Two pirates were playing with golden coins. At the start of the game, the first pirate lost half of his coins to the second pirate, then the second pirate lost half of his coins to the first one, then again the first lost half of his coins to the second pirate. At the end of the game, the first pirate had 15 coins, and the second had 33 coins. How many coins did the first pirate have initially?

Question

Mathematics

Lainey Fitzpatrick

14 December, 02:28

Two pirates were playing with golden coins. At the start of the game, the first pirate lost half of his coins to the second pirate, then the second pirate lost half of his coins to the first one, then again the first lost half of his coins to the second pirate. At the end of the game, the first pirate had 15 coins, and the second had 33 coins. How many coins did the first pirate have initially?

+3

Answers (2)

Know the Answer?

Genevieve Werner · Answer 1

24 coins

Step-by-step explanation:

im to lazy to explain

Miles Cochran · Answer 2

24 coins

Step-by-step explanation:

Start:

first pirate - -> 1/2 coins

second pirate - -> 3/2 coins

Next:

first pirate - -> 1/2 + 3/4 = 5/4 coins

second pirate - -> 3/2 - 3/4 = 3/4 coins

Next:

first pirate - -> 5/4 - 5/8 = 5/8 coins

second pirate - -> 3/4 + 5/8 = 11/8 coins

So:

If 5/8 coins is 15, then 1/8 coins is 3, and 8/8 coins is 24 coins.