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Difference between revisions of "2003 AMC 8 Problems/Problem 10"

(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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== Solution ==
 
== Solution ==
 
Art's cookies have areas of <math> 3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12 </math>. There are 12 cookies in one of Art's batches so everyone used <math> 12 \cdot 12=144 \text{ in}^2 </math> of dough. Trisha's cookies have an area of <math> \frac{3 \cdot 4}{2}=6 </math> so she has <math> \frac{144}{6}=\boxed{\textbf{(E)}\ 24}</math> cookies per batch.
 
Art's cookies have areas of <math> 3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12 </math>. There are 12 cookies in one of Art's batches so everyone used <math> 12 \cdot 12=144 \text{ in}^2 </math> of dough. Trisha's cookies have an area of <math> \frac{3 \cdot 4}{2}=6 </math> so she has <math> \frac{144}{6}=\boxed{\textbf{(E)}\ 24}</math> cookies per batch.
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==See Also==
 +
{{AMC8 box|year=2003|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 23:47, 4 July 2013

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]

How many cookies will be in one batch of Trisha's cookies?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$

Solution

Art's cookies have areas of $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$. There are 12 cookies in one of Art's batches so everyone used $12 \cdot 12=144 \text{ in}^2$ of dough. Trisha's cookies have an area of $\frac{3 \cdot 4}{2}=6$ so she has $\frac{144}{6}=\boxed{\textbf{(E)}\ 24}$ cookies per batch.

See Also

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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