Difference between revisions of "2006 AMC 12A Problems/Problem 1"
(13 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 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>\$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>\mathrm{(A)}\ 31\qquad\mathrm{(B)}\ 32\qquad\mathrm{(C)}\ 33\qquad\mathrm{(D)}\ 34\qquad\mathrm{(E)}\ 35</math> | |
− | |||
− | <math> \mathrm{(A) \ | ||
− | |||
− | |||
== 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=31</math> dollars. The answer is <math>\mathrm{(A)}</math>. | ||
− | + | == See also == | |
+ | {{AMC12 box|year=2006|ab=A|before=First Question|num-a=2}} | ||
+ | {{AMC10 box|year=2006|ab=A|before=First Question|num-a=2}} | ||
+ | {{MAA Notice}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
− |
Revision as of 19:35, 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 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. 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 |
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.