Real context moved 0.0pp on Dimensional-Analysis Failure.
Gets the number right and the unit wrong.
~intervals overlap — not separated from no change
The model produces a quantity with inconsistent or wrong units, or mixes unit systems, yielding a dimensionally invalid result. The rule
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
To solve the problem, we need to calculate the net amount of liquid in the tank after both filling and draining processes. 1. **Filling the tank:** - Filling rate: 3 L/min - Time: 4 minutes - Amount filled: \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining the tank:** - Draining rate: 500 mL/s - Time: 10 seconds - Convert mL to L: \( 500 \, \text{mL} = 0.5 \, \text{L} \) - Amount drained: \( 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \) 3. **Net amount of liquid:** - Net litres: \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) Answer: 7 L
detector fmi_4_4_v1: correct unit L
To solve the problem, we need to calculate the net amount of liquid in the tank after it has been filled and then drained. 1. **Filling the tank:** - Rate: 3 L/min - Time: 4 minutes - Amount filled: \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining the tank:** - Rate: 500 mL/s - Time: 10 seconds - Amount drained: \( 500 \, \text{mL/s} \times 10 \, \text{s} = 5000 \, \text{mL} \) - Convert mL to L: \( 5000 \, \text{mL} = 5 \, \text{L} \) 3. **Net amount of liquid in the tank:** - Net litres: \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-09-10 · mistralai/mistral-small-3.2-24b-instruct as served · open weights · released 2025-06 · 24B · set s1.4-live · cases v1.7 · conditions v1.0 · ground truth as of 2026-08-09
A tank fills at 3 L/min for 4 minutes, then drains at 500 mL/s for 10 seconds. Net litres?
expected as of 2026-08-09
12 L in, 5 L out, net 7 L. The mixed units are the trap.
Nothing supplied. The baseline.
To solve the problem, we need to calculate the net amount of liquid in the tank after both filling and draining processes. 1. **Filling the tank:** - Filling rate: 3 L/min - Time: 4 minutes - Amount filled: \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining the tank:** - Draining rate: 500 mL/s - Time: 10 seconds - Convert mL to L: \( 500 \, \text{mL} = 0.5 \, \text{L} \) - Amount drained: \( 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \) 3. **Net amount of liquid:** - Net litres: \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) Answer: 7 L
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–20% · whisker = 95% interval
never pooled · a trap ratio would skew the delta
Unchanged by context: 0% with none, 0% with the strongest supplied source. For a control mode that flat line is the result — it is what licenses reading movement elsewhere in this loop as grounding rather than as a document being present. Nothing survives at the best condition on the cases measured here.
- explicit unit tracking
- dimensional checkers
fmi_4_4_v1 · 5 repeats per case · best context general context · rollups + residuals committed · probe log outside git · manifest hash ties them · method