1. **State the problem:** Professor Leisurely takes twice as much time to correct 40 exams as Professor Swift takes to correct 60 exams. We want to find how many exams Swift can correct in the time Leisurely takes to correct 60 exams. 2. **Define variables:** Let $t_S$ be the time Swift takes to correct one exam. Let $t_L$ be the time Leisurely takes to correct one exam. 3. **Write the given information as equations:** Leisurely's time to correct 40 exams is $40 t_L$. Swift's time to correct 60 exams is $60 t_S$. Given: $$40 t_L = 2 \times 60 t_S = 120 t_S$$ 4. **Find the ratio of times per exam:** Divide both sides by 40: $$\cancel{40} t_L = \cancel{40} \times 3 t_S \implies t_L = 3 t_S$$ Leisurely takes 3 times as long per exam as Swift. 5. **Find how many exams Swift corrects in Leisurely's time for 60 exams:** Leisurely's time for 60 exams is: $$60 t_L = 60 \times 3 t_S = 180 t_S$$ Swift corrects exams at $t_S$ per exam, so number of exams Swift corrects in that time is: $$\frac{180 t_S}{t_S} = 180$$ **Final answer:** Swift corrects **180 written exams** in the time Leisurely corrects 60 exams.