Below, you will find the mathematical proof that 10 equals 9.99999?. But is that possible or there is something wrong about it? Can you find the error?

x = 9.999999...
10x = 99.999999...
10x - x = 90
9x = 90
x = 10

By including only four matchsticks in below picture, you have to make "Four Triangles And Two Squares"

Prove that the below maths equation is true.

8 + 8 = 91

If,

A = 1

B = 2

C = 3

...

...

Z = 26.

Based on above rule, you need to find an eleven letter word whose letter sum is equal to 52.

You may enter, but you may not come in. I have space, but no room. I have keys, but open no lock. What am I?

Rahul decided to meet Simran so he boards a local train from Bombay station. Just after the station, there is a 1km long tunnel. The train starts and is now accelerating. Rahul is a claustrophobic guy, so what is the best position for him to sit?

There are two dice with empty faces in front of you and a marker. You can mark any number on each of the faces of the two dice, but you have to display all 31 days of the month using the two of them.

Which numbers will you mark on which dice so that you can easily depict all the dates of the month?

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 ?

What does this rebus puzzle means ?

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.