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Difference between revisions of "2003 AMC 8 Problems/Problem 9"

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== Solution ==
 
== Solution ==
  
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The area of one of Art's cookies is <math> 3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12 </math>. As he has <math> 12 </math> cookies in a batch, the amount of dough each person used is <math> 12 \cdot 12=144 </math>. Roger's cookies have an area of <math> \frac{144}{2 \cdot 4}=\frac{144}{8}= 18 </math> cookies in  a batch. In total, the amount of money Art will earn is <math> 12 \cdot 60=720 </math>. Thus, the amount Roger would need to charge per cookie is <math> \frac{720}{18}=\boxed{\textbf{(C)}\ 40} </math>.
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The area of one of Art's cookies is <math> 3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12 </math>. As he has <math> 12 </math> cookies in a batch, the amount of dough each person used is <math> 12 \cdot 12=144 </math>. Roger's cookies have an area of <math> \frac{144}{2 \cdot 4}=\frac{144}{8}= 18 </math> cookies in  a batch. In total, the amount of money Art will earn is <math> 12 \cdot 60=720 </math>. Thus, the amount Roger would need to charge per cookie is <math> \frac{720}{18}=\boxed{\textbf{(C)}\ 40}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=8|num-a=10}}
 
{{AMC8 box|year=2003|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:14, 28 August 2016

Problem

Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures

Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.

$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]

$\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]

$\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]

$\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]

Each friend uses the same amount of dough, and Art makes exactly $12$ cookies. Art's cookies sell for $60$ cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90$

Solution

The area of one of Art's cookies is $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$. As he has $12$ cookies in a batch, the amount of dough each person used is $12 \cdot 12=144$. Roger's cookies have an area of $\frac{144}{2 \cdot 4}=\frac{144}{8}= 18$ cookies in a batch. In total, the amount of money Art will earn is $12 \cdot 60=720$. Thus, the amount Roger would need to charge per cookie is $\frac{720}{18}=\boxed{\textbf{(C)}\ 40}$.

See Also

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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