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.
There is a barrel with no lid and some wine in it. 'This barrel of wine is more than half full,' said Curly. 'No it's not,' says Mo. 'It's less than half full.' Without any measuring implements and without removing any wine from the barrel, how can they easily determine who is correct?
I have two coins.
* One of the coins is a faulty coin having a tail on both sides of it.
* The other coin is a perfect coin (heads on side and tail on other).
I blindfold myself and pick a coin and put the coin on the table. The face of the coin towards the sky is the tail.
What is the probability that another side is also tail?
Using four sevens (7) and a one (1) create the number 100. Except for the five numerals, you can use the usual mathematical operations (+, -, x, :), root and brackets ()