2000 AMC 12 Problem 5

Attempt Problem 5 of the 2000 AMC 12 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 12 solutions, or check the answer key.

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

5.

If ∣x−2∣=p,|x - 2| = p, where x<2,x \lt 2, then what is x−p?x - p?
−2-2
22
2−2p2 - 2p
2p−22p - 2
∣2p−2∣|2p - 2|
Answer: C
Concepts:absolute value
Small hint:
Since x<2,x \lt 2, the quantity x−2x - 2 is negative, so ∣x−2∣=2−x|x - 2| = 2 - x
Big hint:
Solve 2−x=p2 - x = p for x,x, then compute x−px - p
Solution:
Since x<2,x \lt 2, we have ∣x−2∣=2−x=p,|x - 2| = 2 - x = p, so x=2−p.x = 2 - p.
Then x−p=(2−p)−p=2−2p. x - p = (2 - p) - p = 2 - 2p.
Thus, the correct answer is C.
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