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Mixing tank problems

Problem 6.705 · medium

A tank holds \( \displaystyle 100 \) L of water with \( \displaystyle 50 \) kg of salt dissolved. Brine with 1/2 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 \( \displaystyle 10 \) minutes, and the long-run amount.

Rate in − rate out: S′ = 10·1/2 − (10/100)S. The volume stays 100 L.
\[ \frac{d}{d t} 50 = 0 \]
This S(t) satisfies the equation ...✓ Proved
\[ \left. 50 \right|_{\substack{ t=0 }} = 50 \]
... and starts at the given amount.✓ Proved
\[ \left. 50 \right|_{\substack{ t=10 }} = 50 \]
S(10).✓ Proved
\[ \lim_{t \to \infty} 50 = 50 \]
In the long run the tank matches the incoming concentration.✓ Proved

Answer \( S(t) = 50,\ S(10) = 50 \approx 50.000,\ \lim S = 50 \)

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 (misleading) — The solution claims S(t) = 50 is the solution to the differential equation, but S(t) = 50 is a constant function with derivative 0, whereas the RHS of the DE is 5 - 0.1S. At S=50, RHS = 0, so it is an equilibrium, but the initial condition is S(0)=50, so the solution is indeed S(t)=50. Wait, let me re-read carefully. The problem states 50 kg of salt initially. The inflow is 10 L/min * 0.5 kg/L = 5 kg/min. The outflow is (10/100)S = 0.1S. So S' = 5 - 0.1S. If S(0)=50, then S'(0) = 5 - 5 = 0. Since the solution to this linear ODE is unique and S(t)=50 is a constant solution satisfying the IC, S(t)=50 is correct. Why did I think it was wrong? Let me check the algebra again. S' = 5 - 0.1S. If S(t) = 50, S'(t) = 0. RHS = 5 - 0.1(50) = 0. It works. The limit is 50. S(10) is 50. The solution is actually correct. Is there a trick? "Brine with 1/2 kg of salt per liter flows in at 10 L/min". Rate in = 5 kg/min. Volume = 100 L. Rate out = (10/100)S = S/10. S' = 5 - S/10. Equilibrium is S=50. Initial condition S(0)=50. So S(t) stays at 50. The solution is correct. Why would it fail? Maybe the prompt implies a trick? No, it's a standard mixing problem where the initial amount equals the equilibrium amount. The solution is sound.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-10-19 — The solution claims S(t) = 50 is the solution to the differential equation, but S(t) = 50 is a constant function with derivative 0, whereas the RHS of the DE is 5 - 0.1S. At S=50, RHS = 0, so it is an equilibrium, but the initial condition is S(0)=50, so the solution is indeed S(t)=50. Wait, let me re-read carefully. The problem states 50 kg of salt initially. The inflow is 10 L/min * 0.5 kg/L = 5 kg/min. The outflow is (10/100)S = 0.1S. So S' = 5 - 0.1S. If S(0)=50, then S'(0) = 5 - 5 = 0. Since the solution to this linear ODE is unique and S(t)=50 is a constant solution satisfying the IC, S(t)=50 is correct. Why did I think it was wrong? Let me check the algebra again. S' = 5 - 0.1S. If S(t) = 50, S'(t) = 0. RHS = 5 - 0.1(50) = 0. It works. The limit is 50. S(10) is 50. The solution is actually correct. Is there a trick? "Brine with 1/2 kg of salt per liter flows in at 10 L/min". Rate in = 5 kg/min. Volume = 100 L. Rate out = (10/100)S = S/10. S' = 5 - S/10. Equilibrium is S=50. Initial condition S(0)=50. So S(t) stays at 50. The solution is correct. Why would it fail? Maybe the prompt implies a trick? No, it's a standard mixing problem where the initial amount equals the equilibrium amount. The solution is sound.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly sets up the differential equation for a constant-volume mixing problem, identifies the constant solution S(t)=50, and verifies it satisfies the initial condition and the ODE. The conclusions for S(10) and the limit are correct.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the differential equation and verifies that the constant function S(t)=50 is the unique solution satisfying the initial condition. The logic is sound and the calculations are correct.
gpt-oss:20b: pass 2026-10-18

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 2026-10-19 with SymPy 1.14.0.