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 find the net volume, we need to calculate the volume filled and the volume drained. The tank fills at 3 L/min for 4 minutes. To convert this to liters, we multiply the rate by the time: 3 L/min × 4 min = 12 L The tank drains at 500 mL/s for 10 seconds. To convert this to liters, we multiply the rate by the time and then divide by 1000 (since 1 L = 1000 mL): 500 mL/s × 10 s = 5000 mL 5000 mL ÷ 1000 = 5 L Now, we subtract the volume drained from the volume filled to find the net volume: 12 L - 5 L = 7 L Answer: 7 L.
detector fmi_4_4_v1: correct unit L
To solve this problem, we need to calculate the net volume of water in the tank. The tank fills at 3 L/min for 4 minutes. To find the volume of water filled, we multiply the rate of filling (3 L/min) by the time (4 minutes): 3 L/min × 4 min = 12 L The tank then drains at 500 mL/s for 10 seconds. To find the volume of water drained, we need to convert the rate of draining from mL/s to L/s and then multiply by the time: 500 mL/s × (10 s / 60 s) = 500 mL/s × 1/6 L/s = 83.33 mL/s = 0.08333 L/s Now, we multiply the rate of draining (0.08333 L/s) by the time (10 s): 0.08333 L/s × 10 s = 0.8333 L The net volume of water in the tank is the difference between the volume filled and the volume drained: 12 L - 0.8333 L = 11.1667 L Rounding to a reasonable number of decimal places, we get: 11.17 L Answer: 11.17 L.
detector fmi_4_4_v1: correct unit L
loop 2026-08-15 · meta-llama/llama-3.1-8b-instruct as served · open weights · released 2024-07 · 8B · 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.
To find the net volume, we need to calculate the volume filled and the volume drained. The tank fills at 3 L/min for 4 minutes. To convert this to liters, we multiply the rate by the time: 3 L/min × 4 min = 12 L The tank drains at 500 mL/s for 10 seconds. To convert this to liters, we multiply the rate by the time and then divide by 1000 (since 1 L = 1000 mL): 500 mL/s × 10 s = 5000 mL 5000 mL ÷ 1000 = 5 L Now, we subtract the volume drained from the volume filled to find the net volume: 12 L - 5 L = 7 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