Difference between revisions of "2007 AMC 12A Problems/Problem 1"
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| + | {{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #1]] and [[2007 AMC 10A Problems/Problem 1|2007 AMC 10A #1]]}} | ||
== Problem == | == Problem == | ||
| − | One ticket to a show costs | + | One ticket to a show costs <math>\</math><math>20</math> at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan? |
| + | <math>\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math> | ||
| − | == Solution == | + | == Official Solution == |
| − | + | <math>\textbf{Answer: (C)}</math> | |
| − | + | Susan pays <math>(4)(0.75)(20) = 60</math> dollars. Pam pays <math>(5)(0.70)(20) = 70</math> dollars, so she pays <math>70-60=10</math> more dollars than Susan. | |
| − | + | == See also == | |
| − | + | {{AMC12 box|year=2007|ab=A|before=First question|num-a=2}} | |
| + | {{AMC10 box|year=2007|ab=A|before=First question|num-a=2}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | + | {{MAA Notice}} | |
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Latest revision as of 21:13, 22 March 2021
- The following problem is from both the 2007 AMC 12A #1 and 2007 AMC 10A #1, so both problems redirect to this page.
Problem
One ticket to a show costs 
at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?
Official Solution
Susan pays 
dollars. Pam pays 
dollars, so she pays 
more dollars than Susan.
See also
| 2007 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 | |
| 2007 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. 
