Difference between revisions of "2008 AMC 12A Problems/Problem 6"

(Solution 2)
m (Solution 2)
Line 16: Line 16:
  
 
===Solution 2===
 
===Solution 2===
The <math>&#036;</math>90<math> in store </math>A<math> is </math>&#036;<math>15</math> better than the additional <math>10\%</math> off at store <math>B</math>.
+
The <math>&#036;90</math> in store <math>A</math> is <math>&#036;15</math> better than the additional <math>10\%</math> off at store <math>B</math>.
  
Thus the <math>10\%</math> off is equal to <math>&#036;</math>90<math> - </math>&#036;<math>15</math> <math>=</math> <math>&#036;</math>75<math>, and therefore the sticker price is </math>&#036;<math>750</math>.
+
Thus the <math>10\%</math> off is equal to <math>&#036;90</math> - <math>&#036;15</math> <math>=</math> <math>&#036;75</math>, and therefore the sticker price is <math>&#036;750</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:14, 26 January 2015

The following problem is from both the 2008 AMC 12A #6 and 2004 AMC 10A #8, so both problems redirect to this page.

Problem

Heather compares the price of a new computer at two different stores. Store $A$ offers $15\%$ off the sticker price followed by a $$$90$ rebate, and store $B$ offers $25\%$ off the same sticker price with no rebate. Heather saves $$$15$ by buying the computer at store $A$ instead of store $B$. What is the sticker price of the computer, in dollars?

$\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500$

Solution

Solution 1

Let the sticker price be $x$.

The price of the computer is $0.85x-90$ at store $A$, and $0.75x$ at store $B$.

Heather saves $$$15$ at store $A$, so $0.85x-90+15=0.75x$.

Solving, we find $x=750$, and the thus answer is $\mathrm{(A)}$.

Solution 2

The $&#036;90$ in store $A$ is $&#036;15$ better than the additional $10\%$ off at store $B$.

Thus the $10\%$ off is equal to $&#036;90$ - $&#036;15$ $=$ $&#036;75$, and therefore the sticker price is $&#036;750$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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

Invalid username
Login to AoPS