Difference between revisions of "2008 AMC 12A Problems/Problem 6"

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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>\$</math>15<math> at store </math>A<math>, so </math>0.85x-90+15=0.75x<math>.
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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 the thus answer is </math>\mathrm{(A)}<math>.
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Solving, we find <math>x=750</math>, and the thus answer is <math>\mathrm{(A)}</math>.
  
 
===Solution 2===
 
===Solution 2===
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>.
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The <math>\$ 90</math> in store <math>A</math> is <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>\textdollar 90<math> - </math>\textdollar 15<math> </math>=<math> </math>\textdollar 75<math>, and therefore the sticker price is </math>\textdollar 750$.
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 11:13, 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.

Problem

Heather compares the price of a new computer at two different stores. Store $A$ offers $15\%$ off the sticker price followed by a $$90$ rebate, and store $B$ offers $25\%$ off the same sticker price with no rebate. Heather saves $$15$ by buying the computer at store $A$ instead of store $B$. What is the sticker price of the computer, in dollars?

$\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500$

Solution

Solution 1

Let the sticker price be $x$.

The price of the computer is $0.85x-90$ at store $A$, and $0.75x$ at store $B$.

Heather saves $$15$ at store $A$, so $0.85x-90+15=0.75x$.

Solving, we find $x=750$, and the thus answer is $\mathrm{(A)}$.

Solution 2

The $$ 90$ in store $A$ is $$ 15$ better than the additional $10\%$ off at store $B$.

Thus the $10\%$ off is equal to $$ 90$ - $$ 15$ $=$ $$ 75$, and therefore the sticker price is $$ 750$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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