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−$$x$$)=2$$x$$−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(2$$x$$−128)

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

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

8(2$$x$$−128) = 32

$$x=66$$

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

Alternate solution: 

Since, all the players lost in the reverse alphabetical order, below table gives the result of which player lost in each round. 

Now, each player had ₹32 at the end of Round 4. It means that the total money with the friends in the game is $$₹32\times4=₹128$$

Now, Fardeen lost in round 4. It means that Fardeen doubles the amount of money of others. 

Total money with Frank, Faulad and Farhan after Round 4 = $$₹32\times3=₹96$$

Total money with Frank, Faulad and Farhan after Round 3 = $$\dfrac{₹96}{2}=₹48$$

Money with Fradeen after Round 3 = $$₹128-₹48=₹80$$

Now, Farhan lost in Round 3. 

Total money with other 3 friends after Round 2 = $$₹\left(8+8+40\right)=₹56$$

Total money with Farhan after Round 2 = $$₹\left(128-56\right)=₹72$$

Similarly, Faulad lost in Round 2. 

Total money with other 3 friends after Round 1 = $$₹\left(4+36+20\right)=₹60$$

Total money with Faulad after Round 1 = $$₹\left(128-60\right)=₹68$$

Similarly, Frank lost in Round 1.

Total money with other 3 friends at the start = $$₹\left(34+18+10\right)=₹62$$

Total money with Frank at the start = $$₹\left(128-62\right)=₹66$$

The total money with each of the friend after each round is: 

Hence, Frank starts with the most money at the beginning of the game. 

$$\therefore\ $$ The required answer is C.