2003 AMC 10B Problem 12

P11-15~2 minPrint

Linear Equations & Word Problems Systems of Equations Set up the equation/formula and compute; no special trick needed

Al, Betty, and Clare split $\$1000$ 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 $\$1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\$100$ dollars. What was Al's original portion?

Show solution without answering

Key insight

Treat Betty and Clare as one team whose money doubles. If Al starts with a, the team has 1000 - a, and the total 1500 gives a = 400.

Solution

1. Betty and Clare both double their money, so we never need their amounts separately. Treat them as one team.
2. Say Al starts with $a$ dollars. The team starts with the rest: $1000 - a$ dollars.
3. After a year:
- Al has lost $100$ , so he has $a - 100$ .
- The team has doubled, so it has $2(1000 - a) = 2000 - 2a$ .
4. Together they have $1500$ :
$(a - 100) + (2000 - 2a) = 1500.$
The left side simplifies to $1900 - a$ , so $1900 - a = 1500$ and $a = 400$ .
5. Check: the team starts with $600$ and ends with $1200$ . Al ends with $300$ . And $1200 + 300 = 1500$ .

The answer is $\boxed{\textbf{(C)}\ \$400}$ .

Why this works

When two people go through exactly the same change, you can lump them together as one amount. That turns three unknowns into one. The fact that each person starts with a different amount does not affect the answer at all.

Alternative approach

Pretend Al had not lost the $100$ . Then the final total would be $1600$ , which is $600$ more than the $1000$ they started with. All of that growth came from Betty and Clare doubling their money, and doubling adds an amount equal to what you started with. So Betty and Clare started with $600$ together, and Al started with $1000 - 600 = 400$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let Al start with $a$ dollars. Betty and Clare together start with $1000 - a$ dollars. Since each of them doubles her money, together they end with $2(1000 - a)$ , while Al ends with $a - 100$ .

The final total is $1500$ :
$(a - 100) + 2(1000 - a) = 1500.$
$a - 100 + 2000 - 2a = 1500 \quad\Longrightarrow\quad -a + 1900 = 1500 \quad\Longrightarrow\quad a = 400.$

Check: Betty and Clare start with $600$ and end with $1200$ ; Al ends with $300$ ; $1200 + 300 = 1500$ .

The answer is $\boxed{\textbf{(C)}\ \$400}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)