Difference between revisions of "2003 AMC 10B Problems/Problem 12"

(Solution)
(Solution)
Line 9: Line 9:
 
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>\\
+
<cmath>a+b+c=1000</cmath>
<cmath>2a+2b+2c=2000</cmath>\\
+
<cmath>2a+2b+2c=2000</cmath>
\\
+
<cmath>a-100+2b+2c=1500</cmath>
<cmath>a-100+2b+2c=1500\\
+
<cmath>a+2b+2c=1600</cmath>
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>

Revision as of 10:26, 20 July 2019

Problem

Al, Betty, and Clare split $\textdollar 1000$ 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 $\textdollar 1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\textdollar100$ dollars. What was Al's original portion?

$\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500$

Solution

For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let $a, b,$ and $c$ represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have $a-100, 2b,$ and $2c$. From this, we can write two equations.

\[a+b+c=1000\] \[2a+2b+2c=2000\] \[a-100+2b+2c=1500\] \[a+2b+2c=1600\]

Since all we need to find is $a,$ subtract the second equation from the first equation to get $a=400.$

Al's original portion was $\boxed{\textbf{(C)}\ \textdollar 400}$.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS