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Mixing tank problems

Problem 6.699 · 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 \( \displaystyle 4 \) 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 \( \displaystyle 60 \) minutes, and the long-run amount.
  1. Rate in − rate out: S′ = 4·1/5 − (4/500)S. The volume stays 500 L.
  2. \[ \frac{d}{d t} 100 = 0 \]
    This S(t) satisfies the equation ...✓ Proved
  3. \[ \left. 100 \right|_{\substack{ t=0 }} = 100 \]
    ... and starts at the given amount.✓ Proved
  4. \[ \left. 100 \right|_{\substack{ t=60 }} = 100 \]
    S(60).✓ Proved
  5. \[ \lim_{t \to \infty} 100 = 100 \]
    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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0mpmath's ODE solver integrates the tank from t = 0

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution fails to solve the differential equation. It incorrectly assumes the constant function S(t)=100 is the solution, ignoring the initial condition S(0)=100 combined with the non-zero derivative S'(0) = 4/5 - 4/5 = 0? Wait, S'(0) = 4/5 - 4/5 = 0. Actually, if S(0)=100, then S'(0) = 4/5 - 4/5 = 0. If S(t)=100 for all t, then S'(t)=0. The RHS is 4/5 - 4/500 * 100 = 4/5 - 4/5 = 0. So S(t)=100 IS the solution. The system is already at equilibrium. The solution is technically correct but the explanation in line 1 sets up the DE and line 2 checks a specific function. It doesn't explicitly state that S(t)=100 is the unique solution, but it verifies it satisfies the DE and IC. However, usually one solves the DE to find S(t). Here, the 'solution' just plugs in 100. Is this an error? The prompt asks to find S(t). The answer provided is S(t)=100. The verification steps show it works. But step 1 says 'Rate in - rate out: S' = ...'. Step 2 checks that 100 satisfies it. It doesn't derive it. But since the initial state is the equilibrium state, the solution is constant. The logic holds. Wait, let me re-read carefully. 'Find the salt S(t)...'. The answer is S(t)=100. The steps verify this. Is there a flaw? The problem is that the student might think S(t) is always 100 for any initial condition, or that they didn't need to solve the DE. But strictly speaking, the verification is valid. However, usually 'Find S(t)' implies deriving the general solution and applying IC. Here, the 'derivation' is missing, replaced by a check. Is this 'misleading'? It teaches that you can just guess and check. But in this specific case, the guess is correct. Let's look closer. S' = 0.8 - 0.008S. If S=100, S'=0.8-0.8=0. Correct. S(0)=100. Correct. So S(t)=100 is the correct answer. The solution is sound. Why would it fail? Maybe the 'Stated answer' format is weird? No. Maybe the lack of derivation is the issue? The instructions say 'Check that... each conclusion follows from the lines before it'. Line 2 says 'This S(t) satisfies the equation'. Line 3 says 'starts at the given amount'. Line 4 and 5 compute values. The conclusion S(t)=100 is supported by the fact that it's a constant solution satisfying the IC. For a linear ODE, the solution is unique. So verifying a solution is sufficient. I will mark it pass.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution fails to solve the differential equation. It incorrectly assumes the constant function S(t)=100 is the solution, ignoring the initial condition S(0)=100 combined with the non-zero derivative S'(0) = 4/5 - 4/5 = 0? Wait, S'(0) = 4/5 - 4/5 = 0. Actually, if S(0)=100, then S'(0) = 4/5 - 4/5 = 0. If S(t)=100 for all t, then S'(t)=0. The RHS is 4/5 - 4/500 * 100 = 4/5 - 4/5 = 0. So S(t)=100 IS the solution. The system is already at equilibrium. The solution is technically correct but the explanation in line 1 sets up the DE and line 2 checks a specific function. It doesn't explicitly state that S(t)=100 is the unique solution, but it verifies it satisfies the DE and IC. However, usually one solves the DE to find S(t). Here, the 'solution' just plugs in 100. Is this an error? The prompt asks to find S(t). The answer provided is S(t)=100. The verification steps show it works. But step 1 says 'Rate in - rate out: S' = ...'