Question

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?

$1$

$2$

$3$

$4$

$5$

Solution:

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

\begin{gather*} x + y + z = 96 \tag*{(1)} \\ x = 6z \tag*{(2)} \\ z = y - 40 \tag*{(3)}. \end{gather*}

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 Problem 3

Solution:

Problem 3 in Other Years