Difference between revisions of "2006 AMC 12A Problems/Problem 1"
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Sandwiches at Joe's Fast Food cost <math>\$3</math> each and sodas cost <math>\$2</math> each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas? | Sandwiches at Joe's Fast Food cost <math>\$3</math> each and sodas cost <math>\$2</math> each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas? | ||
| − | <math>\ | + | <math>\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35</math> |
== Solution == | == Solution == | ||
| − | The <math>5</math> sandwiches cost <math>5\cdot 3=15</math> dollars. The <math>8</math> sodas cost <math>8\cdot 2=16</math> dollars. In total, the purchase costs <math>15+16= | + | The <math>5</math> sandwiches cost <math>5\cdot 3=15</math> dollars. The <math>8</math> sodas cost <math>8\cdot 2=16</math> dollars. In total, the purchase costs <math>15+16=\boxed{\textbf{(A) }31}</math> dollars. |
== See also == | == See also == | ||
Latest revision as of 16:20, 16 December 2021
- The following problem is from both the 2006 AMC 12A #1 and 2006 AMC 10A #1, so both problems redirect to this page.
Problem
Sandwiches at Joe's Fast Food cost 
each and sodas cost 
each. How many dollars will it cost to purchase 
sandwiches and 
sodas?
Solution
The sandwiches cost 
dollars. The sodas cost 
dollars. In total, the purchase costs 
dollars.
See also
| 2006 AMC 12A (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 12 Problems and Solutions | |
| 2006 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
