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 2
For work package 559 on the freight corridor repair, the project controls engineer performs a contractor change evaluation and classifies fine-grained soil with LL = 59 and PI = 35. Using A-line PI = 0.73(LL - 20), what USCS group is indicated?
- ACH
- BML
- CMH
- DCL
Show the answer and worked solution
Answer: A — CH
Worked solution
- Step 1. Compute A-line
PI_A = 0.73(LL - 20)
PI_A = 0.73(59 - 20) = 28.47
- Step 2. Compare plasticity
PI = 35 is above the A-line, so the soil is clayey.
- Step 3. Use liquid-limit boundary
LL = 59 gives CH.
Why the other choices are wrong
- Choice B
- No single arithmetic or comparison slip lands cleanly on this label; working the method directly, PI = 35 plots above 0.73(59 - 20) = 28.47 and the liquid limit of 59 is at or above 50, so the material classifies as a high-plasticity clay rather than a low-plasticity silt.
- Choice C
- Reversing the A-line comparison and assuming PI = 35 falls below 0.73(59 - 20) = 28.47 shifts the group symbol from clay to silt, even though the plotted point actually sits above the A-line while the liquid limit of 59 correctly signals high plasticity.
- Choice D
- Treating the liquid limit of 59 as though it sat below the 50 cutoff swaps the plasticity label from high to low, even though 59 is at or above 50 and PI = 35 still plots above 0.73(59 - 20) = 28.47, which keeps the soil clayey.