2009 AMC 10A Problem 1 — Solution | LIVE by Po-Shen Loh

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2009 AMC 10A Problem 1

Attempt Problem 1 of the 2009 AMC 10A 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 2009 AMC 10A solutions, or check the answer key.

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

1.

One can holds $12$ ounces of soda. What is the minimum number of cans needed to provide a gallon $(128$ ounces $)$ of soda?

$7$

$8$

$9$

$10$

$11$

Answer: E

Concepts:divisibility estimation

Difficulty rating: 560

Solution:

Since $\dfrac{128}{12} = 10\dfrac{2}{3},$ ten cans hold only $120$ ounces, which is not enough.

Therefore $11$ cans are needed.

Thus, the correct answer is E.

Problem 2#2 →

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