'Ferrari driver' easily beats the 'force driver' in a two-car race. How did Indian newspapers truthfully report so to look as a 'force drive' had outdone the 'Ferrari driver'? Think!!!

The captain of a ship was telling this interesting story: "We travelled the sea far and wide. At one time, two of my sailors were standing on opposite sides of the ship. One was looking west and the other one east. And at the same time, they could see each other clearly." How can that be possible?

What does the below rebus riddle means?

See the given image carefully. What you have to do is move the blue checkers in the position of the black checkers and vice versa. You are only allowed to move the checker to an adjacent empty space. Do it in the least possible moves.

As we know that white starts the game of chess. Can you find the scenario shown in the picture below is possible when all the white pieces are at the original place while the black pawn is not as in the below picture?

A girl is twice as old as her brother and half as old as her father. In 50 years, her brother will be half as old as his father. How old is the daughter now?

Two natural numbers have a sum of less than 100 and are greater than one.

John knows the product of the numbers and Jacob knows the sum of numbers.

The following conversation takes place between them:
John: 'I am not aware of those numbers.'
Jacob: 'I knew you wouldn't be. I am not aware myself.'
John: 'Now I know them!'
Jacob: 'Now I know them, too!'

What are the two numbers?

A man had five children. He had $100 with him to give to his children. He decided to start with the youngest child and then give $2 more than each younger child to his next elder child.

For example, if he gives $x to the youngest child, he will give $(x+2) to the next one, $[(x+2) + 2] to the next one and so on.

Can you find out how much did the youngest one receive?

Rebus Puzzle Definition: A rebus is an illusion device that uses pictures to represent words or parts of words.
What does This Rebus Puzzle Identify?

If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.