Difference between revisions of "2005 AMC 12B Problems/Problem 3"

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== Problem ==
 
== Problem ==
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide. What is the area of the rectangle?
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Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?  
  
 
<math>
 
<math>
\mathrm{(A)}\ \frac14x^2     \qquad
+
\mathrm{(A)}\ \frac15     \qquad
\mathrm{(B)}\ \frac25x^2     \qquad
+
\mathrm{(B)}\ \frac13     \qquad
\mathrm{(C)}\ \frac12x^2     \qquad
+
\mathrm{(C)}\ \frac25     \qquad
\mathrm{(D)}\ x^2     \qquad
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\mathrm{(D)}\ \frac23     \qquad
\mathrm{(E)}\ \frac32x^2
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\mathrm{(E)}\ \frac45
 
</math>
 
</math>
  

Revision as of 21:11, 17 April 2009

Problem

Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?

$\mathrm{(A)}\ \frac15      \qquad \mathrm{(B)}\ \frac13      \qquad \mathrm{(C)}\ \frac25      \qquad \mathrm{(D)}\ \frac23      \qquad \mathrm{(E)}\ \frac45$

Solution

[asy] size(5cm); pen f = fontsize(10); pair A = (0,0); pair B = (0,1); pair C = (2,1); pair DD = (2,0); D(A--B--C--DD--cycle); D(A--C); MP("2w",(A+DD)/2,plain.N,f); MP("w",(C+DD)/2,plain.E,f); MP("x",(A+C)/2,plain.NW,f); [/asy] Using the Pythagorean theorem, we have

$w^2+(2w)^2=x^2 \Rightarrow 5w^2 = x^2 \Rightarrow w^2 = x^2/5.$

$\therefore \mbox{Area} = w \cdot 2w = 2w^2 = \boxed{\frac25x^2}$.

See also