2009 AMC 12B Problem 3

Below is the professionally curated solution for Problem 3 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B 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:percentagefractionlinear equation

Difficulty rating: 1000

3.

Twenty percent less than 6060 is one-third more than what number?

1616

3030

3232

3636

4848

Solution:

Twenty percent less than 6060 is 0.860=48.0.8 \cdot 60 = 48.

One-third more than nn is 43n,\dfrac{4}{3}n, so 43n=48\dfrac{4}{3}n = 48 gives n=36.n = 36.

Thus, the correct answer is D.

Problem 3 in Other Years