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| | + | {{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 <math>A</math> offers <math>15\%</math> off the sticker price followed by a <math>\</ | + | 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>\$</math>90<math> rebate, and store </math>B<math> offers </math>25\%<math> off the same sticker price with no rebate. Heather saves </math>\$<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>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math> | <math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math> | ||
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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>. | 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>\</ | + | Heather saves <math>\$</math>15<math> at store </math>A<math>, so </math>0.85x-90+15=0.75x<math>. |
− | Solving, we find <math>x=750< | + | Solving, we find </math>x=750<math>, and the thus answer is </math>\mathrm{(A)}<math>. |
===Solution 2=== | ===Solution 2=== | ||
− | The <math>\textdollar 90< | + | The </math>\textdollar 90<math> in store </math>A<math> is </math>\textdollar 15<math> better than the additional </math>10\%<math> off at store </math>B<math>. |
− | Thus the <math>10\%< | + | Thus the </math>10\%<math> off is equal to </math>\textdollar 90<math> - </math>\textdollar 15<math> </math>=<math> </math>\textdollar 75<math>, and therefore the sticker price is </math>\textdollar 750$. |
==See Also== | ==See Also== |
Revision as of 11:11, 17 August 2020
- The following problem is from both the 2008 AMC 12A #6 and 2008 AMC 10A #8, so both problems redirect to this page.
Contents
Problem
Heather compares the price of a new computer at two different stores. Store offers off the sticker price followed by a 90B25\%$ 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 15A0.85x-90+15=0.75x$.
Solving, we find$ (Error compiling LaTeX. Unknown error_msg)x=750\mathrm{(A)}$.
===Solution 2=== The$ (Error compiling LaTeX. Unknown error_msg)\textdollar 90A\textdollar 1510\%B$.
Thus the$ (Error compiling LaTeX. Unknown error_msg)10\%\textdollar 90\textdollar 15$$ (Error compiling LaTeX. Unknown error_msg)=$$ (Error compiling LaTeX. Unknown error_msg)\textdollar 75\textdollar 750$.
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.