AMC12 2022 A

AMC12 2022 A · Q2

AMC12 2022 A · Q2. It mainly tests Systems of equations.

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?

三个数的和是$96$。第一个数是第三数的$6$倍，第三数比第二数少$40$。第一个数与第二数之差的绝对值是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$ We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$ Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{\textbf{(E) } 5}.$

设第三数为$x$。则第一个数为$6x$，第二数为$x+40$。我们有 \[6x+(x+40)+x=8x+40=96,\] 从而$x=7$。因此，第一个数是$42$，第二个数是$47$。它们的差的绝对值为$|42-47|=\boxed{\textbf{(E) } 5}$。

Topics

AL.002 Systems of equations