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

Revision as of 20:40, 29 January 2020

One can, can hold $12$ 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 ==
-One elephant, can hold <math>12</math> ounces of human blood, what is the minimum number of elephants needed to provide a gallon (<math>128</math> ounces) of blood?
+One can, can hold <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
@@ Line 6: / Line 6: @@
 \textbf{(C)}\ 9\qquad
 \textbf{(D)}\ 10\qquad
-\textbf{(E)}\ 11</math>\qquad
+\textbf{(E)}\ 11</math>
-\textbf{(F)}\ Option 6$
 == Solution 1 ==