2003 AMC 10B Problems/Problem 12

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

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2003 AMC 10B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also