Question

某天中午，Minneapolis 比 St. Louis 暖和 $N$ 度。到 $4{:}00$ 时，Minneapolis 的气温下降了 $5$ 度，而 St. Louis 的气温上升了 $3$ 度，此时两城气温相差 $2$ 度。所有可能的 $N$ 值的乘积是多少？

At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$

$10$

$30$

$60$

$100$

$120$

答案：C

知识点：绝对值一次方程

难度评级：870

解答：

设中午两城气温差为 $N=m-s$ 。到 $4{:}00$ 时，新温差为 $(m-5)-(s+3)=N-8$ 。

此时两城气温相差 $2$ 度，所以 $|N-8|=2$ 。因此 $N=6$ 或 $N=10$ ，所有可能值的乘积为 $6\cdot10=60$ 。

所以正确答案是 C。

Let $N=m-s$ be the noon temperature difference. At $4{:}00,$ the new difference is $(m-5)-(s+3)=N-8.$

The temperatures then differ by $2$ degrees, so $|N-8|=2.$ Hence $N=6$ or $N=10,$ and the product of all possible values is $6\cdot10=60.$

Thus, the answer is C .

2021 AMC 10B Fall 第 4 题

答案：C

难度评级：870

解答：

其他年份的第 4 题