Where Does Today Come Before Yesterday Riddle
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What Remains When The Roads Are Gone Riddle
A race car driver has completed 12 1/2 laps of a 50-lap race. What fractional part of the race remains?
Hint:
Let's take a look at the explanation of the riddle.
As per the total laps in the race are 50 and the driver has completed 12 1/2 laps. This means, we have to subtract 12 1/2 from the total 50 laps. This equal to 37 1/2 or 37.5
50 - 12 1/2 = 37 1/2 or 37.5
Now, we need to calculate the fractional part of the race remains. For this, we need to divide the remaining laps by total laps, that is, 37 1/2 divide by 50 or 37.5 divided by 50 which will be equal to 0.74 or 3/4.
37 1/2 / 50 = 0.74 or 3/4
Hence, the right answer to the riddle is 3/4 Did you answer this riddle correctly?
YES NO
As per the total laps in the race are 50 and the driver has completed 12 1/2 laps. This means, we have to subtract 12 1/2 from the total 50 laps. This equal to 37 1/2 or 37.5
50 - 12 1/2 = 37 1/2 or 37.5
Now, we need to calculate the fractional part of the race remains. For this, we need to divide the remaining laps by total laps, that is, 37 1/2 divide by 50 or 37.5 divided by 50 which will be equal to 0.74 or 3/4.
37 1/2 / 50 = 0.74 or 3/4
Hence, the right answer to the riddle is 3/4 Did you answer this riddle correctly?
YES NO
A Man Was Doing His Job Riddle
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A Woman Had 2 Sons Riddle
A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins. How could this be so?
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Technical Impaired Elephant
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A Leprechauns Job Riddle
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A Walk In The Desert Riddle
Four men walk into the desert. Suddenly all four are simultaneously knocked out. They awake buried to their heads in the sand unable to look anywhere but straight ahead. They are positioned so that each man sees another's head before him. However between the first and second man there is a separating wall.
So the first man sees only desert. The second man sees only wall. The third man sees another's head and a wall. The fourth man sees two heads and a wall. On top of each mans head is a hat. The underside of each cap is black, but the outside of each cap is either blue or white. Before any of the men can speak, their captors tell them if they speak, they die. However, if any of them can guess the color of their cap on the first try they go free. The captors tell them that there are two blue caps and two white caps.
Being an omniscient observer of the situation, we know that the order of the caps are: blue, white, blue, white. So knowing the perspective of each man in the sand, and that they can only see the color of caps/wall/desert in front of them, which of the four men knows for certain the color of his own cap. More importantly: why?
So the first man sees only desert. The second man sees only wall. The third man sees another's head and a wall. The fourth man sees two heads and a wall. On top of each mans head is a hat. The underside of each cap is black, but the outside of each cap is either blue or white. Before any of the men can speak, their captors tell them if they speak, they die. However, if any of them can guess the color of their cap on the first try they go free. The captors tell them that there are two blue caps and two white caps.
Being an omniscient observer of the situation, we know that the order of the caps are: blue, white, blue, white. So knowing the perspective of each man in the sand, and that they can only see the color of caps/wall/desert in front of them, which of the four men knows for certain the color of his own cap. More importantly: why?
Hint:
The third man. This is because he knows there are only two of each color cap. If the man behind him (the fourth man) saw two caps that were the same color in front of him, he would know that his own must be the opposite. However, because the caps alternate in color. The fourth man has only a 50% chance of getting his hat color correct, so therefore he stays quiet. The third man realizes that the fourth man is quiet because he must not see two caps of the same color in front of him, otherwise the fourth man would say the opposite of the caps in front of him. Therefore, the third man presumes his own cap must be the opposite of the mans in front of him, and his presumption is correct. Under this same logic, after the third man speaks his color hat, the second man, even though he sees only wall, would be the next to go free, because he knows his cap must be the opposite of whichever color the third mans cap was. Did you answer this riddle correctly?
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YES NO
The Awful Date Riddle
A woman and a man go on a date. While on their date, the woman shoots the man, drowns the man, and hangs the man. the next night, the same man and woman go on a date and both of them are 100% fine. how is this possible?
Hint:
The woman shoots a picture of him, drowns the picture to develop it, then she hangs the picture Did you answer this riddle correctly?
YES NO
YES NO
The Day After Tomorrow Riddle
When the day after tomorrow is yesterday, today will be as far from Wednesday as today was from Wednesday when the day before yesterday was tomorrow. What is the day after this day?
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The Dream Job For A Rabbit
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Groups Of Wartime Riddle
I am a group ordinary citizens, called from our jobs to fight and defend our colonies even though I am unskilled with very little training. What wartime group am I?
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A Ponderous House Riddle
I'm a riddle in nine syllables,
An elephant, a ponderous house,
A melon strolling on two tendrils O red fruit,
Ivory, fine timber!
The loaf's big with it's yeasty rising
Money's new minted in this fat purse.
I'm a means, a stage, a cow in calf.
I've eaten a bag of green apples
Boarded the train there's no getting off.
What am I?
An elephant, a ponderous house,
A melon strolling on two tendrils O red fruit,
Ivory, fine timber!
The loaf's big with it's yeasty rising
Money's new minted in this fat purse.
I'm a means, a stage, a cow in calf.
I've eaten a bag of green apples
Boarded the train there's no getting off.
What am I?
Hint:
First Woman Millionaire Riddle
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Solved: 42%
An Island That Has 3 Gods
There is an Island that has 3 gods. One god always tells a lie, and the other always tells the truth. The third god has a random behavior. To top it off, these three gods, being jerks, answer in their own languages such that you are unable to tell which word, between "ja" or "da", means "no" or "yes". You have 3 questions to work out the True god, the false god, and the Random god.
Hint:
Question 1: (To any of the three gods) If I were to ask you "Is that the random god," would your answer be "ja?" (This questions, no matter the answer, will enable you to tell which god is not random i.e. the god who is either False or True)
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" 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?
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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
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