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

Revision as of 13:46, 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 $\mathrm{(E)}$ .

2008 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>\mathrm{(A)}\ 7\qquad
-\mathrm{(A)}\ 7
+\mathrm{(B)}\ 8\qquad
-\qquad
+\mathrm{(C)}\ 9\qquad
-\mathrm{(B)}\ 8
+\mathrm{(D)}\ 10\qquad
-\qquad
+\mathrm{(E)}\ 11</math>
-\mathrm{(C)}\ 9
-\qquad
+== Solution ==
-\mathrm{(D)}\ 10
+<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>.
-\qquad
-\mathrm{(E)}\ 11
+{{AMC10 box|year=2008|ab=A|before=First Question|num-a=2}}
-</math>