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.

I know there are two methods by using three time the same number with a plus(+) operator , you can make sum as 60.

One of them is 20+20+20.

whats the other way ?

You are a cab driver who pools passengers. You pick 3 people from a destination and drop 1 after an hour. 2 people climb aboard at the same time and you drop 3 at the next destination. After some time, you pick 2 passengers only to drop 1 after a short distance where 3 more passengers climb up the cab. You leave the rest of the passengers one by one to their destination and then come back home.

Can you tell the eye color of the cab driver?

In the below-given picture of a pretty girl, there is something wrong. Can you find out what it is?

On a windy rainy night , I was driving in my car.
When i reach the bus stand , i see three people waiting for the bus.

1. An old lady who needs an immediate medical attention
2. My Best Friend
3. Girl whom i love from childhood

My car is a two seat, so can you tell me , what i have done in this situation ?

A B C D E F G H

These are the letters given to you. Now you have to find out the letter that comes two to the right of the letter which is immediately to the left of the letter that comes three to the right of the letter that comes midway between the letter two to the left of the letter C and the letter immediately to the right of the letter F.

Can it be possible to build a house that has all its four walls facing towards the South?

Can you find out the remainder when 3^300 is divided by 5?

Can you identify if it's a murder or a suicide ?

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 ?