Question

一家工厂的工人生产小部件和小装置。对每一种产品，生产时间是固定的，并且对所有工人相同，但两种产品的生产时间不一定相等。在一小时内， $100$ 名工人可以生产 $300$ 个小部件和 $200$ 个小装置。在两小时内， $60$ 名工人可以生产 $240$ 个小部件和 $300$ 个小装置。在三小时内， $50$ 名工人可以生产 $150$ 个小部件和 $m$ 个小装置。求 $m$ 。

The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m.$

答案

答案：450

知识点：速率方程组

难度评级：2020

小提示：

令 $a$ 和 $b$ 分别为生产一个小部件和一个小装置所需的工时；每种情形给出一个线性方程。

Let $a$ and $b$ be the worker-hours needed for one widget and one whoosit; each scenario gives one linear equation

大提示：

前两种情形给出 $3a + 2b = 1$ 和 $4a + 5b = 2$ ；第三种情形总共提供 $150$ 个工时。

The first two scenarios give $3a + 2b = 1$ and $4a + 5b = 2;$ the third supplies $150$ worker-hours in total

解答：

令 $a$ 和 $b$ 分别为生产一个小部件和一个小装置所需的工时。三种情形分别提供 $100$ 、 $120$ 和 $150$ 个工时，所以 $\begin{aligned} 300a + 200b &= 100, \\ 240a + 300b &= 120, \\ 150a + mb &= 150 \end{aligned}\text{。}$

前两个方程化简为 $3a + 2b = 1$ 和 $4a + 5b = 2$ ，解得 $a = \frac{1}{7}$ 且 $b = \frac{2}{7}$ 。代入第三个方程， $\frac{150}{7} + \frac{2m}{7} = 150$ ，所以 $150 + 2m = 1050$ ， $m = 450$ 。

Let $a$ and $b$ be the worker-hours required to make one widget and one whoosit. The three scenarios supply $100,$ $120,$ and $150$ worker-hours, so $\begin{aligned} 300a + 200b &= 100, \\ 240a + 300b &= 120, \\ 150a + mb &= 150. \end{aligned}$

The first two equations simplify to $3a + 2b = 1$ and $4a + 5b = 2,$ giving $a = \frac{1}{7}$ and $b = \frac{2}{7}.$ Substituting into the third, $\frac{150}{7} + \frac{2m}{7} = 150,$ so $150 + 2m = 1050$ and $m = 450.$

2007 AIME II 第 4 题

答案：450

难度评级：2020

小提示：

大提示：

解答：

其他年份的第 4 题