2003 AMC 10B Problems/Problem 12
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).
(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.
~Mathkiddie
Solution 3
Suppose if Al had not lost . The total amount would be . As he has not gained any amount. So, Betty and Clare have collectively gained and as they have doubled their collective fortune, they must have with them at the beginning. This leaves for Al. ~Anshulb
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AMC 10 Problems and Solutions |
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