2001 AMC 10 Problem 10 — Solution | LIVE by Po-Shen Loh

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2001 AMC 10 Problem 10

Attempt Problem 10 of the 2001 AMC 10 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 2001 AMC 10 solutions, or check the answer key.

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10.

If $x,$ $y,$ and $z$ are positive with $xy=24,$ $xz=48,$ and $yz=72,$ then $x+y+z$ is

$18$

$19$

$20$

$22$

$24$

Answer: D

Concepts:system of equations algebraic manipulation

Difficulty rating: 1240

Solution:

Dividing $xz=48$ by $xy=24$ gives $z=2y.$ Then $yz=2y^2=72,$ so $y=6,$ $z=12,$ and $x=24/y=4.$

Hence $x+y+z=22.$ Thus, the correct answer is D.

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