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Klio2033 [76]
3 years ago
13

Two pirates were playing with golden coins. During 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 pirate, then again the first lost half of his coins to the second. At the end of the game, the first pirate had 21 coins, and the second had 45 coins. How many coins did the first pirate have initially?
Mathematics
1 answer:
Simora [160]3 years ago
8 0
Answer: 33
1st Pirate 2nd Pirate
Lost 1/2 Gained 1/2
Gained 1/2 Lost 1/2
Lost 1/2 Gained 1/2
End#: 21 End#: 45

Add 21 and 45
Which equals 66
Then divide 66 by two
Which equals 33
So both pirates started off with 33 coins each

So now we apply the losses and gains to those coins

1st Pirate: 33 2nd Pirate: 33
-1/2= 16.5 +1/2= 49.5
+1/2=41.5 -1/2= 24.75
-1/2= 20.75 +1/2= 45.5

Round 20.75 to the Round 45.5 half down:
nearest integer: = 45
= 21
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