Puzzle: A Visit to the Casino!

There is a casino and it has 4 gates, let say A, B, C and D. 

Now the condition is that every time you enter casino you have to pay $5 and every time you leave the casino, you again have to pay $5. Also, whenever you enter the casino whatever amount you have with you will get double.

Now you enter the casino through gate A and come out through gate B, again you go inside casino from gate C and come out of gate D, at the end of this process you should be left with no money? 

So calculate how much money you should carry with you when you enter the Casino?

THIS should be the amount that you need to carry... 

Solution: Amount Needed for a Casino Visit

What is the puzzle?

Let's suppose you have amount 'x' initially in your wallet.

1. On paying $5 for entry at gate A amount left is x - 5 .

2. After entry into the casino it double and becomes 2 ( x - 5 ) = 2x - 10.

3. For exit at gate B you pay $5 and the amount left with you is 2x - 10 - 5 = 2x - 15.

4. Again, for the entry at gate C, you pay $5 more. So the amount with you will be  - 
    2x - 15 - 5 = 2x - 20.

5. The amount is doubled to become 2(2x - 20) = 4x - 40 after the entry into casino.

6. Now, you have to pay $5 one more time to have exit via gate D. Hence, the amount 
left will be 4x - 40 - 5 = 4x - 45.

7. As per given condition, the amount that you should have on exit must be 0.

Hence, 4x - 45 = 0 i.e. 4x = 45. Therefore, x = 11.25. 

So you should carry $11.25 before you enter into the casino to have $0 after exit out of the casino.

Let's verify this one with the amount at each of stages above.

1. Amount = 11.25 - 5 = 6.25

2. Amount = 2 x 6.25 = 12.50

3. Amount = 12.50 - 5 = 7.50

4. Amount = 7.50 - 5 = 2.50

5. Amount = 2 x 2.50 = 5 

6. Amount = 5 - 5 = 0

7. Amount = 0