Difference between revisions of "2003 AMC 10B Problems/Problem 12"
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− | ==Problem== | + | == Problem == |
Al, Betty, and Clare split <math>\textdollar 1000</math> 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 <math>\textdollar 1500</math> dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose <math>\textdollar100</math> dollars. What was Al's original portion? | Al, Betty, and Clare split <math>\textdollar 1000</math> 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 <math>\textdollar 1500</math> dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose <math>\textdollar100</math> dollars. What was Al's original portion? | ||
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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. | + | 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 \implies 2a+2b+2c=2000</cmath> |
− | 2a+2b+2c=2000 | + | <cmath>(a-100)+2b+2c=1500 \implies a+2b+2c=1600</cmath> |
− | |||
− | a-100+2b+2c=1500\ | ||
− | a+2b+2c=1600</cmath> | ||
− | 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> Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>. |
− | + | == Solution 2 == | |
− | ==See Also== | + | Suppose Betty and Clare <math>1000-x</math> dollars and Al <math>x</math> dollars originally. Then, <math>(x-100)+2(1000-x)=1500 \implies x=400</math>, so Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>. |
+ | |||
+ | ~Mathkiddie | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | If Al had not lost <math>\textdollar 100</math>, the total amount would be <math>\textdollar 1600</math>. So, Betty and Clare collectively gained <math>\textdollar 600</math> and as they have doubled their collective fortune, | ||
+ | they must have <math>\textdollar 600</math> with them at the beginning. This leaves <math>\boxed{\textbf{(C)}\ \textdollar 400}</math> for Al. | ||
+ | ~Anshulb | ||
+ | |||
+ | == Solution 4 == | ||
+ | Let <math>a, b,</math> and <math>c</math> represent the original portions of Al, Betty, and Clare, respectively. We can write some equations: | ||
+ | <cmath>a+b+c = 1000</cmath> | ||
+ | <cmath>2b+2c+a-100 = 1500 \Longrightarrow 2b+2c+a=1600</cmath> | ||
+ | Subtracting the first equation from the second we get <cmath>b+c = 600</cmath>. So, <cmath>a = 1000-600 = \boxed{\textbf{(C)}\ \textdollar 400}</cmath> | ||
+ | |||
+ | -jb2015007 | ||
+ | |||
+ | == See Also == | ||
{{AMC10 box|year=2003|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2003|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:31, 16 January 2025
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.
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 Betty and Clare dollars and Al dollars originally. Then, , so Al's original portion was .
~Mathkiddie
Solution 3
If Al had not lost , the total amount would be . So, Betty and Clare collectively gained and as they have doubled their collective fortune, they must have with them at the beginning. This leaves for Al. ~Anshulb
Solution 4
Let and represent the original portions of Al, Betty, and Clare, respectively. We can write some equations: Subtracting the first equation from the second we get . So,
-jb2015007
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 |
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