Question

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?

Answer 1

Answer:

Faulad started with Rs. 66

Step-by-step explanation:

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 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, other 3 partners will lose and each time, loser doubles the amount of money that the others have.

i.e., after 2nd game, Fardeen 's money =2(2x−128)

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

i.e., after 4th game, Farhan 's money =8(2x−128)

At start of game Faulad started with 66rs.