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
    

Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. 

The wheel began operation at 10 in the morning and shut down 30 minutes later. 

What's the maximum number of people that could have taken a ride on the wheel in that time period?

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride

What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A. 

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.