This one seems hard, but it's simple! There is a casino and it has 4 gates (let's name them as gate A, B, C and D). Every time you come inside the casino, you have to pay \$5 and every time you leave the casino, you again have to pay \$5. Also know that, whatever amount of money you carry with you inside the casino, it doubles. For example if you carry \$5, it will become \$10. You have to go inside the casino through gate A, come out of B, again go inside the casino through gate C and finally come out of gate D. Now, how much money should you carry inside the casino so that when you finally come out of the gate D, you should be left with no (zero amount) money?

Working backwards: You need \$0 when exiting gate D, so you had \$0 + $5 = $5 after paying at D. Before doubling at gate C exit, you had $5/2 = $2.50. After paying at C entry, you had $2.50 + $5 = \$7.50. Before doubling at B exit, you had \$7.50/2 = \$3.75. After paying at B exit, you had \$3.75 + $5 = $8.75. Before doubling at A entry, you had $8.75/2 = $4.375. After paying at A entry, you started with $4.375 + $5 = \$9.375. Let me verify: Start with \$11.25, pay $5 = $6.25, double = $12.50, pay $5 = $7.50, double = $15, pay $5 = $10, double = $20, pay $5 = \$15. That's not zero. Let me recalculate with \$11.25: After A entry: $11.25 - $5 = \$6.25. After doubling inside: \$12.50. After B exit: $12.50 - $5 = \$7.50. After C entry doubling: \$15. After D exit: $15 - $5 = $10. Not zero. The correct answer requires working backward more carefully.