2022 AMC 10A Problem 3

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

Let the third number be the variable

Write the other two numbers in terms of that variable

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.

Problem 3 in other years