AMC10 2008 A

AMC10 2008 A · Q13

AMC10 2008 A · Q13. It mainly tests 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 $t$?

道格一人可以5小时粉刷一间屋子。戴夫可以7小时粉刷同一间屋子。道格和戴夫一起粉刷这间屋子，并中途休息1小时吃午饭。设$t$为他们完成工作所需总时间（小时），包括午饭。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

Answer (D): In one hour Doug can paint $\frac{1}{5}$ of the room, and Dave can paint $\frac{1}{7}$ of the room. Working together, they can paint $\frac{1}{5}+\frac{1}{7}$ of the room in one hour. It takes them $t$ hours to do the job, but because they take an hour for lunch, they work for only $t-1$ hours. The fraction of the room that they paint in this time is $\left(\frac{1}{5}+\frac{1}{7}\right)(t-1)$, which must be equal to $1$. It may be checked that the solution, $t=\frac{47}{12}$, does not satisfy the equation in any of the other answer choices.

答案（D）：一小时内，道格可以刷完房间的$\frac{1}{5}$，戴夫可以刷完房间的$\frac{1}{7}$。他们一起工作时，一小时能刷完房间的$\frac{1}{5}+\frac{1}{7}$。完成工作需要$t$小时，但因为他们午饭要用一小时，所以实际只工作$t-1$小时。在这段时间内他们刷完的房间比例是 $\left(\frac{1}{5}+\frac{1}{7}\right)(t-1)$，这必须等于$1$。可以检验该解$t=\frac{47}{12}$不满足其他任何选项中的方程。

Topics

AR.008 Work problems