AMC10 2022 A

AMC10 2022 A · Q3

AMC10 2022 A · Q3. 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的差的绝对值是多少？

(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 $|42-47|=\boxed{\textbf{(E) } 5}.$

设第三个数为 $x$。则第一个数为 $6x$，第二个数为 $x+40$。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