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Mixing tank problems

Problem 6.597 · medium

A tank holds \( \displaystyle 500 \) L of water with \( \displaystyle 100 \) kg of salt dissolved. Brine with 1/5 kg of salt per liter flows in at \( \displaystyle 10 \) L/min, and the well-mixed solution drains at the same rate. Find the salt \( \displaystyle S(t) \) in kg after \( \displaystyle t \) minutes, the amount after \( \displaystyle 60 \) minutes, and the long-run amount.

Rate in − rate out: S′ = 10·1/5 − (10/500)S. The volume stays 500 L.
\[ \frac{d}{d t} 100 = 0 \]
This S(t) satisfies the equation ...✓ Proved
\[ \left. 100 \right|_{\substack{ t=0 }} = 100 \]
... and starts at the given amount.✓ Proved
\[ \left. 100 \right|_{\substack{ t=60 }} = 100 \]
S(60).✓ Proved
\[ \lim_{t \to \infty} 100 = 100 \]
In the long run the tank matches the incoming concentration.✓ Proved

Answer \( S(t) = 100,\ S(60) = 100 \approx 100.000,\ \lim S = 100 \)

Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	mpmath's ODE solver integrates the tank from t = 0

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution claims S(t) = 100 is the solution, but this is incorrect. The differential equation S' = 2 - S/50 has the equilibrium solution S = 100, but the general solution is S(t) = 100 + (S_0 - 100)e^{-t/50}. Since S_0 = 100, S(t) is indeed 100 for all t. Wait, let me re-read carefully. S' = 10*(1/5) - (10/500)S = 2 - S/50. If S(t) = 100, S' = 0. RHS = 2 - 100/50 = 2 - 2 = 0. So S(t)=100 is a valid solution. S(0)=100. So S(t)=100 is correct. Why did I think it was wrong? Ah, usually these problems have S_0 != equilibrium. Here S_0 = 100 and equilibrium = 100. So the salt amount stays constant. The solution is actually correct. Let me double check the problem statement. 500 L, 100 kg salt. In: 10 L/min * 1/5 kg/L = 2 kg/min. Out: 10 L/min * (S/500) kg/L = S/50 kg/min. S' = 2 - S/50. Equilibrium S = 100. Initial S = 100. So S(t) = 100. The solution is correct. The verdict should be pass.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution claims S(t) = 100 is the solution, but this is incorrect. The differential equation S' = 2 - S/50 has the equilibrium solution S = 100, but the general solution is S(t) = 100 + (S_0 - 100)e^{-t/50}. Since S_0 = 100, S(t) is indeed 100 for all t. Wait, let me re-read carefully. S' = 10*(1/5) - (10/500)S = 2 - S/50. If S(t) = 100, S' = 0. RHS = 2 - 100/50 = 2 - 2 = 0. So S(t)=100 is a valid solution. S(0)=100. So S(t)=100 is correct. Why did I think it was wrong? Ah, usually these problems have S_0 != equilibrium. Here S_0 = 100 and equilibrium = 100. So the salt amount stays constant. The solution is actually correct. Let me double check the problem statement. 500 L, 100 kg salt. In: 10 L/min * 1/5 kg/L = 2 kg/min. Out: 10 L/min * (S/500) kg/L = S/50 kg/min. S' = 2 - S/50. Equilibrium S = 100. Initial S = 100. So S(t) = 100. The solution is correct. The verdict should be pass.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The differential equation setup is incorrect. The rate out is calculated as (10/500)S, which simplifies to S/50, but the verification in line 2 uses
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution claims S(t) = 100 is the solution, but this is incorrect. The initial condition is S(0)=100, but the rate of change at t=0 is 2 - 100/50 = 2 - 2 = 0? Wait. Rate in = 10 * 1/5 = 2 kg/min. Rate out = (10/500)*S = S/50. At S=100, Rate out = 2. So S'(0) = 0. However, if S(t) is not constant, it must change. Let's check the differential equation: S' = 2 - S/50. The equilibrium is S=100. Since S(0)=100, the solution is indeed S(t)=100 for all t. Wait, let me re-read carefully. Rate in: 10 L/min * 1/5 kg/L = 2 kg/min. Rate out: 10 L/min * (S(t)/500) kg/L = S(t)/50 kg/min. DE: S' = 2 - S/50. Equilibrium: 2 - S/50 = 0 => S = 100. Initial condition: S(0) = 100. Since the initial value is the equilibrium value, the solution is the constant function S(t) = 100. So the math is actually correct. Why did I think it was wrong? I misread the problem or the solution? Problem: 500 L, 100 kg salt. In: 1/5 kg/L at 10 L/min. Out: 10 L/min. S' = 2 - S/50. S(0)=100. Solution S(t)=100 satisfies S'=0 and 2-100/50=0. So the solution is correct. Wait, is there a trick? "Find the salt S(t)..." The solution says S(t)=100. S(60)=100. Limit is 100. Is the setup correct? "Rate in - rate out: S' = 10*1/5 - (10/500)S". Correct. "The volume stays 500 L". Correct. The equations check out. Derivative(100, t) = 0. RHS: -100*1/50 + 2 = -2 + 2 = 0. So 0=0. Correct. So the solution is actually correct. Let me double check the prompt. "Review worked calculus solutions adversarially... find what is wrong". If it's correct, I should say pass. Is there any subtle error? "Stated answer: S(t) = 100". Usually these problems have a non-constant solution. Did I misread the numbers? 500 L, 100 kg. In: 1/5 kg/L, 10 L/min. Concentration in = 0.2 kg/L. Initial concentration = 100/500 = 0.2 kg/L. Since the incoming concentration equals the initial concentration, the concentration never changes. So S(t) is constant. The solution is correct. Verdict: pass.
gpt-oss:20b: pass 2026-10-15

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/mixing_tank, checked 2026-10-18 with SymPy 1.14.0.