Original question written by FE Exam AI Prep from the public FE Civil exam specification. Before release, a solver model re-derived the answer twice from the question alone, without its answer key, and a critic model checked the result. No licensed engineer reviews these questions. It is not an NCEES question and does not come from any NCEES practice exam. FE Exam AI Prep is an independent study tool and is not affiliated with, endorsed by, or sponsored by NCEES.
Atterberg Limits and the A-Line: FE Civil practice problem 3
For work package 554 on the culvert scour repair, the assistant resident engineer performs an independent check calculation and classifies fine-grained soil with LL = 44 and PI = 23. Using A-line PI = 0.73(LL - 20), what USCS group is indicated?
- AML
- BMH
- CSM
- DCL
Show the answer and worked solution
Answer: D — CL
Worked solution
- Step 1. Compute A-line
PI_A = 0.73(LL - 20)
PI_A = 0.73(44 - 20) = 17.52
- Step 2. Compare plasticity
PI = 23 is above the A-line, so the soil is clayey.
- Step 3. Use liquid-limit boundary
LL = 44 gives CL.
Why the other choices are wrong
- Choice A
- Reversing the A-line comparison and assuming PI = 23 falls below 0.73(44 - 20) = 17.52 shifts the group symbol from clay to silt, even though the plotted point actually sits above the A-line while the liquid limit of 44 correctly signals low plasticity.
- Choice B
- No single arithmetic or comparison slip lands cleanly on this label; working the method directly, PI = 23 plots above 0.73(44 - 20) = 17.52 and the liquid limit of 44 is under 50, so the material classifies as a low-plasticity clay rather than a high-plasticity silt.
- Choice C
- This label borrows the coarse-grained dual-symbol procedure, which is built from sieve gradation and fines content rather than the plasticity chart, so it ignores that the material was already identified as fine-grained with LL = 44 and PI = 23 plotting above the A-line as a clay.