2022 AMC 10A Problem 3

Attempt Problem 3 of the 2022 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

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3.

The sum of three numbers is 96.96. The first number is 66 times the third number, and the third number is 4040 less than the second number. What is the absolute value of the difference between the first and second numbers?

11

22

33

44

55

Answer: E
Concepts:system of equationslinear equation
Difficulty rating: 900
Solution:

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

x+y+z=96(1)x=6z(2)z=y40(3).\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}

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

From this, we get that x=6z=67=42 x = 6 \cdot z = 6 \cdot 7 = 42 and y=z+40=7+40=47. y = z + 40 = 7 + 40 = 47.

Therefore, yx=4742=5.y - x = 47 - 42 = 5.

Thus, E is the correct answer.

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