Question

三个数的和为 $96$ 。第一个数是第三个数的 $6$ 倍，第三个数比第二个数少 $40$ 。第一个数与第二个数之差的绝对值是多少？

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?

$1$

$2$

$3$

$4$

$5$

答案：E

知识点：方程组一次方程

难度评级：900

解答：

设三个数分别为 $x, y$ 和 $z$ 。由题意可列出相应关系。

$\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}$

这三个关系为将 $(3)$ 改写为 $y = z + 40$ 再把这个等式和 $(2)$ 代入 $(1)$ 得 $6z + z + 40 + z = 96$ $8z + 40 = 96$ $8z = 56 \Rightarrow z = 7.$

因此且 $x = 6 \cdot z = 6 \cdot 7 = 42$ $y = z + 40 = 7 + 40 = 47.$

$y - x = 47 - 42 = 5$ 。

所以正确答案是 E。

Let $x, y,$ and $z$ be the three numbers. The conditions from the problem give us the following relations:

$\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}$

Rearranging $(3),$ we get $y = z + 40.$ Plugging this new equation and $(2)$ into $(1),$ we get $6z + z + 40 + z = 96$ $8z + 40 = 96$ $8z = 56 \Rightarrow z = 7.$

From this, we get that $x = 6 \cdot z = 6 \cdot 7 = 42$ and $y = z + 40 = 7 + 40 = 47.$

Therefore, $y - x = 47 - 42 = 5.$

Thus, E is the correct answer.

2022 AMC 10A 第 3 题

答案：E

难度评级：900

解答：

其他年份的第 3 题