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Mixing tank problems

Problem 6.588 · medium

A tank holds \( \displaystyle 100 \) L of water with \( \displaystyle 20 \) kg of salt dissolved. Brine with 1/5 kg of salt per liter flows in at \( \displaystyle 4 \) 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 \( \displaystyle 30 \) minutes, and the long-run amount.

Rate in − rate out: S′ = 4·1/5 − (4/100)S. The volume stays 100 L.
Reviewed
\[ \frac{d}{d t} 20 = 0 \]
This S(t) satisfies the equation ...✓ Proved
\[ \left. 20 \right|_{\substack{ t=0 }} = 20 \]
... and starts at the given amount.✓ Proved
\[ \left. 20 \right|_{\substack{ t=30 }} = 20 \]
S(30).✓ Proved
\[ \lim_{t \to \infty} 20 = 20 \]
In the long run the tank matches the incoming concentration.✓ Proved

Answer \( S(t) = 20,\ S(30) = 20 \approx 20.000,\ \lim S = 20 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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 that the incoming concentration (1/5 kg/L) matches the initial concentration in the tank (20 kg / 100 L = 1/5 kg/L), leading to a constant solution S(t) = 20. The setup, verification, and limits are all correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies that the incoming concentration (1/5 kg/L) matches the initial concentration in the tank (20 kg / 100 L = 1/5 kg/L), leading to a constant solution S(t) = 20. The setup, verification, and limits are all correct.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies that the incoming concentration (1/5 kg/L) equals the initial concentration (20 kg / 100 L = 1/5 kg/L), implying the salt amount remains constant at 20 kg. The differential equation setup and subsequent verification are correct.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15
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.