2009 AMC 10A Problem 1

Below is the professionally curated solution for Problem 1 of the 2009 AMC 10A, 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.

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Concepts:divisibilityestimation

Difficulty rating: 560

1.

One can holds 1212 ounces of soda. What is the minimum number of cans needed to provide a gallon (128(128 ounces)) of soda?

77

88

99

1010

1111

Solution:

Since 12812=1023,\dfrac{128}{12} = 10\dfrac{2}{3}, ten cans hold only 120120 ounces, which is not enough.

Therefore 1111 cans are needed.

Thus, the correct answer is E.

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