2000 AMC 12 Problem 5

Below is the professionally curated solution for Problem 5 of the 2000 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:absolute value

Difficulty rating: 1150

5.

If x2=p,|x - 2| = p, where x<2,x \lt 2, then what is xp?x - p?

2-2

22

22p2 - 2p

2p22p - 2

2p2|2p - 2|

Solution:

Since x<2,x \lt 2, we have x2=2x=p,|x - 2| = 2 - x = p, so x=2p.x = 2 - p.

Then xp=(2p)p=22p. x - p = (2 - p) - p = 2 - 2p.

Thus, the correct answer is C.

Problem 5 in Other Years