Little Airy Creatures
We are little airy Creatures,
All of diffrent Voice and Features,
One of us in Glass is set,
One of us youll find in Jet,
Tother you may see in Tin,
And the fourth a Box within,
If the fifth you should pursue
It can never fly from you.
What are we?
All of diffrent Voice and Features,
One of us in Glass is set,
One of us youll find in Jet,
Tother you may see in Tin,
And the fourth a Box within,
If the fifth you should pursue
It can never fly from you.
What are we?
Hint:
The Air You Breathe Riddle
I am the element found in just about
everything. I am the air you breathe, and support the growth of all living things. What am I?
everything. I am the air you breathe, and support the growth of all living things. What am I?
Hint:
Prince Age Riddle
What are their ages?
Hint:
Current Future Past
Princess x 2z (x+y)/2
Prince y x z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x y = d
2z x = d
x/2 + y/2 z = d
I then created a matrix and solved it using row reduction.
x y z
1 -1 0 d
-1 0 2 d
.5 .5 -1 d
It reduced to:
x y z
1 0 0 4d
0 1 0 3d
0 0 1 5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess:
4
8
16
24
32
40
48
56
64
72
80
Prince:
3
6
12
18
24
30
36
42
48
54
60 Did you answer this riddle correctly?
YES NO
Princess x 2z (x+y)/2
Prince y x z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x y = d
2z x = d
x/2 + y/2 z = d
I then created a matrix and solved it using row reduction.
x y z
1 -1 0 d
-1 0 2 d
.5 .5 -1 d
It reduced to:
x y z
1 0 0 4d
0 1 0 3d
0 0 1 5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess:
4
8
16
24
32
40
48
56
64
72
80
Prince:
3
6
12
18
24
30
36
42
48
54
60 Did you answer this riddle correctly?
YES NO
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
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
Richest Air Riddle
Hint:
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
Tell Us What You See
Still can't see it? Look harder!
Hint: Stare at the white contrast.
3 Gallon Jug And 5 Gallon Jug
The problem is to fill one of the jugs with exactly 4 gallons of water. How do you do it?
You've got to defuse a bomb by placing exactly 4 gallons (15 L) of water on a sensor. The problem is, you only have a 5 gallon (18.9 L) jug and a 3 gallons (11 L) jug on hand! This classic riddle, made famous in Die Hard 3.
Hint:
Fill the 5-jug up completely. There will be, of course, 5 gallons in the 5-jug. You must fill all the gallons up to the top, otherwise you don't actually know how much you have.
Use the water from the 5-jug to fill up the 3-jug. You're left with 3 gallons in the 3-jug and 2 gallons in the 5-jug.
Pour out the 3-gallon jug. You're left with nothing in the 3-jug and 2 gallons in the 5-jug.
Transfer the water from the 5-jug to the three jug. You're left with 2 gallons in the 3-jug. And nothing in the 5-jug.
Fill up the 5-jug completely. You now have 2 gallons in the 3-jug and 5 in the 5-jug. This means that there is 1 gallon (3.8 L) of space left in the 3-jug.
Use the water from the 5-jug to fill up the 3-jug. Fill up the last gallon of space in the 3-jug with the water from the 5-jug. This leaves you with 3 gallons in the 3-jug, and 4 gallons in the 5-jug.
Fill the 3-jug completely with water. You now have 3 gallons (11.4 L) of water.
Transfer this water into the 5-jug. You now have nothing in the 3-jug, and 3 gallons (11.4 L) in the 5-jug.
Re-fill the 3-jug with water. You now have 3 gallons (11.4 L) in the 3-jug and 3 gallons in the 5-jug.
Fill the 5-jug with water from your 3-jug. You now have 1 gallon (3.8 L) in the 3-jug and 5 gallons (18.9 L) in the 5-jug. This is because, in the last step, you only had 2 gallons (7.6 L) of space left over, so you could only pour 2 gallons.
Pour out the 5-jug and refill it with your 1 gallon. You now have nothing in the 3-jug and 1 gallon in the 5-jug
Fill up the 3-jug. You now have 3 gallons (11.4 L) in the 3-jug and 1 in the 5-jug.
Transfer the 3 gallons (11.4 L) of water into the 5-jug to end up with 4 gallons (15.1 L). Simply pour over your three gallons into the 5-jug, which only had 1 gallon (3.8 L) in it previously. 1+3=4, and a successfully defused bomb. Did you answer this riddle correctly?
YES NO
Use the water from the 5-jug to fill up the 3-jug. You're left with 3 gallons in the 3-jug and 2 gallons in the 5-jug.
Pour out the 3-gallon jug. You're left with nothing in the 3-jug and 2 gallons in the 5-jug.
Transfer the water from the 5-jug to the three jug. You're left with 2 gallons in the 3-jug. And nothing in the 5-jug.
