Difference between revisions of "2009 AMC 10A Problems/Problem 1"
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== Problem == | == Problem == | ||
− | One can | + | One can holds <math>12</math> ounces of soda, what is the minimum number of cans needed to provide a gallon (<math>128</math> ounces) of soda? |
<math>\textbf{(A)}\ 7\qquad | <math>\textbf{(A)}\ 7\qquad | ||
Line 9: | Line 9: | ||
== Solution 1 == | == Solution 1 == | ||
− | <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{\boxed{(E)}}</math>. |
== Solution 2 == | == Solution 2 == | ||
− | We want to find <math>\left\ | + | We want to find <math>\left\lceil\frac{128}{12}\right\rceil</math> because there are a whole number of cans. |
− | < | + | <math>\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.</math> |
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | {{AMC10 box|year=2009|ab=A|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:00, 8 June 2021
Contents
Problem
One can holds ounces of soda, what is the minimum number of cans needed to provide a gallon ( ounces) of soda?
Solution 1
cans would hold ounces, but , so cans are required. Thus, the answer is .
Solution 2
We want to find because there are a whole number of cans.
See Also
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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