AMC12 2008 A

AMC12 2008 A · Q10

AMC12 2008 A · Q10. It mainly tests Linear equations, Work problems.

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 $t$?

Doug 刷完一间房间需要 $5$ 小时。Dave 刷完同一间房间需要 $7$ 小时。Doug 和 Dave 一起刷这间房间，并在中途午休 1 小时。设 $t$ 为他们完成工作所需的总时间（小时），包括午休。下列哪个方程由 $t$ 满足？

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

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

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

(E) (5+7)t = 1 (5+7)t = 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

Doug can paint $\frac{1}{5}$ of a room per hour, Dave can paint $\frac{1}{7}$ of a room per hour, and the time they spend working together is $t-1$. Since rate multiplied by time gives output, $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$ If one person does a job in $a$ hours and another person does a job in $b$ hours, the time it takes to do the job together is $\frac{ab}{a+b}$ hours. Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in $\frac{5\times7}{5+7} = \frac{35}{12}$ hours. They also take 1 hour for lunch, so the total time $t = \frac{35}{12} + 1$ hours. Looking at the answer choices, $(D)$ is the only one satisfied by $t = \frac{35}{12} + 1$.

Doug 每小时能刷 $\frac{1}{5}$ 间房，Dave 每小时能刷 $\frac{1}{7}$ 间房，他们一起工作的时间为 $t-1$。由于“效率 $\times$ 时间 = 完成量”，有 $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$ 若一人用 $a$ 小时完成工作，另一人用 $b$ 小时完成工作，则两人合作完成所需时间为 $\frac{ab}{a+b}$ 小时。 Doug 用 5 小时刷完，Dave 用 7 小时刷完，因此两人合作用时 $\frac{5\times7}{5+7} = \frac{35}{12}$ 小时。再加上 1 小时午休，总时间 $t = \frac{35}{12} + 1$ 小时。查看选项，只有 $(D)$ 满足 $t = \frac{35}{12} + 1$。

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