Real context moved 0.0pp on Dimensional-Analysis Failure.
~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 determine the net amount of liquid in the tank after the filling and draining processes, we'll calculate each process separately and then find the difference. 1. **Filling the tank:** - **Rate:** 3 L/min - **Time:** 4 minutes - **Volume filled:** \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining the tank:** - **Rate:** 500 mL/s - **Time:** 10 seconds - **Volume drained:** \( 500 \, \text{mL/s} \times 10 \, \text{s} = 5000 \, \text{mL} \) - Convert mL to L: \( 5000 \, \text{mL} = 5 \, \text{L} \) 3. **Net volume:** - **Net volume:** \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) 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. We'll do this by calculating the amount of liquid added during the filling phase and subtracting the amount of liquid removed during the draining phase. 1. **Filling Phase:** - Rate: 3 L/min - Time: 4 minutes - Amount filled: \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining Phase:** - Rate: 500 mL/s - Time: 10 seconds - First, convert the rate to litres per second: \( 500 \, \text{mL/s} = 0.5 \, \text{L/s} \) - Amount drained: \( 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \) 3. **Net Amount:** - Net litres: \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · mistralai/mistral-small-3.2-24b-instruct as served · open weights · released 2025-06 · 24B · set s1.3-live · cases v1.6 · 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 determine the net amount of liquid in the tank after the filling and draining processes, we'll calculate each process separately and then find the difference. 1. **Filling the tank:** - **Rate:** 3 L/min - **Time:** 4 minutes - **Volume filled:** \( 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \) 2. **Draining the tank:** - **Rate:** 500 mL/s - **Time:** 10 seconds - **Volume drained:** \( 500 \, \text{mL/s} \times 10 \, \text{s} = 5000 \, \text{mL} \) - Convert mL to L: \( 5000 \, \text{mL} = 5 \, \text{L} \) 3. **Net volume:** - **Net volume:** \( 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \) 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