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Difference between revisions of "2007 AMC 12A Problems/Problem 1"
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<math>\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math>
 
<math>\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math>
  
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== Solution ==
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== Official Solution ==
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<math>P</math> = the amount Pam spent
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<math>\textbf{Answer: (C)}</math>  
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<math>S</math> = the amount Susan spent
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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.
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* <math>P=5 \cdot (20 \cdot .7) = 70</math>
 
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* <math>S=4 \cdot (20 \cdot .75) = 60</math>
 
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Pam pays 10 more dollars than Susan <math>\Rightarrow\fbox{C}</math>
 
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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 $$$20$ 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?

$\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20$

Official Solution

$\textbf{Answer: (C)}$ Susan pays $(4)(0.75)(20) = 60$ dollars. Pam pays $(5)(0.70)(20) = 70$ dollars, so she pays $70-60=10$ 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. AMC logo.png

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