2014 AMC 10A 第 15 题

Question

David 从家开车去机场赶飞机。第一小时他开了

35

英里，但意识到如果继续按这个速度行驶会迟到

1

小时。之后他把速度提高

15

英里每小时，并提前

30

分钟到达。机场离他家多少英里？

David drives from his home to the airport to catch a flight. He drives

35

miles in the first hour, but realizes that he will be

1

hour late if he continues at this speed. He increases his speed by

15

miles per hour for the rest of the way to the airport and arrives

30

minutes early. How many miles is the airport from his home?

140

175

210

245

280

比较原来过慢的计划和提速后的实际行程。

Compare the original too-slow schedule with the faster schedule

提速后的行程总共节省了

1.5

小时。

The faster trip saves

1.5

hours overall

第一小时后，David 的速度为

50

英里每小时。

若机场离家

x

英里，则两种行程时间相差一个半小时，因此

\dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) = \dfrac 32\text{。}

解这个方程：

\begin{aligned} \dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) &= \dfrac 32\\ \dfrac x{35} - \dfrac{x-35+50}{50} &= \dfrac 32\\ 10x - 7(x+15)&= 525\\ 10x - 7x-105&= 525\\ 3x&= 630\\ x&=210 \end{aligned}

所以机场离家

x=210

英里。

所以正确答案是 C。

Note that David drives at

50

miles per hour after one hour.

Then, if the airport is

x

miles from David’s house, we know that:

\dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) = \dfrac 32

We solve this equation as follows:

\begin{aligned} \dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) &= \dfrac 32\\ \dfrac x{35} - \dfrac{x-35+50}{50} &= \dfrac 32\\ 10x - 7(x+15)&= 525\\ 10x - 7x-105&= 525\\ 3x&= 630\\ x&=210 \end{aligned}

Therefore, the airport is

x=210

miles from David’s house.

Thus, C is the correct answer.

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