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Difference between revisions of "2003 AMC 8 Problems/Problem 8"
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==Solution==
 
==Solution==
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<math> \textbf{(A)}</math> Art is the right answer because out of all the cookies, Art's had an area of <math>12 \text{ in}^2</math>, which was the greatest area out of all the cookies' areas. Roger's cookie had an area of <math>8 \text{ in} ^2</math>, and both Paul and Trisha's cookies had an area of <math>6 \text{ in}^2</math>. This means Art makes less cookies, since his cookie area is the greatest. The answer is not that there is a tie between Paul and Trisha because they can make the most cookies with a given amount of cookie dough, not the least.
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<math>\boxed{\textbf{(A)}\ \text{Art}}</math> is the right answer because out of all the cookies, Art's had an area of <math>12 \text{ in}^2</math>, which was the greatest area out of all the cookies' areas. Roger's cookie had an area of <math>8 \text{ in} ^2</math>, and both Paul and Trisha's cookies had an area of <math>6 \text{ in}^2</math>. This means Art makes less cookies, since his cookie area is the greatest. The answer is not that there is a tie between Paul and Trisha because they can make the most cookies with a given amount of cookie dough, not the least.
  
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==See Also==
 
{{AMC8 box|year=2003|num-b=7|num-a=9}}
 
{{AMC8 box|year=2003|num-b=7|num-a=9}}

Revision as of 02:44, 24 December 2012

Problem

$\textbf{Bake Sale}$

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. Who gets the fewest cookies from one batch of cookie dough?

$\textbf{(A)}\ \text{Art}\qquad\textbf{(B)}\ \text{Roger}\qquad\textbf{(C)}\ \text{Paul}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.}$

Solution

$\boxed{\textbf{(A)}\ \text{Art}}$ is the right answer because out of all the cookies, Art's had an area of $12 \text{ in}^2$, which was the greatest area out of all the cookies' areas. Roger's cookie had an area of $8 \text{ in} ^2$, and both Paul and Trisha's cookies had an area of $6 \text{ in}^2$. This means Art makes less cookies, since his cookie area is the greatest. The answer is not that there is a tie between Paul and Trisha because they can make the most cookies with a given amount of cookie dough, not the least.

See Also

2003 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions
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