Difference between revisions of "2009 AMC 10A Problems/Problem 1"

Revision as of 17:03, 25 December 2017

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

$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 9\qquad \mathrm{(D)}\ 10\qquad \mathrm{(E)}\ 11$

$10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\mathrm{(E)}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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 {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}}
 {{MAA Notice}}
-== Solution 2 ==
-We can divide <math>128/12</math> and round up because there are a whole number of cans.
-<math>128/12 = 10R8\longrightarrow 11\longrightarrow \fbox{E}.</math>