Fill up the 5-jug completely. You now have 2 gallons in the 3-jug and 5 in the 5-jug. This means that there is 1 gallon (3.8 L) of space left in the 3-jug.
Use the water from the 5-jug to fill up the 3-jug. Fill up the last gallon of space in the 3-jug with the water from the 5-jug. This leaves you with 3 gallons in the 3-jug, and 4 gallons in the 5-jug.
Fill the 3-jug completely with water. You now have 3 gallons (11.4 L) of water.
Transfer this water into the 5-jug. You now have nothing in the 3-jug, and 3 gallons (11.4 L) in the 5-jug.
Re-fill the 3-jug with water. You now have 3 gallons (11.4 L) in the 3-jug and 3 gallons in the 5-jug.
Fill the 5-jug with water from your 3-jug. You now have 1 gallon (3.8 L) in the 3-jug and 5 gallons (18.9 L) in the 5-jug. This is because, in the last step, you only had 2 gallons (7.6 L) of space left over, so you could only pour 2 gallons.
Pour out the 5-jug and refill it with your 1 gallon. You now have nothing in the 3-jug and 1 gallon in the 5-jug
Fill up the 3-jug. You now have 3 gallons (11.4 L) in the 3-jug and 1 in the 5-jug.
Transfer the 3 gallons (11.4 L) of water into the 5-jug to end up with 4 gallons (15.1 L). Simply pour over your three gallons into the 5-jug, which only had 1 gallon (3.8 L) in it previously. 1+3=4, and a successfully defused bomb. Did you answer this riddle correctly?
YES NO
12 Islanders Teeter Totter Riddle
You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.
How can you find out which islander is the one that has a different weight?
Hint:
Six on one side - six on the other = one side is heavier.
Take the heavier six men, divide them into three and three (random).
Three on one side - three on the other = one side will one heavier.
Divide that three men from the heavier side side, have one on one side - one on the other.
Two results can determine which of the last three men weight is a different weight than each other.
With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.
Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.
You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."
Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less. Did you answer this riddle correctly?
YES NO
Take the heavier six men, divide them into three and three (random).
Three on one side - three on the other = one side will one heavier.
Divide that three men from the heavier side side, have one on one side - one on the other.
Two results can determine which of the last three men weight is a different weight than each other.
With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.
Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.
You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."
Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less. Did you answer this riddle correctly?
YES NO
Lighter Than Air Riddle
Hint:
Two Zero And Two Four Riddle
Hint:
B) 2024
When you pronounce a number say, 3006, it is pronounced as three thousand six. But, it is not pronounced as three two zero and six. Because, it will result in 3206. It might be grammatically but mathematically wrong or vice versa.
The pronunciation does not say how many but what the number is at the particular position.
Hence, 2024 has two zero and two four. Did you answer this riddle correctly?
YES NO
When you pronounce a number say, 3006, it is pronounced as three thousand six. But, it is not pronounced as three two zero and six. Because, it will result in 3206. It might be grammatically but mathematically wrong or vice versa.
The pronunciation does not say how many but what the number is at the particular position.
Hence, 2024 has two zero and two four. Did you answer this riddle correctly?
YES NO
What Comes Immediately Before Today Riddle
Hint: Determine what "Today" is, then work backwards until you determine the day in question.
MONDAY. "Today" is Sunday. Now, starting at Sunday and working backwards (from "the day before yesterday"), we have the following: - 2 + 3 -2 + 1 + 2 - 1 ...So, we go from Sunday to Fri. to Mon. to Sat. to Sun. to Tues. to MONDAY. Did you answer this riddle correctly?
YES NO
YES NO
12 Pills Riddle
Hint:
E = easier in "1", H = heavier in "1". 1: Weight 4:4. If they balance go to "2", if they don't balance, go to "3". 2: Balance 1:1 of the pills you didn't weight yet. Then weight one you didn't weight and one you did weight. If they balanced in the first weighing, and balanced in the second weighing, the last pill is the right one. If they balanced in the first weighing and didn't balance in the second, the one you didn't use before is the right pill. If they didn't balance at all, it's the pill you weighed twice. If they didn't balance in the first weighing, but balanced in the second, it is the first pill. 3: Weight EHH : EHH. If they balance, weight one you already weighed, with an unweighed and go to "4". If they don't balance go to "5". 4: If they balance, the one you didn't weight at all is the right pill. If they don't balance, the one you only weighed once is the right one. 5: Give away every pill that was once easier AND once heavier. You should only have EHH left. Weight H:H. If they balance, E is the right one. If the don't balance, the one which was only heavier the whole time, is the right pill. Did you answer this riddle correctly?
YES NO
YES NO
I Can Be Hot I Can Be Cold Riddle
Hint:
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