Difference between revisions of "2005 AMC 12B Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | 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{\ | + | 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{\textbf{(C) }\frac{2}{5}}</math>. |
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This was just a simple manipulation of the equation. No solving was needed! | This was just a simple manipulation of the equation. No solving was needed! | ||
Latest revision as of 13:14, 14 December 2021
- 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?
Solution
Let 
Brianna's money. We have 
. Thus, the money left over is 
, so the answer is 
.
This was just a simple manipulation of the equation. No solving was needed!
See also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
