Difference between revisions of "2008 AMC 12A Problems/Problem 6"
m (→Solution 2) |
(→Solution 2) |
||
| Line 18: | Line 18: | ||
The <math>\$ 90</math> in store <math>A</math> is <math>\$ 15</math> better than the additional <math>25\%-15\% = 10\%</math> off at store <math>B</math>. | The <math>\$ 90</math> in store <math>A</math> is <math>\$ 15</math> better than the additional <math>25\%-15\% = 10\%</math> off at store <math>B</math>. | ||
| − | Thus the <math>10\%</math> off is equal to <math>\$ 90</math> - <math>\$ 15</math> <math>=</math> <math>\$ 75</math>, and therefore the sticker price is <math>\$ 750</math>. | + | Thus the <math>10\%</math> off is equal to <math>\$ 90</math> <math>-</math> <math>\$ 15</math> <math>=</math> <math>\$ 75</math>, and therefore the sticker price is <math>\$ 750</math>. |
==See Also== | ==See Also== | ||
Revision as of 16:36, 4 June 2021
- The following problem is from both the 2008 AMC 12A #6 and 2008 AMC 10A #8, so both problems redirect to this page.
Problem
Heather compares the price of a new computer at two different stores. Store 
offers 
off the sticker price followed by a 
rebate, and store 
offers 
off the same sticker price with no rebate. Heather saves 
by buying the computer at store instead of store . What is the sticker price of the computer, in dollars?
Solution
Solution 1
Let the sticker price be 
.
The price of the computer is 
at store and 
at store .
Heather saves at store , so 
.
Solving, we find 
, and thus the answer is 
.
Solution 2
The 
in store is 
better than the additional 
off at store .
Thus the 
off is equal to 
, and therefore the sticker price is 
.
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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. 
