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?
First, let's calculate the amount of water filled in the tank: The tank fills at a rate of 3 liters per minute for 4 minutes. \[ 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, let's calculate the amount of water drained from the tank: The tank drains at a rate of 500 milliliters per second for 10 seconds. \[ 500 \, \text{mL/s} \times 10 \, \text{s} = 5000 \, \text{mL} \] Convert milliliters to liters: \[ 5000 \, \text{mL} = 5 \, \text{L} \] Now, calculate the net amount of water in the tank: \[ 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] Answer: 7 L
detector fmi_4_4_v1: correct unit L
To solve the problem, we need to calculate the total amount of water added to the tank and subtract the amount of water drained from it. 1. Calculate the amount of water filled: - The tank fills at a rate of 3 liters per minute (L/min). - It fills for 4 minutes. - Total water filled = 3 L/min * 4 min = 12 liters. 2. Calculate the amount of water drained: - The tank drains at a rate of 500 milliliters per second (mL/s). - It drains for 10 seconds. - Total water drained = 500 mL/s * 10 s = 5000 mL. - Convert milliliters to liters: 5000 mL = 5 liters. 3. Calculate the net amount of water in the tank: - Net water = Total water filled - Total water drained - Net water = 12 liters - 5 liters = 7 liters. Answer: 7 liters
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · qwen/qwen-2.5-72b-instruct as served · open weights · released 2024-09 · 72B · 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.
First, let's calculate the amount of water filled in the tank: The tank fills at a rate of 3 liters per minute for 4 minutes. \[ 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, let's calculate the amount of water drained from the tank: The tank drains at a rate of 500 milliliters per second for 10 seconds. \[ 500 \, \text{mL/s} \times 10 \, \text{s} = 5000 \, \text{mL} \] Convert milliliters to liters: \[ 5000 \, \text{mL} = 5 \, \text{L} \] Now, calculate the net amount of water in the tank: \[ 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] 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