Frank, Fardeen, Faulad and Farhan are playing a game where the loser doubles the amount of money that the others have. They manage to play four games. Each player loses a game each as per reverse alphabetical orders of their names. At the end of the game each player was left with ₹32 each. Who started with the most money at the beginning of the game?

At end of the game each player holds = 32

Total money is 4×32=Rs. 128

Let Frank start with Rs. x

Then the total money other 3 have =Rs. (128−x)

Frank lost the first game. He gives Rs. (128−x) to the other 3 players.

Therefore, after the 1st game, money that Frank has

=x−(128−x)=2x−128

Total money other 3 players have =2(128−x)

In the second, third, and fourth games, the other 3 partners will lose, and each time, the loser doubles the amount of money that the others have.

i.e., after 2nd game, Frank money =2(2x−128)

i.e., after 3rd game, Frank money =4(2x−128)

i.e., after the 4th game, Frank money =8(2x−128)

At the start of the game, Frank started with 66rs.