Difference between revisions of "2008 AMC 12A Problems/Problem 6"
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− | The <math> | + | The <math>$</math>90<math> in store </math>A<math> is </math>$<math>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> | + | Thus the <math>10\%</math> off is equal to <math>$</math>90<math> - </math>$<math>15</math> <math>=</math> <math>$</math>75<math>, and therefore the sticker price is </math>$<math>750</math>. |
==See Also== | ==See Also== |
Revision as of 18:37, 25 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 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 the thus answer is .
Solution 2
The 90A$ better than the additional off at store .
Thus the off is equal to 90$ 75$.
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.