1. **State the problem:** We have an inverted square pyramid tank with height $h=7$ m and top edge length $a=5$ m. Initially, water depth is 3 m. Water is pumped in at 17 gallons per minute. We want to find how many hours it takes to fill the tank. 2. **Formula and concepts:** The volume $V$ of a square pyramid is given by $$V=\frac{1}{3} \times \text{base area} \times \text{height}.$$ Here, the base is a square with side length $s$ that varies with water depth $x$. 3. **Relate side length to water depth:** Since the pyramid is similar at all heights, side length $s$ at depth $x$ is proportional to $x$: $$s=\frac{a}{h}x=\frac{5}{7}x.$$ 4. **Volume of water at depth $x$:** $$V(x)=\frac{1}{3} s^2 x=\frac{1}{3} \left(\frac{5}{7}x\right)^2 x=\frac{1}{3} \times \frac{25}{49} x^3=\frac{25}{147} x^3.$$ 5. **Calculate total volume of tank:** For $x=7$ m, $$V_{total}=\frac{25}{147} \times 7^3=\frac{25}{147} \times 343=\frac{25 \times 343}{147}=\frac{8575}{147} \approx 58.37 \text{ m}^3.$$ 6. **Calculate initial volume of water:** For $x=3$ m, $$V_{initial}=\frac{25}{147} \times 3^3=\frac{25}{147} \times 27=\frac{675}{147} \approx 4.59 \text{ m}^3.$$ 7. **Volume to fill:** $$V_{fill}=V_{total}-V_{initial}=58.37-4.59=53.78 \text{ m}^3.$$ 8. **Convert volume to gallons:** 1 cubic meter = 264.172 gallons, $$V_{fill} = 53.78 \times 264.172 \approx 14207.5 \text{ gallons}.$$ 9. **Calculate time to fill:** Pump rate = 17 gallons/min, $$t=\frac{14207.5}{17} \approx 835.15 \text{ minutes}.$$ 10. **Convert time to hours:** $$\frac{835.15}{60} \approx 13.92 \text{ hours}.$$ **Final answer:** It will take approximately 13.92 hours to fill the tank.