A game is played by three players. The loser has to triple the money of each of the other players has. Three games are played and each one loses a game. At the end all have the same amount namely Rs.

Question

A game is played by three players. The loser has to triple the money of each of the other players has. Three games are played and each one loses a game. At the end all have the same amount namely Rs. 54. The amount the first loser has at the beginning is

TestHub · Accepted Answer

Let the players' names be A, B, and C. They lose the games in the order of A, B, C.

$A B C A m o u n t a f t er 3^{r d} g am e 545454 A m o u n t b e f ore 3^{r d} g am e \frac{54}{3} = 18 \frac{54}{3} = 18 54 + \frac{2}{3} (54) + \frac{2}{3} (54) = 126 A m o u n t b e f ore 2^{n d} g am e \frac{18}{3} = 6 18 + \frac{2}{3} (18) + \frac{2}{3} (126) = 114 \frac{126}{3} = 42 A m o u n t b e f ore 1^{s t} g am e 6 + \frac{2}{3} (42) + \frac{2}{3} (114) = 110 \frac{114}{3} = 38 \frac{42}{3} = 14$

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