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

(Created page with "== 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...")
 
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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 area 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 a cookie is <math> \frac{720}{18}=\boxed{\textbf{(C)}\ 40} </math>.
 
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 area 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 a cookie is <math> \frac{720}{18}=\boxed{\textbf{(C)}\ 40} </math>.
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{{AMC8 box|year=2003|num-b=8|num-a=10}}

Revision as of 09:00, 25 November 2011

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 area 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 a cookie is $\frac{720}{18}=\boxed{\textbf{(C)}\ 40}$.

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
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