Real context moved -2.0pp 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 this problem, we need to calculate the total volume of water added to the tank and the total volume of water drained from the tank, and then find the net volume. 1. **Filling the Tank:** - The tank fills at a rate of 3 liters per minute for 4 minutes. - Volume added = Rate × Time = \(3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L}\). 2. **Draining the Tank:** - The tank drains at a rate of 500 mL/s for 10 seconds. - First, convert the draining rate to liters per second: \(500 \, \text{mL/s} = 0.5 \, \text{L/s}\). - Volume drained = Rate × Time = \(0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L}\). 3. **Net Volume:** - Net volume = Volume added - Volume drained = \(12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L}\). Answer: \(7 \, \text{L}\)
detector fmi_4_4_v1: correct unit L
To solve this problem, we need to calculate the total volume of liquid added to the tank and then subtract the volume that drains out. 1. **Filling the Tank:** - The tank fills at a rate of 3 liters per minute for 4 minutes. - Total volume added = \(3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L}\). 2. **Draining the Tank:** - The tank drains at a rate of 500 mL/s for 10 seconds. - First, convert the draining rate to liters per second: \(500 \, \text{mL/s} = 0.5 \, \text{L/s}\). - Total volume drained = \(0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L}\). 3. **Net Volume:** - Net volume in the tank = Volume added - Volume drained - Net volume = \(12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L}\). Answer: 7 L
detector fmi_4_4_v1: correct unit L
loop 2026-09-09 · microsoft/phi-4 as served · open weights · released 2024-12 · 14B · 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 this problem, we need to calculate the total volume of water added to the tank and the total volume of water drained from the tank, and then find the net volume. 1. **Filling the Tank:** - The tank fills at a rate of 3 liters per minute for 4 minutes. - Volume added = Rate × Time = \(3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L}\). 2. **Draining the Tank:** - The tank drains at a rate of 500 mL/s for 10 seconds. - First, convert the draining rate to liters per second: \(500 \, \text{mL/s} = 0.5 \, \text{L/s}\). - Volume drained = Rate × Time = \(0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L}\). 3. **Net Volume:** - Net volume = Volume added - Volume drained = \(12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L}\). Answer: \(7 \, \text{L}\)
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–40% · 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