2003 AMC 10B Problem 12

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

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.

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 $1000 - 600 = 400$.

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