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Mixing tank problems

Problem 6.593 · 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 \( \displaystyle 20 \) 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 \( \displaystyle 20 \) minutes, and the long-run amount.

Rate in − rate out: S′ = 20·1/5 − (20/100)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
\[ \left. 20 \right|_{\substack{ t=20 }} = 20 \]
S(20).✓ Proved
\[ \lim_{t \to \infty} 20 = 20 \]
In the long run the tank matches the incoming concentration.✓ Proved

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

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 that the rate of change is zero because the incoming salt concentration equals the initial concentration, leading to a constant solution. The algebraic checks confirm this constant satisfies the ODE and initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies that the rate of change is zero because the incoming salt concentration equals the initial concentration, leading to a constant solution. The algebraic checks confirm this constant satisfies the ODE and initial conditions.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The differential equation in line 1 is incorrect: the rate in is 4 kg/min and rate out is 0.2S, so S' = 4 - 0.2S, not 4 - 2S. Consequently, the constant solution S(t)=20 is wrong; the correct solution approaches 20 kg but starts at 20 kg only if the initial condition matched the equilibrium, which it does, but the derivation implies a different dynamics. Wait, if S(0)=20 and equilibrium is 20, then S(t)=20 is actually the correct solution. Let me re-read carefully. Rate in = 20 L/min * 1/5 kg/L = 4 kg/min. Rate out = 20 L/min * (S/100) kg/L = 0.2S kg/min. DE: S' = 4 - 0.2S. Equilibrium: 4 - 0.2S = 0 => S = 20. Initial condition S(0) = 20. Since the initial amount equals the equilibrium amount, S(t) = 20 for all t is indeed the correct solution. The sentence in line 1 says 'S' = 20*1/5 - (20/100)S'. 20*1/5 is 4. (20/100)S is 0.2S. So the equation written is S' = 4 - 0.2S. This is correct. The solution S(t)=20 satisfies this. The notes in the equation lines are just checking the constant function. The logic holds. Why did I think it was wrong? Ah, I misread '20*1/5' as something else or miscalculated. 20/5 = 4. 20/100 = 1/5 = 0.2. The DE is correct. The solution is correct. The limit is correct. The answer is correct. Is there a subtle error? 'Rate in - rate out: S' = ...'. This is standard. The volume stays 100 L. Correct. The solution checks out. Wait, looking at line 2: `Derivative(20, t) = -20*1/5 + (1/5)*20`. The LHS is 0. The RHS is -4 + 4 = 0. This confirms S(t)=20 is a solution. Line 1 sets up the DE. Line 2 verifies the candidate solution. Line 3 verifies IC. Line 4 calculates S(20). Line 5 calculates limit. Everything seems correct. Why would this fail? Maybe the phrasing 'Rate in - rate out' implies a specific sign convention? No, that's standard. Maybe the 'Stated answer' format is weird? No, I'm reviewing the solution steps. The solution is sound.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The differential equation setup is incorrect: the rate in is 4 kg/min, but the rate out is (20/100)S = 0.2S, so S' = 4 - 0.2S. The solution S(t)=20 is a constant equilibrium, but the initial condition is S(0)=20, which happens to match the equilibrium. However, the stated answer implies S(t) is always 20, which is technically correct for this specific initial condition, but the derivation in line 2 is nonsensical (checking derivative of constant 20 against the ODE). Wait, if S(t)=20, S'=0. RHS = 4 - 0.2(20) = 0. So S(t)=20 is indeed the solution. The notes in line 2 are confusing but the math holds. Let's re-read carefully. S(0)=20. Equilibrium is S_eq = 4/0.2 = 20. Since S(0) = S_eq, S(t) = 20 for all t. The solution is actually correct. Why did I think it was wrong? The prompt asks to find S(t). The solution provides S(t)=20. This is correct. The 'unchecked' equations are just verification steps. Is there a subtle error? Line 1 says S' = 20*1/5 - (20/100)S. This is correct. Line 2 checks if S(t)=20 satisfies it. It does. Line 3 checks IC. It does. Line 4 and 5 check values. It seems correct. Wait, is the 'stated answer' format weird? No. Is there a trap? 'Brine with 1/5 kg of salt per liter flows in at 20 L/min'. Rate in = 4 kg/min. Volume = 100 L. Rate out = (S/100)*20 = S/5. S' = 4 - S/5. Equilibrium S = 20. Initial S = 20. So S(t) = 20. The solution is correct. Why would it fail? Maybe the 'unchecked' lines are not sentences? The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Lines 2-5 are marked [equation, unchecked]. The prompt says 'do not re-check the algebra of a line marked proved'. These are marked 'unchecked'. Does that mean I should check them? Or does it mean the CAS didn't check them? The prompt says 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. Lines 2-5 are NOT marked proved. They are marked 'unchecked'. This implies I might need to verify them? Or perhaps the format is just indicating they are equations. Let's look at the 'Stated answer'. S(t) = 20. This is correct. Is there any misleading part? The solution is trivial because the initial condition equals the steady state. A student might miss the general solution form. But the question asks for S(t), and S(t)=20 is the correct function. I will mark it 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.