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

Latest revision as of 08:00, 8 June 2021

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

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

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

We want to find $\left\lceil\frac{128}{12}\right\rceil$ because there are a whole number of cans.

$\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.$

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

@@ Line 1: / Line 1: @@
-== Problem 1 ==
+== Problem ==
-One can holds <math>12</math> ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?
+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>
+<math>\textbf{(A)}\ 7\qquad
-\mathrm{(A)}\ 7
+\textbf{(B)}\ 8\qquad
-\qquad
+\textbf{(C)}\ 9\qquad
-\mathrm{(B)}\ 8
+\textbf{(D)}\ 10\qquad
-\qquad
+\textbf{(E)}\ 11</math>
-\mathrm{(C)}\ 9
-\qquad
-\mathrm{(D)}\ 10
-\qquad
-\mathrm{(E)}\ 11
-</math>
-[[2009 AMC 10A Problems/Problem 1|Solution]]
+== 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{\boxed{(E)}}</math>.
+== Solution 2 ==
+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}}
+{{MAA Notice}}