Difference between revisions of "2007 AMC 12A Problems/Problem 1"

m (rv)
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
One ticket to a show costs <math>20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickers using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?
+
One ticket to a show costs <math>20</math> at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickers 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>
+
<math>\mathrm{(A)}\ 2\qquad \mathrm{(B)}\ 5\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 15\qquad \mathrm{(E)}\ 20</math>
  
 
== Solution ==
 
== Solution ==
</math>P<math> = the amount Pam spent
+
<math>P</math> = the amount Pam spent
</math>S<math> = the amount Susan spent
+
<math>S</math> = the amount Susan spent
  
* </math>P=5 \cdot (20 \cdot .7) = 70<math>
+
* <math>P=5 \cdot (20 \cdot .7) = 70</math>
* </math>S=4 \cdot (20 \cdot .75) = 60$
+
* <math>S=4 \cdot (20 \cdot .75) = 60</math>
  
 
Pam pays 10 more dollars than Susan.
 
Pam pays 10 more dollars than Susan.

Revision as of 19:37, 19 September 2007

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 tickers using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?

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

Solution

$P$ = the amount Pam spent $S$ = the amount Susan spent

  • $P=5 \cdot (20 \cdot .7) = 70$
  • $S=4 \cdot (20 \cdot .75) = 60$

Pam pays 10 more dollars than Susan.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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