2008 AMC 12A Problems/Problem 6

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 $A$ 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 $, and therefore the sticker price is$ $ $750$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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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.

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2008 AMC 12A Problems/Problem 6

Contents

Problem

Solution

Solution 1

Solution 2

See Also