Difference between revisions of "2003 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500</math> | <math>\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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, marked by (1) and (2). | 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, marked by (1) and (2). | ||
| Line 19: | Line 19: | ||
Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>. | Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Suppose the total amount of money Betty and Clare has is <math>1000-x</math> and Al has <math>x</math> dollars. Then, <math>(x-100)+2(1000-x)=1500</math>, so Al has <math>\boxed{\textbf{(C)}\ \textdollar 400}</math> dollars. | ||
==See Also== | ==See Also== | ||
Revision as of 17:10, 3 October 2022
Contents
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 1
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, marked by (1) and (2).
![\[a+b+c=1000\]](http://latex.artofproblemsolving.com/7/5/8/758a113c4680a0bbc3ce28fefd61a2e47a8db399.png)
![\[(1). \text{ }2a+2b+2c=2000\]](http://latex.artofproblemsolving.com/4/7/b/47b435cba1e52e1776a9001017791304b42638e4.png)
![\[a-100+2b+2c=1500\]](http://latex.artofproblemsolving.com/a/e/8/ae8ff2820c248dd9515df1709037eccc86be5e26.png)
![\[(2). \text{ }a+2b+2c=1600\]](http://latex.artofproblemsolving.com/1/2/6/126041f6e69ade4b098b1a4a752abd8081da1690.png)
(Equations (1) and (2) are derived from each equation above them.)
Since all we need to find is 
subtract the second equation from the first equation to get 
Al's original portion was 
.
Solution 2
Suppose the total amount of money Betty and Clare has is 
and Al has 
dollars. Then, 
, so Al has dollars.
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. 
