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Mixing tank problems

Problem 6.449 · medium

A tank holds \( \displaystyle 100 \) L of water with \( \displaystyle 10 \) kg of salt dissolved. Brine with 1/10 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′ = 10·1/10 − (10/100)S. The volume stays 100 L.
Reviewed
\[ \frac{d}{d t} 10 = 0 \]
This S(t) satisfies the equation ...✓ Proved
\[ \left. 10 \right|_{\substack{ t=0 }} = 10 \]
... and starts at the given amount.✓ Proved
\[ \left. 10 \right|_{\substack{ t=60 }} = 10 \]
S(60).✓ Proved
\[ \lim_{t \to \infty} 10 = 10 \]
In the long run the tank matches the incoming concentration.✓ Proved

Answer \( S(t) = 10,\ S(60) = 10 \approx 10.000,\ \lim S = 10 \)

Lines: 4 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The solution correctly identifies the differential equation and verifies that the constant function S(t)=10 satisfies both the ODE and the initial condition. The reasoning is sound.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the differential equation and verifies that the constant function S(t)=10 satisfies both the ODE and the initial condition. The reasoning is sound.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The differential equation setup in line 1 has the wrong sign for the outflow term (it should be minus, not plus, or rather the equation written is S'
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The differential equation setup in line 1 is incorrect. The rate out is (S(t)/100) * 10 = S(t)/10, but the equation written is S' = 1 - S/10, which
gpt-oss:20b: pass 2026-10-14

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 2026-10-17 with SymPy 1.14.0.