耕地紧急：最少增派多少农夫

4 人 5 小时共 20 人时耕 2 块地，故单个农夫速率为 $2/20=1/10$ 块/小时。

若增派 $n$ 人，则 12 小时可耕地数为

$$12(4+n)\cdot\frac{1}{10} \ge 9\Rightarrow 4+n\ge 7.5\Rightarrow n\ge 3.5.$$

因此最少需要增派 4 人。

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Original Explanation

Let $x$ be the rate at which an individual farmer tills per hour. There were $20$ "farmer hours" of work to till $2$ plots of land, as there were $4$ farmers working for $5$ hours. Therefore, $20x = 2$, so $x = \frac{1}{10}$.

Our job is to find the minimal integer $n$ such that $\frac{1}{10} \cdot (4 + n) \cdot 12 \geq 9$.

This is since there would be $4+n$ farmers if $n$ farmers are sent and they work for $12$ hours, so we would be $12(4+n)$ farmer hours. Solving this, we get that $n \geq \frac{7}{2}$. This means that $n = 4$, as $n$ must be an integer.

To confirm this answer, we see that with $8$ total farmers, they would work at a rate of $\frac{1}{10} \cdot 8 = \frac{4}{5}$ of a plot per hour, so it would take them $\frac{45}{4} < 12$ hours to till $9$ plots.