2009 AMC 10B Problem 5 — Solution | LIVE by Po-Shen Loh

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2009 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B 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:percentage linear equation

Difficulty rating: 870

5.

Twenty percent less than $60$ is one-third more than what number?

$16$

$30$

$32$

$36$

$48$

Solution:

Twenty percent less than $60$ is $\dfrac45\cdot60=48.$

If $n$ is the unknown number, one-third more than $n$ is $\dfrac43 n,$ so $\dfrac43 n=48$ $n=36.$

Thus, the correct answer is D.

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