Difference between revisions of "2003 AMC 10B Problems/Problem 12"
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For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let <math>a, b,</math> and <math>c</math> represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have <math>a-100, 2b,</math> and <math>2c</math>. From this, we can write two equations. | For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let <math>a, b,</math> and <math>c</math> represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have <math>a-100, 2b,</math> and <math>2c</math>. From this, we can write two equations. | ||
| − | <cmath>a+b+c=1000\\ | + | <cmath>a+b+c=1000</cmath>\\ |
| − | 2a+2b+2c=2000\\ | + | <cmath>2a+2b+2c=2000</cmath>\\ |
\\ | \\ | ||
| − | a-100+2b+2c=1500\\ | + | <cmath>a-100+2b+2c=1500\\ |
a+2b+2c=1600</cmath> | a+2b+2c=1600</cmath> | ||
Revision as of 10:25, 20 July 2019
Problem
Al, Betty, and Clare split 
among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of 
dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose 
dollars. What was Al's original portion?
Solution
For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let 
and 
represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have 
and 
. From this, we can write two equations.
![\[a+b+c=1000\]](http://latex.artofproblemsolving.com/7/5/8/758a113c4680a0bbc3ce28fefd61a2e47a8db399.png)
\\
![\[2a+2b+2c=2000\]](http://latex.artofproblemsolving.com/c/1/6/c16ecef53b1429803d7d1018c387e1e88a6ff3d1.png)
\\
\\
![\[a-100+2b+2c=1500\\ a+2b+2c=1600\]](http://latex.artofproblemsolving.com/7/d/9/7d9cd69d83e1eb5e6931c00aae4264c2f18454c6.png)
Since all we need to find is 
subtract the second equation from the first equation to get 
Al's original portion was 
.
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
