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.

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.

Replace the question mark with the correct number.

2 + 2= 4
3 + 3= 18
4 + 4= 48
5 + 5= ?

The chance of Mr John winning the lottery is 10%. All participants lined up and Mr John is 4th in the row. The first three participants lose the lottery.

What is the chance of Mr John now?

When a clock is observed, the hour hand is at a minute mark and the minutes hand is six minutes ahead of it.

When the clock is observed again after some time, the hour hand is precisely on a different minute mark and the minute hand is seven minutes ahead of it.

Can you calculate how much time has elapsed between the two observations?

A bomb exploded in a room full of married couples and no one died. How?

You stand in front of two doors. A guard stands next to each door. You know the following things: one path leads to paradise, the other leads to death. You cannot distinguish between the two doors. You also know that one of the two guards always tells the truth and the other always lies. You have permission to ask one guard one question to discover which door leads to paradise. What one question would you ask to guarantee you enter the door to paradise?

Do You See It?

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 ?

Keyboard words are not in alphabetical order, Why?