AMC8 2009

AMC8 2009 · Q14

AMC8 2009 · Q14. It mainly tests Fractions, Rates (speed).

Austin and Temple are 50 miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging 60 miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged 40 miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

奥斯汀和坦普沿 35 号州际公路相距 50 英里。邦妮开车从奥斯汀到坦普女儿家，平均时速 60 英里。把车留给女儿后，邦妮乘坐巴士沿同一路线返回奥斯汀，返回行程平均时速 40 英里。往返行程的平均速度是多少英里每小时？

(A) 46 46

(B) 48 48

(D) 52 52

(E) 54 54

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Find the time traveling to Temple by dividing the distance, 50 miles, by the rate, 60 miles per hour: $\frac{50}{60}=\frac{5}{6}$ hours. Find the time returning by dividing the distance, 50 miles, by the rate, 40 miles per hour: $\frac{50}{40}=\frac{5}{4}$ hours. Find the average speed for the round trip by dividing the total distance, $2\cdot 50=100$ miles, by the total time, $\frac{5}{6}+\frac{5}{4}=\frac{10}{12}+\frac{15}{12}=\frac{25}{12}$ hour. The average speed is $\frac{100}{\frac{25}{12}}=100\left(\frac{12}{25}\right)=48$ miles per hour. NOTE: The harmonic mean $h$ of 2 numbers $a$ and $b$ is found using the formula $h=\frac{2ab}{a+b}$. The harmonic mean is the average rate if the same distance is traveled at two different rates. If $a=60$ and $b=40$, then $h=\frac{2\cdot 60\cdot 40}{60+40}=\frac{4800}{100}=48$ miles per hour.

答案（B）：去 Temple 的行程时间可用距离 50 英里除以速度 60 英里/小时：$\frac{50}{60}=\frac{5}{6}$ 小时。返程时间可用距离 50 英里除以速度 40 英里/小时：$\frac{50}{40}=\frac{5}{4}$ 小时。往返平均速度等于总路程除以总时间；总路程为 $2\cdot 50=100$ 英里，总时间为 $\frac{5}{6}+\frac{5}{4}=\frac{10}{12}+\frac{15}{12}=\frac{25}{12}$ 小时。平均速度为 $\frac{100}{\frac{25}{12}}=100\left(\frac{12}{25}\right)=48$ 英里/小时。注：两个数 $a$ 和 $b$ 的调和平均数 $h$ 的公式为 $h=\frac{2ab}{a+b}$。当以两种不同速度走相同距离时，调和平均数就是平均速度。若 $a=60$ 且 $b=40$，则 $h=\frac{2\cdot 60\cdot 40}{60+40}=\frac{4800}{100}=48$ 英里/小时。

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