Difference between revisions of "2008 AMC 12A Problems/Problem 6"
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| + | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #6]] and [[2008 AMC 10A Problems/Problem 8|2008 AMC 10A #8]]}} | ||
==Problem == | ==Problem == | ||
| − | Heather compares the price of a new computer at two different stores. Store A offers <math>15\%</math> off the sticker price followed by a | + | Heather compares the price of a new computer at two different stores. Store <math>A</math> offers <math>15\%</math> off the sticker price followed by a <math>\$90</math> rebate, and store <math>B</math> offers <math>25\%</math> off the same sticker price with no rebate. Heather saves <math>\$15</math> by buying the computer at store <math>A</math> instead of store <math>B</math>. What is the sticker price of the computer, in dollars? |
| − | <math>\ | + | <math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math> |
==Solution== | ==Solution== | ||
| − | Let <math> | + | ===Solution 1=== |
| + | Let the sticker price be <math>x</math>. | ||
| − | + | The price of the computer is <math>0.85x-90</math> at store <math>A</math> and <math>0.75x</math> at store <math>B</math>. | |
| − | |||
| − | + | Heather saves <math>\$15</math> at store <math>A</math>, so <math>0.85x-90+15=0.75x</math>. | |
| + | |||
| + | Solving, we find <math>x=750</math>, and thus the answer is <math>\mathrm{(A)}</math>. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | The <math>\$ 90</math> at store <math>A</math> is <math>\$ 15</math> greater 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>-</math> <math>\$ 15</math> <math>=</math> <math>\$ 75</math>, and therefore the sticker price is <math>\$ 750</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=5|num-a=7}} | {{AMC12 box|year=2008|ab=A|num-b=5|num-a=7}} | ||
| + | {{AMC10 box|year=2008|ab=A|num-b=7|num-a=9}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:53, 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 
at store is 
greater 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. 
