1985 AMC 12 第 4 题

先试着解答 1985 AMC 12 第 4 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1985 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

4.

一个大袋子里装有一美分、十美分和二十五美分硬币。十美分硬币的数量是一美分硬币的两倍,二十五美分硬币的数量是十美分硬币的三倍。袋中可能装有的钱数是
A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
$306\$306
$333\$333
$342\$342
$348\$348
$360\$360
答案:C
知识点:整除性钱币比与比例
小提示:
设一美分硬币的数量为 pp
Let the number of pennies be pp
大提示:
用 pp 的倍数表示硬币总价值(单位:美分)
Express the total value in cents as a multiple of pp
解答:
若有 pp 枚一美分硬币,则有 2p2p 枚十美分硬币和 6p6p 枚二十五美分硬币。总价值为 p+10(2p)+25(6p)=171pp+10(2p)+25(6p)=171p 美分。所列美元数中,34200=171⋅20034200=171\cdot200 美分是可能的。
所以正确答案是 C。
If there are pp pennies, there are 2p2p dimes and 6p6p quarters. Their total value is p+10(2p)+25(6p)=171pp+10(2p)+25(6p)=171p cents. Of the listed dollar amounts, 34200=171⋅20034200=171\cdot200 cents is possible.
Thus the correct answer is C.
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