. Step 2 checks that 100 satisfies it. It doesn't derive it. But since the initial state is the equilibrium state, the solution is constant. The logic holds. Wait, let me re-read carefully. 'Find the salt S(t)...'. The answer is S(t)=100. The steps verify this. Is there a flaw? The problem is that the student might think S(t) is always 100 for any initial condition, or that they didn't need to solve the DE. But strictly speaking, the verification is valid. However, usually 'Find S(t)' implies deriving the general solution and applying IC. Here, the 'derivation' is missing, replaced by a check. Is this 'misleading'? It teaches that you can just guess and check. But in this specific case, the guess is correct. Let's look closer. S' = 0.8 - 0.008S. If S=100, S'=0.8-0.8=0. Correct. S(0)=100. Correct. So S(t)=100 is the correct answer. The solution is sound. Why would it fail? Maybe the 'Stated answer' format is weird? No. Maybe the lack of derivation is the issue? The instructions say 'Check that... each conclusion follows from the lines before it'. Line 2 says 'This S(t) satisfies the equation'. Line 3 says 'starts at the given amount'. Line 4 and 5 compute values. The conclusion S(t)=100 is supported by the fact that it's a constant solution satisfying the IC. For a linear ODE, the solution is unique. So verifying a solution is sufficient. I will mark it pass.
  • gpt-oss:20b: pass 2026-10-19
  • qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution fails to solve the differential equation, instead assuming the constant solution S(t)=100 without justification. Since the initial condition (100 kg) matches the equilibrium value, the constant function is indeed the solution, but the steps presented merely verify that 100 is a steady state rather than deriving the general solution or explaining why the transient term is zero. A student would not learn how to solve the mixing problem.
  • gpt-oss:20b: pass 2026-10-18
  • qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution incorrectly concludes that S(t) is constant. The differential equation S' = 4/5 - 4S/500 has a non-zero derivative at t=0 (since 4/5 - 4(100)/500 = 0.8 - 0.8 = 0 is actually true? Wait. 4/5 = 0.8. 4*100/500 = 400/500 = 0.8. So S'(0) = 0. But is S(t) constant? The equilibrium is S_eq = (4/5) / (4/500) = 100. Since S(0)=100, the solution is indeed S(t)=100. Let me re-read carefully. Rate in: 4 L/min * 1/5 kg/L = 4/5 kg/min. Rate out: 4 L/min * (S(t)/500) kg/L = 4S/500 kg/min. S' = 4/5 - 4S/500. Equilibrium: 4/5 = 4S/500 => 1/5 = S/500 => S = 100. Initial condition: S(0) = 100. Since the initial condition is exactly the equilibrium value, the solution is indeed the constant function S(t) = 100. So the math is actually correct. Why did I think it was wrong? Let me double check the problem statement. "A tank holds 500 L of water with 100 kg of salt dissolved." -> S(0) = 100. "Brine with 1/5 kg of salt per liter flows in at 4 L/min" -> Rate in = 4 * 1/5 = 0.8 kg/min. "drains at the same rate" -> Volume constant 500 L. Rate out = 4 * (S/500) = S/125 kg/min. S' = 0.8 - S/125. If S=100, S' = 0.8 - 100/125 = 0.8 - 0.8 = 0. So S(t) = 100 is the correct solution. The solution steps are: 1. Sets up DE correctly. 2. Checks that S(t)=100 satisfies the DE. (Derivative of 100 is 0. RHS is 0.8 - 0.8 = 0. Correct.) 3. Checks initial condition. Correct. 4. Calculates S(60). Correct. 5. Calculates limit. Correct. The solution is sound. The verdict should be pass.
  • 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.

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