2008 AMC 12A 第 10 题

Question

Doug 粉刷一个房间需要

5

小时。Dave 粉刷同一个房间需要

7

小时。Doug 和 Dave 一起粉刷房间，并午休一小时。设

t

为他们完成工作所需总时间（小时），包括午饭时间。

t

满足下列哪个方程？

Doug can paint a room in

5

hours. Dave can paint the same room in

7

hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let

t

be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by

t?

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t + 1) = 1

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t + 1 = 1

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)t = 1

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1

(5 + 7)t = 1

答案：D

知识点：速率一次方程

难度评级：1380

小提示：

他们一起每小时粉刷

\dfrac{1}{5} + \dfrac{1}{7}

个房间。

Together they paint

\dfrac{1}{5} + \dfrac{1}{7}

of the room each hour

大提示：

总时间

t

中有一小时是午饭，所以实际工作时间为

t - 1

小时。

Of the

t

hours, one hour is lunch, so they actually work

t - 1

hours

解答：

Doug 每小时粉刷

\tfrac{1}{5}

个房间，Dave 每小时粉刷

\tfrac{1}{7}

个房间，所以合速度为

\tfrac{1}{5} + \tfrac{1}{7}

个房间每小时。

总时间为

t

，其中

1

小时用于午饭，所以实际工作

t - 1

小时。完成一个房间给出

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1\text{。}

所以正确答案是 D。

In one hour Doug paints

\tfrac{1}{5}

of the room and Dave paints

\tfrac{1}{7},

so together they paint

\tfrac{1}{5} + \tfrac{1}{7}

of the room per hour.

Of the total time

t,

one hour is spent at lunch, so they work for

t - 1

hours. The fraction painted must equal

1,

giving

\left(\dfrac{1}{5} + \dfrac{1}{7}\right)(t - 1) = 1.

Thus, D is the correct answer.

2008 AMC 12A 第 10 题

答案：D

难度评级：1380

小提示：

大提示：

解答：

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