2000 AMC 10 Problem 9 — Solution | LIVE by Po-Shen Loh

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2000 AMC 10 Problem 9

Attempt Problem 9 of the 2000 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 2000 AMC 10 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

If $|x - 2| = p,$ where $x \lt 2,$ then $x - p =$

$-2$

$2$

$2 - 2p$

$2p - 2$

$|2p - 2|$

Answer: C

Concepts:absolute value algebraic manipulation

Difficulty rating: 1170

Small Hint:

Since $x \lt 2,$ the quantity $x - 2$ is negative, so $|x - 2| = 2 - x$

Big Hint:

Solve $2 - x = p$ for $x,$ then substitute into $x - p$

Solution:

Because $x \lt 2,$ we have $|x - 2| = 2 - x = p,$ so $x = 2 - p.$

Then $x - p = (2 - p) - p = 2 - 2p.$

Thus, the correct answer is C.

Problem 9 in Other Years

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