1. **State the problem:** Betty paints twice as fast as Dan. Together, Dan and Betty paint 2400 square feet in 4 hours. Sue joins them, and all three paint 3600 square feet in 3 hours. We want to find how many square feet Sue can paint alone in 4 hours and 27 minutes. 2. **Define variables:** Let Dan's painting rate be $d$ square feet per hour. 3. **Express Betty's rate:** Betty paints twice as fast as Dan, so her rate is $2d$ square feet per hour. 4. **Calculate combined rate of Dan and Betty:** Together, they paint 2400 square feet in 4 hours, so their combined rate is: $$\frac{2400}{4} = 600 \text{ square feet per hour}$$ 5. **Set up equation for Dan and Betty's rates:** $$d + 2d = 3d = 600$$ 6. **Solve for Dan's rate:** $$d = \frac{600}{3} = 200 \text{ square feet per hour}$$ 7. **Calculate Betty's rate:** $$2d = 2 \times 200 = 400 \text{ square feet per hour}$$ 8. **Calculate combined rate of Dan, Betty, and Sue:** They paint 3600 square feet in 3 hours, so their combined rate is: $$\frac{3600}{3} = 1200 \text{ square feet per hour}$$ 9. **Set up equation for Sue's rate $s$:** $$d + 2d + s = 1200$$ $$3d + s = 1200$$ 10. **Substitute $d=200$ into the equation:** $$3 \times 200 + s = 1200$$ $$600 + s = 1200$$ 11. **Solve for Sue's rate:** $$s = 1200 - 600 = 600 \text{ square feet per hour}$$ 12. **Calculate time Sue works alone:** 4 hours and 27 minutes = 4.45 hours (since 27 minutes = 27/60 = 0.45 hours). 13. **Calculate area Sue can paint alone:** $$\text{Area} = s \times \text{time} = 600 \times 4.45 = 2670 \text{ square feet}$$ **Final answer:** Sue can paint **2670** square feet alone in 4 hours and 27 minutes.