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

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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>
  
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==Solution==
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==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, marked by (1) and (2).  
 
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).  
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Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>.
 
Al's original portion was <math>\boxed{\textbf{(C)}\ \textdollar 400}</math>.
  
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==Solution 2==
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Suppose the total amount of money Betty and Clare has is <math>1000-x</math> and Al has <math>x</math> dollars. Then, <math>(x-100)+2(1000-x)=1500</math>, so Al has <math>\boxed{\textbf{(C)}\ \textdollar 400}</math> dollars.
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~Mathkiddie
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==Solution 3==
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Suppose if Al had not lost <math>\textdollar 100</math>. The total amount would be <math>\textdollar 1600</math>. As he has not gained any amount. So, Betty and Clare have collectively gained <math>\textdollar 600</math> and as they have doubled their collective fortune,
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they must have <math>\textdollar 600</math> with them at the beginning. This leaves <math>\boxed{\textbf{(C)}\ \textdollar 400}</math> for Al.
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~Anshulb
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:11, 2 January 2023

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 1

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, marked by (1) and (2).

\[a+b+c=1000\] \[(1). \text{ }2a+2b+2c=2000\] \[a-100+2b+2c=1500\] \[(2). \text{ }a+2b+2c=1600\]

(Equations (1) and (2) are derived from each equation above them.)

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 $\boxed{\textbf{(C)}\ \textdollar 400}$.

Solution 2

Suppose the total amount of money Betty and Clare has is $1000-x$ and Al has $x$ dollars. Then, $(x-100)+2(1000-x)=1500$, so Al has $\boxed{\textbf{(C)}\ \textdollar 400}$ dollars.

~Mathkiddie

Solution 3

Suppose if Al had not lost $\textdollar 100$. The total amount would be $\textdollar 1600$. As he has not gained any amount. So, Betty and Clare have collectively gained $\textdollar 600$ and as they have doubled their collective fortune, they must have $\textdollar 600$ with them at the beginning. This leaves $\boxed{\textbf{(C)}\ \textdollar 400}$ for Al. ~Anshulb

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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