Difference between revisions of "2003 AMC 10B Problems/Problem 12"
(→Solution) |
m (→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | 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, marked by (1) and (2). |
<cmath>a+b+c=1000</cmath> | <cmath>a+b+c=1000</cmath> | ||
− | <cmath>2a+2b+2c=2000</cmath> | + | <cmath>(1). \text{ }2a+2b+2c=2000</cmath> |
<cmath>a-100+2b+2c=1500</cmath> | <cmath>a-100+2b+2c=1500</cmath> | ||
− | <cmath>a+2b+2c=1600</cmath> | + | <cmath>(2). \text{ }a+2b+2c=1600</cmath> |
+ | |||
+ | (Equations (1) and (2) are derived from each equation above them.) | ||
Since all we need to find is <math>a,</math> subtract the second equation from the first equation to get <math>a=400.</math> | Since all we need to find is <math>a,</math> subtract the second equation from the first equation to get <math>a=400.</math> |
Latest revision as of 11:30, 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, marked by (1) and (2).
(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 .
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.