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, calculate the total volume of water filled into the tank: - The tank fills at a rate of 3 liters per minute. - It fills for 4 minutes. \[ \text{Total volume filled} = 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, calculate the total volume of water drained from the tank: - The tank drains at a rate of 500 milliliters per second. - It drains for 10 seconds. Convert milliliters to liters: \[ 500 \, \text{mL} = 0.5 \, \text{L} \] \[ \text{Total volume drained} = 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \] Now, calculate the net volume of water in the tank: \[ \text{Net volume} = \text{Total volume filled} - \text{Total volume drained} = 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] Answer: 7 L
detector fmi_4_4_v1: correct unit L
To solve the problem, we need to calculate the net amount of water in the tank after the filling and draining processes. 1. **Filling the tank:** - The tank fills at a rate of 3 liters per minute. - It fills for 4 minutes. - Total volume filled = 3 L/min * 4 min = 12 liters. 2. **Draining the tank:** - The tank drains at a rate of 500 milliliters per second. - It drains for 10 seconds. - Total volume drained = 500 mL/s * 10 s = 5000 mL. - Convert milliliters to liters: 5000 mL = 5 liters. 3. **Net volume in the tank:** - Net volume = Volume filled - Volume drained - Net volume = 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 s1.1-live · cases v1.3 · 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, calculate the total volume of water filled into the tank: - The tank fills at a rate of 3 liters per minute. - It fills for 4 minutes. \[ \text{Total volume filled} = 3 \, \text{L/min} \times 4 \, \text{min} = 12 \, \text{L} \] Next, calculate the total volume of water drained from the tank: - The tank drains at a rate of 500 milliliters per second. - It drains for 10 seconds. Convert milliliters to liters: \[ 500 \, \text{mL} = 0.5 \, \text{L} \] \[ \text{Total volume drained} = 0.5 \, \text{L/s} \times 10 \, \text{s} = 5 \, \text{L} \] Now, calculate the net volume of water in the tank: \[ \text{Net volume} = \text{Total volume filled} - \text{Total volume drained} = 12 \, \text{L} - 5 \, \text{L} = 7 \, \text{L} \] Answer: 7 L
fmi_4_4_v1
detector fmi_4_4_v1: correct unit L
0–40% · 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