[Algebra]
Problem: Let $T=100$. Nellie has a flight from Rome to Athens that is scheduled to last for $T+30$ minutes. However, owing to a tailwind, her flight only lasts for $T$ minutes. The plane's speed is 1.5 miles per minute faster than what it would have been for the originally scheduled flight. Compute the distance (in miles) that the plane travels.

Solution: Let $D$ be the distance in miles traveled by the plane. The given conditions imply that $\frac{D}{T}-\frac{D}{T+30}=1.5 \Rightarrow \frac{30 D}{T(T+30)}=1.5 \Rightarrow D=\frac{T(T+30)}{20}$. With $T=100, D=5 \cdot 130=\mathbf{6 5 0}$.

Final answer: 650