2019 AMC 8 第 23 题

先试着解答 2019 AMC 8 第 23 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 2019 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。

23.

Euclid 高中最后一场篮球赛后,统计得知全队总分的 14\dfrac{1}{4} 由 Alexa 得到,27\dfrac{2}{7} 由 Brittany 得到。Chelsea 得了 1515 分。其余 77 名队员中没有人得分超过 22 分。其余 77 名队员一共得了多少分?

After Euclid High School's last basketball game, it was determined that 14\dfrac{1}{4} of the team's points were scored by Alexa and 27\dfrac{2}{7} were scored by Brittany. Chelsea scored 1515 points. None of the other 77 team members scored more than 22 points. What was the total number of points scored by the other 77 team members?

1010

1111

1212

1313

1414

答案:B
知识点:整除性一次方程
难度评级:1490
视频讲解:
解答视频缩略图
Play video

Click to load, then click again to play

文字解答:

设总得分为 xx,所求答案为 yy

由题意可得 x4+2x7+15+y=x. \dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x.

化简得 y+15=13x28. y + 15 = \dfrac{13x}{28}.

因为其余 77 名队员每人的得分不超过 22 分,所以 y14y \leq 14

此外,xx 必须是 28.28. 的倍数。若 x=28,x = 28,y+15=13y=2, y + 15 = 13 \Rightarrow y = -2, 不合要求。

x=228=56,x = 2 \cdot 28 = 56,y+15=26y=11, y + 15 = 26 \Rightarrow y = 11, 符合要求。

下一个可能的总分是 x=84,x=84,这会得到 y=24>14.y=24>14.更大的 2828 的倍数会使 yy 更大,所以 y=11y=11 是唯一可能值。

所以正确答案是 B

Let xx be the total number of points and yy be the desired answer.

Then from the problem statement we get that x4+2x7+15+y=x. \dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x.

Simplifying yields y+15=13x28. y + 15 = \dfrac{13x}{28}.

We know that y14y \leq 14 since none of the 77 team members scored more than 22 points.

We also know that xx must be a multiple of 28.28. If x=28,x = 28, then we get that y+15=13y=2, y + 15 = 13 \Rightarrow y = -2, which is not allowed.

If x=228=56,x = 2 \cdot 28 = 56, then we have that y+15=26y=11, y + 15 = 26 \Rightarrow y = 11, which works.

The next possible total is x=84,x=84, which would give y=24>14.y=24>14. Every larger multiple of 2828 makes yy still larger, so y=11y=11 is the only possible value.

Thus, the correct answer is B.

← 第 22 题#22
完整试卷

其他年份的第 23 题

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8