1951 AMC 12 第 50 题

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50.

汤姆、迪克和哈里一起开始一段 100100 英里的旅程。汤姆与哈里乘汽车,以每小时 2525 英里的速度前进;迪克则以每小时 55 英里的速度步行。行驶一段距离后,哈里下车,改以每小时 55 英里的速度继续步行;汤姆掉头去接迪克,并使迪克与哈里同时到达终点。整个旅程所需的小时数为:

Tom, Dick, and Harry started out on a 100100-mile journey. Tom and Harry went by automobile at the rate of 2525 mph, while Dick walked at the rate of 55 mph. After a certain distance, Harry got off and walked on at 55 mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:

55

66

77

88

以上答案均不正确

None of these answers

答案:D
知识点:路程、速度与时间方程组
难度评级:2100
小提示:

将汽车的行程分为最初向前行驶、返回接迪克以及最后再次向前行驶三段

Separate the car’s motion into the first forward trip, the return for Dick, and the final forward trip

大提示:

若三段时间为 t1,t2,t3t_1,t_2,t_3,分别为汽车、迪克和哈里写出一个路程为 100100 英里的方程

If those times are t1,t2,t3t_1,t_2,t_3, write one 100100-mile equation for the car, Dick, and Harry

解答:

t1,t2,t3t_1,t_2,t_3 分别为哈里下车前汽车行驶的时间、汽车返回接迪克的时间以及汽车载着迪克向前行驶的时间。汽车、迪克和哈里各自行进 100100 英里,因此 25t125t2+25t3=100,5t1+5t2+25t3=100,25t1+5t2+5t3=100 \begin{aligned} 25t_1-25t_2+25t_3&=100,\\ 5t_1+5t_2+25t_3&=100,\\ 25t_1+5t_2+5t_3&=100 \end{aligned}\text{。}各式除以 55 并求解,得到 t1=3, t2=2, t3=3t_1=3,\ t_2=2,\ t_3=3。因此,他们共同的行程时间为 t1+t2+t3=8t_1+t_2+t_3=8 小时。

因此,正确答案是 D

Let t1,t2,t3t_1,t_2,t_3 be the car’s times before Harry leaves it, while it returns for Dick, and while it carries Dick forward. The car, Dick, and Harry each cover 100100 miles, giving 25t125t2+25t3=100,5t1+5t2+25t3=100,25t1+5t2+5t3=100. \begin{aligned} 25t_1-25t_2+25t_3&=100,\\ 5t_1+5t_2+25t_3&=100,\\ 25t_1+5t_2+5t_3&=100. \end{aligned} Dividing by 55 and solving yields t1=3, t2=2, t3=3.t_1=3,\ t_2=2,\ t_3=3. Therefore the common travel time is t1+t2+t3=8t_1+t_2+t_3=8 hours.

Thus, the correct answer is D.

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