WHATA LONG AND HARD R R RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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?
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Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.


Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.

It works like this ('-' means 'pass'):

Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose

Result: 75% chance of winning! Did you answer this riddle correctly?
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He Philosopher winds the grandfather clock to a random time right before leaving, 9:00 for example. Although this is not the right time, the clock can now be used to measure elapsed time. As soon as he arrives at his friend's house, the Philosopher looks at the time on his friend's clock. Let's say the time is 7:15. He stays overnight and then, before leaving in the morning, he looks at the clock one more time. Let's say the time is now 10:15 (15 hours later). When the Philosopher arrives home, he looks at his grandfather clock. Let's say his clock reads 12:40. By subtracting the time he set it to when he left (9:00) from the current time (12:40) he knows that he has been gone for 15 hours and 40 minutes. He knows that he spent 15 hours at his friends house, so that means he spent 40 minutes walking. Since he walked at the same speed both ways, it took him 20 minutes to walk from his friend's home back to his place. So the correct time to set the clock to in this example would therefore be 10:15 (the time he left his friend's house) + 20 minutes (the time it took him to walk home) = 10:35. Did you answer this riddle correctly?
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He shot his reflection in the bathroom mirror. Did you answer this riddle correctly?
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Alone.

Behead me and I am Lone.
Behead me again and I am One. Did you answer this riddle correctly?
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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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Just imagine a 5 pointed star, and then plant one tree at each point, and one tree where the sides intersect.

There are actually several distinct solutions. All of them can be constructed as follows:

Draw a nice long straight line.
Draw a second straight line that intersects the first.
Draw three more straight lines making sure each line intersects all the lines youve already drawn and avoiding any of the previous points of intersection. That is, no three lines should intersect at the same point.

With the first four lines, theres only one topologically distinct configuration, but by varying the position of the fifth line, several different distinct configurations can be created. Did you answer this riddle correctly?
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Total Money taken = $100($50+$50)

Now,

Bag's Price = $ 97

Remaining Amount = $100 - $97

= $ 3

Returned = $ 1 + $ 1

=$2

In pocket = $1

Total money owed = $100- ( Returned amount)

= $98( Bag's amount and reserved amount)

So, it was a calculation mistake. Did you answer this riddle correctly?
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You don't knock on your own hotel door and the man did. Did you answer this riddle correctly?
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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?
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Ask one of the gaurds 'Which door would the other guard say leads to paradise?' If you ask the truthful one he will say the lying guard would say the wrong door. If you ask the lying guard he would say the truthful gaurd would pick the wrong door as well (since he is lying). So no matter what you could choose the door opposite of what they say and end up in paradise. Did you answer this riddle correctly?
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The third man is in a Wheel Chair. So he rolls out on Wheel Chair, instead of Walking out. Did you answer this riddle correctly?
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The liars are as follows:

1.Julia
2.Jack

The rest are telling the truth

1.Jane
2.Joey
3.John

Jack says out of the 3 names listed one is lying. That was a lie. Therefore those are the three that can not tell a lie... Did you answer this riddle correctly?
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Dice all look the same. On a die, the 1, 3, and 5 all have a dot in the center. The 3 has 2 dots on either side of the center dot, and the 5 has 4 dots around the center dot. Johnny simply counted the number of dots around the outside. A "3" has 2 "petals around the rose, or polar bears around an ice hole." The "5" has 4 "petals" or "polar bears." Roll some dice and it will become clear!! Did you answer this riddle correctly?
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I am a bad conductor! Did you answer this riddle correctly?
YES  NO