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2006 AMC 12A Problems/Problem 5

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The following problem is from both the 2006 AMC 12A #5 and 2008 AMC 10A #5, so both problems redirect to this page.

Problem

Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was 8 dollars, and there was an additional cost of 2 dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?

\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ }  5

Solution

Dave and Doug paid 8+2=10 dollars in total. Doug paid for three slices of plain pizza, which cost \frac{3}{8}\cdot 8=3. Dave paid 10-3=7 dollars. Dave paid 7-3=4 more dollars than Doug. The answer is \mathrm{(D)}.

See also

2006 AMC 12A (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
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
2006 AMC 10A (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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