AMC10 2002 A

AMC10 2002 A · Q12

AMC10 2002 A · Q12. It mainly tests Linear equations, Rates (speed).

Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

Earl E. Bird 先生每天早上正好 8:00 离开家去上班。当他平均速度为 40 英里/小时时，到达工作地点晚了 3 分钟。当平均速度为 60 英里/小时时，早到了 3 分钟。Bird 先生应该以多少平均速度（英里/小时）开车才能准时到达工作地点？

(A) 45 45

(B) 48 48

(D) 55 55

(E) 58 58

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let $t$ be the number of hours Mr. Bird must travel to arrive on time. Since three minutes is the same as $0.05$ hours, $40(t+0.05)=60(t-0.05)$. Thus, $40t+2=60t-3$, so $t=0.25$. The distance from his home to work is $40(0.25+0.05)=12$ miles. Therefore, his average speed should be $12/0.25=48$ miles per hour. OR Let $d$ be the distance from Mr. Bird’s house to work, and let $s$ be the desired average speed. Then the desired driving time is $d/s$. Since $d/60$ is three minutes too short and $d/40$ is three minutes too long, the desired time must be the average, so $\dfrac{d}{s}=\dfrac{1}{2}\left(\dfrac{d}{60}+\dfrac{d}{40}\right).$ This implies that $s=48$.

（B）设 $t$ 为伯德先生为了准时到达必须行驶的小时数。由于 3 分钟等于 $0.05$ 小时，$40(t+0.05)=60(t-0.05)$。因此， $40t+2=60t-3$，所以 $t=0.25$。他从家到工作的距离是 $40(0.25+0.05)=12$ 英里。因此，他的平均速度应为 $12/0.25=48$ 英里/小时。或者设 $d$ 为伯德先生从家到工作的距离，设 $s$ 为所需的平均速度。则所需的行驶时间为 $d/s$。由于 $d/60$ 比所需时间短 3 分钟，而 $d/40$ 比所需时间长 3 分钟，所需时间必须是两者的平均值，所以 $\dfrac{d}{s}=\dfrac{1}{2}\left(\dfrac{d}{60}+\dfrac{d}{40}\right)。$ 由此可得 $s=48$。

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