2008 AMC 12A Problems/Problem 6: Difference between revisions
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==Solution== | ==Solution== | ||
===Solution 1=== | |||
Let the sticker price be <math>x</math>. | Let the sticker price be <math>x</math>. | ||
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Solving, we find <math>x=750</math>, and the thus answer is <math>\mathrm{(A)}</math>. | Solving, we find <math>x=750</math>, and the thus answer is <math>\mathrm{(A)}</math>. | ||
===Solution 2=== | |||
The <math>\</math><math>90</math> in store <math>A</math> is <math>\</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>\</math><math>90</math> <math>-</math> <math>\</math><math>15</math> <math>=</math> <math>\</math><math>75</math>, and therefore the sticker price is <math>\</math><math>750</math>. | |||
==See Also== | ==See Also== | ||
Revision as of 15:42, 5 January 2009
- 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 ![]()
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 | ||