1994 AMC 12 第 16 题

先试着解答 1994 AMC 12 第 16 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 1994 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。

16.

一个袋子里有一些红色弹珠,其余都是蓝色弹珠。若取走一颗红色弹珠,剩余弹珠中红色弹珠占七分之一;若不取走红色弹珠而改为取走两颗蓝色弹珠,剩余弹珠中红色弹珠占五分之一。袋子里原来有多少颗弹珠?
Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
88
2222
3636
5757
7171
答案:B
知识点:systems of equationsproportions
小提示:
设红色弹珠数为 RR,弹珠总数为 TT
Let RR be the red count and TT the total count
大提示:
把两种操作分别写成 R−1T−1=17\frac{R-1}{T-1}=\frac{1}{7} 和 RT−2=15\frac{R}{T-2}=\frac{1}{5}
Translate the two experiments into R−1T−1=17\frac{R-1}{T-1}=\frac{1}{7} and RT−2=15\frac{R}{T-2}=\frac{1}{5}
解答:
两个条件给出 R−1T−1=17\frac{R-1}{T-1}=\frac17 和 RT−2=15\frac{R}{T-2}=\frac15。因此 T=7R−6T=7R-6 且 T=5R+2T=5R+2。于是 R=4R=4,T=22T=22。所以正确答案是 B。
The two conditions give R−1T−1=17\frac{R-1}{T-1}=\frac17 and RT−2=15.\frac{R}{T-2}=\frac15. Thus T=7R−6T=7R-6 and T=5R+2.T=5R+2. Hence R=4R=4 and T=22.T=22. Thus the correct answer is B.
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