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

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

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

解答：

其他年份的第 10 题