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

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== Problem ==
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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #3]] and [[2005 AMC 10B Problems|2005 AMC 10B #5]]}}
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>
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==Problem==
\mathrm{(A)}\ \frac15      \qquad
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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?
\mathrm{(B)}\ \frac13      \qquad
 
\mathrm{(C)}\ \frac25      \qquad
 
\mathrm{(D)}\ \frac23      \qquad
 
\mathrm{(E)}\ \frac45
 
</math>
 
  
== Solution ==
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<math>\mathrm{(A)}\ \frac15 \qquad\mathrm{(B)}\ \frac13 \qquad\mathrm{(C)}\ \frac25 \qquad\mathrm{(D)}\ \frac23 \qquad\mathrm{(E)}\ \frac45 </math>
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>.
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==Solution==
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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{\text{(C)}\frac{2}{5}}</math>.
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This was just a simple manipulation of the equation. No solving was needed!
  
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=4|num-a=6}}
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{{AMC12 box|year=2005|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 23:18, 16 March 2020

The following problem is from both the 2005 AMC 12B #3 and 2005 AMC 10B #5, so both problems redirect to this page.

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 $\boxed{\text{(C)}\frac{2}{5}}$. This was just a simple manipulation of the equation. No solving was needed!

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 AMC 10 Problems and Solutions
2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AMC 12 Problems and Solutions

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