Best riddles for teens

simplecleanlogicinterview

You are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which. How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.
Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile. How this will work? lets take an example. So initially there are 5 heads, so suppose you divide it in 2 piles. Case: P1 : H H T T T P2 : H H H T T Now when P1 will be flipped P1 : T T H H H P1(Heads) = P2(Heads) Another case: P1 : H T T T T P2 : H H H H T Now when P1 will be flipped P1 : H H H H T P1(Heads) = P2(Heads)
79.54 %
40 votes
logicmathtrickystoryclever

Three people check into a hotel room. The bill is $30 so they each pay $10. After they go to the room, the hotel's cashier realizes that the bill should have only been $25. So he gives $5 to the bellhop and tells him to return the money to the guests. The bellhop notices that $5 can't be split evenly between the three guests, so he keeps $2 for himself and then gives the other $3 to the guests. Now the guests, with their dollars back, have each paid $9 for a total of $27. And the bellhop has pocketed $2. So there is $27 + $2 = $29 accounted for. But the guests originally paid $30. What happened to the other dollar?
This riddle is just an example of misdirection. It is actually nonsensical to add $27 + $2, because the $27 that has been paid includes the $2 the bellhop made. The correct math is to say that the guests paid $27, and the bellhop took $2, which, if given back to the guests, would bring them to their correct payment of $27 - $2 = $25.
79.52 %
59 votes
logicmathsimple

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? Distances are unspecified. Speeds are unspecified, but constant. Give a number which represents the answer in minutes.
The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes. Jack must have been walking for 45 minutes.
79.52 %
59 votes
logicmathclean

You are visiting NYC when a man approaches you. "Not counting bald people, I bet a hundred bucks that there are two people living in New York City with the same number of hairs on their heads," he tells you. "I'll take that bet!" you say. You talk to the man for a minute, after which you realize you have lost the bet. What did the man say to prove his case?
This is a classic example of the pigeonhole principle. The argument goes as follows: assume that every non-bald person in New York City has a different number of hairs on their head. Since there are about 9 million people living in NYC, let's say 8 million of them aren't bald. So 8 million people need to have different numbers of hairs on their head. But on average, people only have about 100,000 hairs. So even if there was someone with 1 hair, someone with 2 hairs, someone with 3 hairs, and so on, all the way up to someone with 100,000 hairs, there are still 7,900,000 other people who all need different numbers of hairs on their heads, and furthermore, who all need MORE than 100,000 hairs on their head. You can see that additionally, at least one person would need to have at least 8,000,000 hairs on their head, because there's no way to have 8,000,000 people all have different numbers of hairs between 1 and 7,999,999. But someone having 8,000,000 is an essential impossibility (as is even having 1,000,000 hairs), So there's no way this situation could be the case, where everyone has a different number of hairs. Which means that at least two people have the same number of hairs.
79.52 %
59 votes
logicsimple

How much dirt would be in a hole 6 feet deep and 6 feet wide that has been dug with a square edged shovel?
None. No matter how big a hole is, it's still a hole: the absence of dirt. And those of you who said 36 cubic feet are wrong for another reason, too. You would have needed the length measurement too. So you don't even know how much air is in the hole.
79.52 %
59 votes
logicmathclever

You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square. But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board. Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square. Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
79.52 %
59 votes
logicsimplecleanclever

Your friend pulls out a perfectly circular table and a sack of quarters, and proposes a game. "We'll take turns putting a quarter on the table," he says. "Each quarter must lay flat on the table, and cannot sit on top of any other quarters. The last person to successfully put a quarter on the table wins." He gives you the choice to go first or second. What should you do, and what should your strategy be to win?
You should go first, and put a quarter at the exact center of the table. Then, each time your opponent places a quarter down, you should place your next quarter in the symmetric position on the opposite side of the table. This will ensure that you always have a place to set down our quarter, and eventually your oppponent will run out of space.
79.50 %
89 votes