AMC10 2002 B

AMC10 2002 B · Q21

AMC10 2002 B · Q21. It mainly tests Rates (speed), Arithmetic misc.

Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?

安迪的草坪面积是贝丝草坪面积的两倍，是卡洛斯草坪面积的三倍。卡洛斯的割草机割草速度是贝丝割草机的一半，是安迪割草机三分之一。他们同时开始修剪草坪，谁会最先完成？

(A) Andy Andy

(B) Beth Beth

(D) Andy and Carlos tie for first. Andy和Carlos并列第一

(E) All three tie. 三者并列

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let $A$ be the number of square feet in Andy’s lawn. Then $A/2$ and $A/3$ are the areas of Beth’s lawn and Carlos’ lawn, respectively, in square feet. Let $R$ be the rate, in square feet per minute, that Carlos’ lawn mower cuts. Then Beth’s mower and Andy’s mower cut at rates of $2R$ and $3R$ square feet per minute, respectively. Thus, Andy takes $\frac{A}{3R}$ minutes to mow his lawn, Beth takes $\frac{A/2}{2R}=\frac{A}{4R}$ minutes to mow hers, and Carlos takes $\frac{A/3}{R}=\frac{A}{3R}$ minutes to mow his. Since $\frac{A}{4R}<\frac{A}{3R}$, Beth will finish first.

(B) 设 $A$ 为 Andy 的草坪面积（单位：平方英尺）。则 Beth 的草坪面积为 $A/2$，Carlos 的草坪面积为 $A/3$（单位：平方英尺）。设 $R$ 为 Carlos 的割草机割草速率（单位：平方英尺/分钟）。则 Beth 和 Andy 的割草机速率分别为 $2R$ 和 $3R$ 平方英尺/分钟。因此， Andy 割完草坪需要 $\frac{A}{3R}$ 分钟， Beth 割完草坪需要 $\frac{A/2}{2R}=\frac{A}{4R}$ 分钟，并且 Carlos 割完草坪需要 $\frac{A/3}{R}=\frac{A}{3R}$ 分钟。由于 $\frac{A}{4R}<\frac{A}{3R}$，所以 Beth 会最先完成。

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