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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]]}}
 
{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #6]] and [[2008 AMC 10A Problems/Problem 8|2008 AMC 10A #8]]}}
 
==Problem ==
 
==Problem ==
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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>\$</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?
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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?
  
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</math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500$
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<math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math>
  
 
==Solution==
 
==Solution==
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Let the sticker price be <math>x</math>.
 
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>.
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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>.
  
−
Heather saves <math>\$</math>15<math> at store </math>A<math>, so </math>0.85x-90+15=0.75x<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 the thus answer is </math>\mathrm{(A)}<math>.
+
Solving, we find <math>x=750</math>, and thus the answer is <math>\mathrm{(A)}</math>.
  
 
===Solution 2===
 
===Solution 2===
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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>.
+
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>\textdollar 90<math> - </math>\textdollar 15<math> </math>=<math> </math>\textdollar 75<math>, and therefore the sticker price is </math>\textdollar 750$.
+
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==

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 $A$ 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 thus the answer is $\mathrm{(A)}$.

Solution 2

The $$ 90$ at store $A$ is $$ 15$ greater than the additional $25\%-15\% = 10\%$ 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 (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

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