Water Tank Refill Plan
A community garden stores rainwater in a cylindrical tank of radius 1.2 m and height 2.5 m. A pump fills the tank at a rate that is not constant: it delivers 18 litres per minute for the first 20 minutes, then 12 litres per minute until the tank holds 5,000 litres, and finally 8 litres per minute. The garden needs 4,800 litres for a weekend irrigation schedule. A sensor reports the tank currently holds 1,750 litres. The pump must not run more than 240 minutes total, and the garden wants a 15% safety margin above the 4,800-litre target before stopping. One litre equals 0.001 cubic metres, and the tank's usable capacity is 90% of its geometric volume.
1. Determine whether the pump can reach the target with the safety margin within the 240-minute limit. Show the volume and time calculations.
Read the worked answer
Target volume. 4,800 litres plus a 15% margin: 4,800 × 1.15 = 5,520 litres.
Usable capacity check. Geometric volume = π × 1.2² × 2.5 ≈ 3.1416 × 1.44 × 2.5 ≈ 11.31 m³ = 11,310 litres. Usable = 0.9 × 11,310 ≈ 10,179 litres, so 5,520 litres fits.
Time to add 5,520 − 1,750 = 3,770 litres.
- Stage 1 (0–20 min): 18 × 20 = 360 litres. Remaining: 3,410 litres.
- Stage 2 runs until the tank holds 5,000 litres, i.e. until 5,000 − 1,750 = 3,250 litres have been added. After stage 1, 3,250 − 360 = 2,890 litres remain at 12 L/min: 2,890 ÷ 12 ≈ 240.8 min. This already exceeds the 240-minute limit.
Conclusion. The pump cannot reach 5,520 litres within 240 minutes; it would need roughly 20 + 241 + (5,520 − 5,000) ÷ 8 ≈ 20 + 241 + 65 ≈ 326 minutes. The plan fails the time constraint.
2. The garden can lower the safety margin or accept a smaller irrigation amount. Find the largest safety-margin percentage (to the nearest whole percent) that keeps the total pumping time at or below 240 minutes, assuming the irrigation target stays 4,800 litres.
Read the worked answer
Set up the time function. Let the required volume be V litres, with V ≥ 5,000 so the final stage is used. Time = 20 + (5,000 − 1,750 − 360) ÷ 12 + (V − 5,000) ÷ 8 = 20 + 2,890 ÷ 12 + (V − 5,000) ÷ 8 ≈ 20 + 240.83 + (V − 5,000) ÷ 8.
Solve for V at 240 minutes. 240 − 260.83 = (V − 5,000) ÷ 8 gives a negative value, so no V ≥ 5,000 works. Check V between 3,250 and 5,000 (stage 2 only): time = 20 + (V − 1,750 − 360) ÷ 12 ≤ 240 → V − 2,110 ≤ 2,640 → V ≤ 4,750 litres.
Convert to margin. Margin = (4,750 − 4,800) ÷ 4,800 < 0, so even 4,800 litres is unreachable in 240 minutes. Maximum deliverable volume: 20 + (V − 2,110) ÷ 12 = 240 → V = 4,750 litres. That is below the irrigation target, so the margin would have to be negative (about −1%). The garden must extend the time limit or reduce the irrigation target; no nonnegative margin works.
3. Suppose the pump's first-stage rate is unknown but the second and third stages are as stated. If the pump reaches exactly 5,520 litres in exactly 240 minutes, what first-stage rate (L/min) is required? Round to one decimal place and judge whether it is physically plausible given the other rates.
Read the worked answer
Write the time equation. Let r be the stage-1 rate for 20 minutes. Stage 1 adds 20r litres. Stage 2 must bring the tank from 1,750 + 20r to 5,000, adding 3,250 − 20r litres at 12 L/min. Stage 3 adds 520 litres at 8 L/min, taking 65 minutes.
Total time. 20 + (3,250 − 20r) ÷ 12 + 65 = 240 → (3,250 − 20r) ÷ 12 = 155 → 3,250 − 20r = 1,860 → 20r = 1,390 → r = 69.5 L/min.
Plausibility judgment. 69.5 L/min is nearly four times the stated stage-1 rate of 18 L/min and far above the later rates, which is inconsistent with a pump that slows as back-pressure builds. The value is arithmetically correct but physically implausible; a real fix is a longer time window or a larger initial volume, not a faster first stage.