2004 AMC 10B Problem 7

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 $\boxed{\textbf{(A)}\ 5}$.

An exchange rate is a proportion, so the received amount is the original amount times a fixed ratio. The story then reduces to a single linear equation. The final step, converting $d$ into a digit sum, exists to stop guessing from the choices; always reread what is actually asked before answering.

Each block of $7$ U.S. dollars turns into $10$ Canadian, a gain of $3$. The $60$ she spent is exactly the total gain (she ends with the same number $d$), so there were $60/3 = 20$ blocks and $d = 7 \cdot 20 = 140$.

• Applying the rate backwards (7 Canadian per 10 U.S.), which produces a negative d, or reporting 140 itself instead of its digit sum.
• Writing the equation as $\frac{10}{7}(d - 60) = d$, spending the $60$ before the exchange instead of after.