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

Revision as of 13:47, 15 February 2009

One can holds $12$ 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
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All AMC 10 Problems and Solutions

Revision as of 13:46, 15 February 2009 (view source) I like pie (talk \| contribs) (Added solution) ← Older edit		Revision as of 13:47, 15 February 2009 (view source) I like pie (talk \| contribs) m (Fixed year in AMC box) Newer edit →
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	<math>10</math> cans would hold <math>120</math> ounces, but <math>128>120</math>, so <math>11</math> cans are required. Thus, the answer is <math>\mathrm{(E)}</math>.		<math>10</math> cans would hold <math>120</math> ounces, but <math>128>120</math>, so <math>11</math> cans are required. Thus, the answer is <math>\mathrm{(E)}</math>.

−	{{AMC10 box\|year=~~2008~~\|ab=A\|before=First Question\|num-a=2}}	+	{{AMC10 box\|year=2009\|ab=A\|before=First Question\|num-a=2}}