John and his team plan to rob a safe. They got just one chance to break the code else the local police will be informed. Below are clues:
A) Exactly one number is perfectly placed: 9 8 1
B) Everything is incorrect: 9 2 4
C) Two numbers are part of the code of the safe but are wrongly placed: 0 9 3
D) One number is part of the code of the safe but is wrongly placed: 1 4 7
E) One number is part of the code of the safe but is wrongly placed: 7 8 3

Joseph buys three kinds of chocolates for 100 rupees. The first one is priced at 5 rupees, second one at 3 rupees and third one at 0.5 rupees.

If he bought 100 chocolates in total, how many pieces do you think he bought each chocolate?

A new clothing store has a unique method of pricing items. A vest costs $20, a tie costs $15, a blouse costs $30, and underwear costs $45. How much would pants cost?

What word is pronounced the same if you take away four of its five letters?

It's a 7-letter word.
If we remove 1 letter from it, it remains the same.
If we remove 2 letters from it, it remains the same.
If we remove 3 letters from it, it remains the same.
If we remove all the letters from it, still it remains the same.
What is it?

A very interesting trivia question for all.

Why do men's shirts have their buttons on the right side and women's have buttons on the left-hand side?

Arrange the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 above and below the division line in a manner that the thus formed fractions equal to 1/3.

(You can use one number only once)

You have $100 with you and you have to buy 100 balls with it. 100 is the exact figure and you can't go below or above the numbers and you have to use the entire $100. If there is no kind of tax applied how many of each of the following balls will you be able to buy:

Green Balls costing $6
Yellow Balls costing $3
Black Balls costing $0.10

Now, how many of each must you buy to fulfil the condition given?

what's the probability of getting a king or a queen from a pack of 52 cards?

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.