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

Revision as of 21:13, 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

Let $m=$ Brianna's money. We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ . Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{\frac25}$ .

@@ Line 11: / Line 11: @@
 == Solution ==
-<asy>
+Let <math>m=</math>Brianna's money.  We have <math>\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})</math>.  Thus, the money left over is <math>m-\frac35m = \frac25m</math>, so the answer is <math>\boxed{\frac25}</math>.
-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
-<math>w^2+(2w)^2=x^2 \Rightarrow 5w^2 = x^2 \Rightarrow w^2 = x^2/5.</math>
-<math>\therefore \mbox{Area} = w \cdot 2w = 2w^2 = \boxed{\frac25x^2}</math>.
 == See also ==
 * [[2005 AMC 12B Problems]]

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

Revision as of 21:13, 17 April 2009

Problem

Solution

See also