2004 AMC 10B Problem 7

1. Try a small amount first to see how the exchange works. $7$ U.S. dollars become $10$ Canadian dollars. $14$ U.S. dollars become $20$ Canadian. So $d$ U.S. dollars become $\frac{10}{7}d$ Canadian dollars.
2. She spends $60$ and has $d$ left:
$$\frac{10}{7}d - 60 = d.$$
3. Move the $d$ terms together. $\frac{10}{7}d - d = \frac{3}{7}d$, so
$$\frac{3}{7}d = 60, \qquad d = 60 \times \frac{7}{3} = 140.$$
4. Check: $140$ U.S. dollars is $20$ groups of $7$, which becomes $20$ groups of $10$, or $200$ Canadian. Spending $60$ leaves $140$. It works.
5. The question asks for the sum of the digits of $d$: $1 + 4 + 0 = 5$.

The answer is $\boxed{\textbf{(A)}\ 5}$.

An exchange rate means the new amount is the old amount times a fixed fraction. Once you write that down, the story is one equation. Always reread what the question asks at the end: here it wants the digit sum, not $d$.

Think in groups of $7$ U.S. dollars. Each group becomes $10$ Canadian dollars, so each group "gains" $3$ dollars. She ends with the same number $d$ she started with, so the $60$ she spent must be exactly the total gain. That is $60 \div 3 = 20$ groups, so $d = 20 \times 7 = 140$, with digit sum $5$.

Every $7$ U.S. dollars becomes $10$ Canadian dollars, so $d$ U.S. dollars become $\frac{10}{7} d$ Canadian dollars. After spending $60$ she has $\frac{10}{7} d - 60$, and this is supposed to equal $d$:
$$\frac{10}{7} d - 60 = d \quad\Longrightarrow\quad \frac{3}{7} d = 60 \quad\Longrightarrow\quad d = 140.$$

Check: $140$ U.S. dollars give $200$ Canadian; spending $60$ leaves $140$. The digits of $140$ sum to $1 + 4 + 0 = 5$.

The answer is $\boxed{\textbf{(A)}\ 5}$.