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

(Created page with "==Problem== Al, Betty, and Clare split <math> \ </math><math>1000</math> among them to be invested in different ways. Each begins with a different amount. At the end of one year...")
 
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==Problem==
 
==Problem==
  
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Al, Betty, and Clare split <math> \ </math><math>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> \ </math><math>1500</math> dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose <math> \ </math><math>100</math> dollars. What was Al's original portion?
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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) }\ </math><math>250 \qquad\textbf{(B) }\ </math> <math>350 \qquad\textbf{(C) }\ </math><math>400\qquad\textbf{(D) }\ </math> <math>450\qquad\textbf{(E) }\ </math><math>500 </math>
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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>
  
 
==Solution==
 
==Solution==
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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.
  
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<cmath>a+b+c=1000</cmath>
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<cmath>a+b+c=1000\\
 +
2a+2b+2c=2000\\
 +
\\
 +
a-100+2b+2c=1500\\
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a+2b+2c=1600</cmath>
  
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<cmath>a-100+2b+2c=1500</cmath>
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Since all we need to find is <math>a,</math> subtract the second equation from the first equation to get <math>a=400.</math>
  
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Since all we need to find is <math>a,</math> substitute <math>b+c</math> in terms of <math>a</math> and solve for <math>a.</math>
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Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>.
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<cmath>b+c=1000-a</cmath>
 
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<cmath>\begin{align*}
 
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a-100+2(b+c)&=1500\\
 
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a+2(1000-a)&=1600\\
 
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a+2000-2a&=1600\\
 
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a&=400\end{align*}</cmath>
 
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Al's original portion was <math>\boxed{\mathrm{(C) \ } 400}</math>.
 
  
 
==See Also==
 
==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}}

Revision as of 18:12, 26 November 2011

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 (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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