You want to boil a two-minute egg. If you only have a three-minute timer (hourglass), a four-minute timer and a five-minute timer, how can you boil the egg for only two minutes?

You have two buckets - one holds exactly 5 gallons and the other 3 gallons. How can you measure 4 gallons of water into the 5 gallon bucket?

(Assume you have an unlimited supply of water and that there are no measurement markings of any kind on the buckets.)

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

You can use any mathematical symbols in the space provided to make all above algebraic expressions true.

In an interview, a boy was asked an unusual question 'How two persons sitting with a table in between them can't see each other?' He was unable to reply. Can you?

I am the fastest animal but cannot climb the tree. What am I?

On my way to St. Ives I saw a man with 7 wives. Each wife had 7 sacks. Each sack had 7 cats. Each cat had 7 kittens. Kittens, cats, sacks, wives. How many were going to St. Ives?

13 decks of cards have been mixed. What is the minimum number of cards that must be taken out from the above-mixed cards to guarantee at least one 'four of a kind?

Solve the rebus riddle in the picture below.

Three cars are driving on a track that forms a perfect circle and is wide enough that multiple cars can pass anytime. The car that is leading in the race right now is driving at 55 MPH and the car that is trailing at the last is going at 45 MPH. The car that is in the middle is somewhere between these two speeds.
Right now, you can assume that there is a distance of x miles between the leading car and the middle car and x miles between the middle car and the last car and also, x is not equal to 0 or 1.
The cars maintain their speed till the leading car catches up with the last car and then every car stops. In this scenario, do you think of any point when the distance between any two pairs will again be x miles i.e. the pairs will be x distance apart at the same time ?

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.