工程问题

考什么 · What it tests

需要先会 · Prerequisites

• 分数运算 / Fraction arithmetic
• 比与比例 / Ratios and proportions
• 一元一次方程 / One-variable linear equations

常见套路 · Common moves

1. 总量设为 1，效率变分数：一个人 6 天完工，效率就是每天 $\displaystyle \frac{1}{6}$，合作时效率相加——工程题的通用起手式，全程不用知道具体工作量。

   Set the whole job to 1, and rates become fractions: someone who finishes in 6 days works at $\displaystyle \frac{1}{6}$ per day, and cooperating rates add — the universal opening, with no need to know the actual amount of work.
2. 注水排水就是正负效率叠加：进水管是正效率、出水管是负效率，同时开就相加，和多人合作是一个模型。

   Filling and draining are just signed rates added: an inflow pipe is a positive rate, an outflow pipe negative, and opening both adds them — the same model as workers cooperating.
3. 中途加入分段算：有人半路加入或退出，就按人员变化把时间切成几段，每段效率不同分别算工作量。

   Split into stages when someone joins midway: if a worker enters or leaves partway, cut the time at each change and compute the work done in each stretch at its own rate.

易错点 · Common pitfalls

• 把两人的「天数」直接平均或相加当合作时间：6 天和 3 天合作不是 4.5 天也不是 9 天，要用效率相加 $\displaystyle \frac{1}{6}+\frac{1}{3}$ 再取倒数。

  Averaging or adding the two "days" for the joint time: 6 days and 3 days together is neither 4.5 nor 9 — add the rates $\displaystyle \frac{1}{6}+\frac{1}{3}$ and take the reciprocal.
• 效率和时间对不上：设了总量为 1 之后，时间 = 工作量 ÷ 效率，别拿天数直接加减。

  Mismatching rate and time: once the job is 1, time = work ÷ rate — don't add or subtract the days directly.