Rainwater Tank Fill Time
A 1,600-liter rainwater tank at a community garden currently holds 400 liters. From t = 0 minutes a supply line adds water at a constant 60 liters per minute. The overflow pump is off at the start; it switches on automatically the moment the tank reaches 700 liters and then removes 15 liters per minute while the supply line keeps running at 60. The tank is considered full at 1,600 liters. A volunteer writes the single equation 400 + 45t = 1,600 and concludes the tank fills in about 26.7 minutes.
1. Write a correct model for the tank volume V(t) and determine when the tank is actually full.
Read the worked answer
The rate is not constant for the whole fill, so the model is piecewise.
Phase 1 (pump off): net rate 60 L/min, starting at 400 L. The tank reaches 700 L when $400 + 60t = 700 \Rightarrow t = 5$ minutes.
$$V(t) = 400 + 60t, \quad 0 \le t \le 5$$
Phase 2 (pump on): net rate $60 - 15 = 45$ L/min, starting from 700 L at t = 5.
$$V(t) = 700 + 45(t - 5), \quad t \ge 5$$
Full means $V = 1600$: $700 + 45(t-5) = 1600 \Rightarrow 45(t-5) = 900 \Rightarrow t - 5 = 20 \Rightarrow t = 25$.
The tank is full at t = 25 minutes. Check: 900 L gained at 45 L/min = 20 min, plus 5 min of the fast phase.
2. At what time does the tank hold 1,150 liters?
Read the worked answer
1,150 L is above 700 L, so it occurs in Phase 2.
$$700 + 45(t - 5) = 1150$$ $$45(t-5) = 450 \Rightarrow t - 5 = 10 \Rightarrow t = 15$$
The tank holds 1,150 L at t = 15 minutes. A quick plausibility check: at t = 5 the tank has 700 L, and 10 more minutes at 45 L/min adds 450 L, giving 1,150 L. Consistent.
3. Explain what the volunteer's equation 400 + 45t = 1,600 actually describes, and why it gives a time that is too long.
Read the worked answer
The equation charges the 15 L/min outflow against the tank from t = 0, i.e., it assumes the pump was already running during the first minutes. That describes a different situation: a tank with both lines open from the start.
Because the real tank fills at the faster 60 L/min for the first 5 minutes, the volunteer's constant-slow-rate model understates the volume gained early and therefore overstates the time:
- Volunteer: $45t = 1200 \Rightarrow t \approx 26.7$ min.
- Correct: 25 min.
The 1.7-minute gap equals the extra water gained during the fast phase divided by the slow rate: $15 \text{ L/min} \times 5 \text{ min} = 75$ L, and $75/45 \approx 1.67$ min. The lesson is that a single linear equation is only valid while the rate is constant.