AMC10 2018 A

AMC10 2018 A · Q8

AMC10 2018 A · Q8. It mainly tests Systems of equations, Money / coins.

Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?

Joe 有 23 枚硬币，包括 5 分币、10 分币和 25 分币。他的 10 分币比 5 分币多 3 枚，总价值 320 分。请问 Joe 的 25 分币比 5 分币多多少枚？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $n$ be the number of 5-cent coins Joe has, and let $x$ be the requested value—the number of 25-cent coins Joe has minus the number of 5-cent coins he has. Then Joe has $(n+3)$ 10-cent coins and $(n+x)$ 25-cent coins. The given information leads to the equations $$ n+(n+3)+(n+x)=23 $$ $$ 5n+10(n+3)+25(n+x)=320. $$ These equations simplify to $3n+x=20$ and $8n+5x=58$. Solving these equations simultaneously yields $n=6$ and $x=2$. Joe has 2 more 25-cent coins than 5-cent coins. Indeed, Joe has 6 5-cent coins, 9 10-cent coins, and 8 25-cent coins.

答案（C）：设 $n$ 为乔拥有的 5 分硬币数量，设 $x$ 为所求值——乔拥有的 25 分硬币数量减去他拥有的 5 分硬币数量。则乔有 $(n+3)$ 枚 10 分硬币和 $(n+x)$ 枚 25 分硬币。由题目信息可得方程 $$ n+(n+3)+(n+x)=23 $$ $$ 5n+10(n+3)+25(n+x)=320. $$ 化简得 $3n+x=20$ 和 $8n+5x=58$。联立解得 $n=6$，$x=2$。乔的 25 分硬币比 5 分硬币多 2 枚。确实，乔有 6 枚 5 分硬币、9 枚 10 分硬币和 8 枚 25 分硬币。

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