2014 AMC 10B 第 6 题

先试着解答 2014 AMC 10B 第 6 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 2014 AMC 10B 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。

6.

Orvin 去商店时带的钱刚好够买 3030 个气球。到店后,他发现气球正在促销:按原价买 11 个气球,第二个可减价 13\frac{1}{3}。Orvin 最多可以买多少个气球?
Orvin went to the store with just enough money to buy 3030 balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 11 balloon at the regular price and get a second at 13\frac{1}{3} off the regular price. What is the greatest number of balloons Orvin could buy?
3333
3434
3636
3838
3939
答案:C
知识点:比与比例钱币配对与分组
小提示:
把每个原价气球和一个折扣气球配成一组。
Pair each full-price balloon with its discounted companion
大提示:
两个气球的促销组合花费等于 53\frac53 个原价气球。
A two-balloon sale pair costs 53\frac53 regular balloons
解答:
每一对气球的价格是一个气球原价的 1+23=531+\frac23=\frac53 倍。
因此,带着刚好够买 3030 个原价气球的钱,Orvin 可以买 30⋅65=3630\cdot\frac65=36 个气球。
所以正确答案是 C。
Each pair of balloons costs 1+23=531+\frac23=\frac53 times the regular price of one balloon.
Thus, with the money for 3030 regular-price balloons, Orvin can buy 30⋅65=3630\cdot\frac65=36 balloons.
Thus, the correct answer is C .
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