2015 AMC 8 第 20 题

先试着解答 2015 AMC 8 第 20 题,然后核对你的答案与视频与精心整理的解答,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 2015 AMC 8 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。

20.

拉尔夫去商店买了 1212 双袜子,总价 $24\$24。他买的袜子中有些每双 $1\$1,有些每双 $3\$3,有些每双 $4\$4。如果每种袜子他至少买一双,那么拉尔夫买了多少双 $1\$1 的袜子?
Ralph went to the store and bought 1212 pairs of socks for a total of $24\$24. Some of the socks he bought cost $1\$1 a pair, some of the socks he bought cost $3\$3 a pair, and some of the socks he bought cost $4\$4 a pair. If he bought at least one pair of each type, how many pairs of $1\$1 socks did Ralph buy?
44
55
66
77
88
答案:D
知识点:方程组模运算
小提示:
设价格为 $1\$1、$3\$3 和 $4\$4 的袜子双数为变量
Let the numbers of $1,\$1, $3,\$3, and $4\$4 pairs be variables.
大提示:
用总价方程减去总双数方程
Subtract the pair-count equation from the cost equation.
视频讲解:
解答视频缩略图
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文字解答:
设 aa、bb、cc 分别为 $1\$1、$3\$3、$4\$4 的袜子双数。则 a+b+c=12,a+b+c=12\text{,}a+3b+4c=24。a+3b+4c=24\text{。}
两式相减得 2b+3c=122b+3c=12。因为每种至少买一双,所以 b>0b>0、c>0c>0。又 3c<123c<12,所以 c<4c<4。从方程模 22 看,cc 必须是偶数,所以 c=2c=2。
于是 2b+6=122b+6=12,得 b=3b=3,所以 a=12−3−2=7a=12-3-2=7。
所以正确答案是 D。
Let a,a, b,b, and cc be the numbers of $1,\$1, $3,\$3, and $4\$4 pairs, respectively. Then a+b+c=12,a+b+c=12, a+3b+4c=24.a+3b+4c=24.
Subtracting gives 2b+3c=12.2b+3c=12. Since at least one pair of each type was bought, b>0b>0 and c>0.c>0. Also 3c<12,3c<12, so c<4.c<4. Modulo 2,2, the equation gives cc even, so c=2.c=2.
Then 2b+6=12,2b+6=12, so b=3,b=3, and a=12−3−2=7.a=12-3-2=7.
Thus, D is the correct answer.
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