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 5
For the Clear Lake utility relocation corridor, a design checker preparing a design validation memo classifies fine-grained soil with liquid limit LL = 39 and plasticity index PI = 19. Using A-line PI = 0.73(LL - 20), what USCS group is indicated?
- AMH
- BSM
- CCL
- DML
Show the answer and worked solution
Answer: C — CL
Worked solution
- Step 1. Compute A-line value
PI_A = 0.73(LL - 20)
PI_A = 0.73(39 - 20) = 13.87
- Step 2. Compare to A-line
PI = 19 is above the A-line, so the soil is clayey.
- Step 3. Apply liquid-limit boundary
LL = 39 gives CL.
Why the other choices are wrong
- Choice A
- No single arithmetic or comparison slip lands cleanly on this label; working the method directly, PI = 19 plots above 0.73(39 - 20) = 13.87 and the liquid limit of 39 is under 50, so the material classifies as a low-plasticity clay rather than a high-plasticity silt.
- Choice B
- 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 = 39 and PI = 19 plotting above the A-line as a clay.
- Choice D
- Reversing the A-line comparison and assuming PI = 19 falls below 0.73(39 - 20) = 13.87 shifts the group symbol from clay to silt, even though the plotted point actually sits above the A-line while the liquid limit of 39 correctly signals low plasticity.