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2018 AMC 10A Problem 8

Problem 8 of 25EasierAlgebraArithmetic

Joe has a collection of 2323 coins, consisting of 55-cent coins, 1010-cent coins, and 2525-cent coins. He has 33 more 1010-cent coins than 55-cent coins, and the total value of his collection is 320320 cents. How many more 2525-cent coins does Joe have than 55-cent coins?

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Solution

Let xx be the number of 55-cent coins that Joe has. Then the number of 1010-cent coins he has is x+3.x + 3. Therefore, Joe has 23−x−(x+3)=20−2x 23 - x - (x + 3) = 20 - 2x 2525-cent coins. The total value of all these coins is 5x+10(x+3)+25(20−2x) 5x + 10(x + 3) + 25(20 - 2x) =530−35x.= 530 - 35x. We know that 530−35x=320⇒x=6. 530 - 35x = 320 \Rightarrow x = 6. This means that Joe has 20−2⋅6=820 - 2 \cdot 6 = 8 2525-cent coins. Therefore, he has 8−6=28 - 6 = 2 more 2525-cent coins than 55-cent coins. Thus, C is the correct answer.
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