are equal, weigh the remaining two against each other. The heavier one is the counterfeit. If one of the groups of three is heavier, weigh two of those coins against each other. If one is heavier, its the counterfeit. If theyre equal weight, the third coin is the counterfeit. Did you answer this riddle correctly?
YES NO
YES NO
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