Difference between revisions of "2006 AMC 12A Problems/Problem 5"
m (→See also: box) |
Dairyqueenxd (talk | contribs) (→Solution) |
||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #5]] and [[2006 AMC 10A Problems/Problem 5|2006 AMC 10A #5]]}} | ||
== Problem == | == Problem == | ||
| − | |||
Doug and Dave shared a pizza with <math>8</math> equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was <math>8</math> dollars, and there was an additional cost of <math>2</math> 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? | Doug and Dave shared a pizza with <math>8</math> equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was <math>8</math> dollars, and there was an additional cost of <math>2</math> 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? | ||
| − | <math> \ | + | <math> \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math> |
== Solution == | == Solution == | ||
| − | + | Dave and Doug paid <math>8+2=10</math> dollars in total. Doug paid for three slices of plain pizza, which cost <math>\frac{3}{8}\cdot 8=3</math>. Dave paid <math>10-3=7</math> dollars. Dave paid <math>7-3=\boxed{\textbf{(D) }4}</math> more dollars than Doug. | |
| − | Dave and Doug paid <math>8+2=10</math> dollars in total. Doug paid for three slices of plain pizza, which cost <math>\frac{3}{8}\cdot 8=3</math>. Dave paid <math>10-3=7</math> dollars. Dave paid <math>7-3=4</math> more dollars than Doug | ||
== See also == | == See also == | ||
| − | |||
| − | |||
{{AMC12 box|year=2006|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2006|ab=A|num-b=4|num-a=6}} | ||
| + | {{AMC10 box|year=2006|ab=A|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 16:57, 16 December 2021
- The following problem is from both the 2006 AMC 12A #5 and 2006 AMC 10A #5, so both problems redirect to this page.
Problem
Doug and Dave shared a pizza with 
equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was dollars, and there was an additional cost of 
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?
Solution
Dave and Doug paid 
dollars in total. Doug paid for three slices of plain pizza, which cost 
. Dave paid 
dollars. Dave paid 
more dollars than Doug.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
| All AMC 12 Problems and Solutions | |
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
