Difference between revisions of "2006 AMC 12A Problems/Problem 1"
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #1]] and [[2006 AMC 10A Problems|2006 AMC 10A #1]]}} | {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #1]] and [[2006 AMC 10A Problems|2006 AMC 10A #1]]}} | ||
== Problem == | == Problem == | ||
| − | Sandwiches at Joe's Fast Food cost <math>\ | + | Sandwiches at Joe's Fast Food cost <math>\$3 each and sodas cost </math>\$2 each. How many dollars will it cost to purchase <math>5</math> sandwiches and <math>8</math> sodas? |
<math>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math> | <math>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math> | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2006|ab=A|before=First Question|num-a=2}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 18:33, 5 March 2016
- 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 
$2 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. The answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
