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ASTM SOILS
TECHNICIAN PRACTICE EXAM
2026/2027 EDITION
150 Original Questions | Integrated Answers & Detailed Rationales
Classification • Sampling • Compaction • Strength • Field Testing
SOIL MECHANICS • LABORATORY & FIELD METHODS
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,ASTM SOILS TECHNICIAN | PRACTICE EXAM
Section 1: Brief Introduction
This 150-question ASTM soils technician practice exam reviews soil classification, sampling,
laboratory testing, field methods, compaction control, and geotechnical interpretation. It
includes calculations and scenarios covering moisture, specific gravity, gradation, Atterberg
limits, permeability, consolidation, shear strength, CBR, equipment safety, quality control, and
reporting.
Section 2: The Complete Exam
Select the single best response. The keyed answer and a concise rationale follow each item. Questions mix standards
selection, procedure, calculation, data interpretation, safety, and applied soil mechanics.
1. A soil specimen has 62% by dry mass passing the No. 200 sieve. Under the USCS, which broad
group is indicated before plasticity-based symbol assignment?
A. Clean gravel, because the No. 4 sieve controls the fine/coarse split.
B. Fine-grained soil, because at least half of the material passes the No. 200 sieve.
C. Organic soil, because fines content alone identifies organic matter.
D. Coarse-grained soil, because most particles are visible without magnification.
Answer: B. Fine-grained soil, because at least half of the material passes the No. 200 sieve.
Rationale: ASTM D2487 divides soils by whether 50% or more passes the 75-µm No. 200 sieve:
those at or above that threshold are fine-grained. Liquid limit, plasticity index, and organic
behavior then determine the group symbol.
2. A coarse-grained soil has most of its coarse fraction passing the No. 4 sieve but retained
above the No. 200 sieve. Which primary fraction is indicated?
A. Clay, because particle shape is not measured by sieving.
B. Sand, because more than half of the coarse fraction passes the No. 4 sieve.
C. Gravel, because every particle retained on No. 200 is gravel.
D. Silt, because No. 4 is a fine-grained sieve.
Answer: B. Sand, because more than half of the coarse fraction passes the No. 4 sieve.
Rationale: In USCS, gravel versus sand is assigned by whether the coarse fraction is
predominantly retained on or passes the 4.75-mm No. 4 sieve. The No. 200 sieve distinguishes
fines from the coarse fraction (D2487, D6913/D6913M).
3. A gravelly soil contains 8% fines and meets the USCS dual-symbol fines range. Which
classification approach is appropriate?
A. Classify it as CH because all soils with 8% fines are high-plasticity clay.
B. Use only an AASHTO group index in place of a USCS symbol.
C. Ignore the fines and report only GW.
D. Use a dual symbol that reflects both the gravel gradation and the fines type, such as GW-GM
or GW-GC when supported by test results.
Answer: D. Use a dual symbol that reflects both the gravel gradation and the fines type,
such as GW-GM or GW-GC when supported by test results.
Rationale: For a coarse-grained soil with about 5–12% fines, D2487 uses a dual symbol to
communicate gradation and fines characteristics. The second symbol depends on plasticity
testing, not visual appearance alone.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 2
,ASTM SOILS TECHNICIAN | PRACTICE EXAM
4. A clean sand has D10 = 0.20 mm, D30 = 0.60 mm, and D60 = 1.60 mm. What do its gradation
coefficients indicate?
A. Cu = 4 and Cc = 0.75, automatically classifying the sample as SW.
B. Cu = 0.125 and Cc = 8, indicating poorly graded gravel.
C. Cu = 8 and Cc = 1.125, satisfying the usual USCS well-graded sand coefficient criteria if other
requirements are met.
D. Cu = 8 and Cc = 4.8, which fails the curvature criterion.
Answer: C. Cu = 8 and Cc = 1.125, satisfying the usual USCS well-graded sand coefficient
criteria if other requirements are met.
Rationale: Cu = D60/D10 = 1.60/0.20 = 8; Cc = D30²/(D10D60) = 0.36/0.32 = 1.125. For clean
sand, the usual SW coefficient checks are Cu ≥ 6 and 1 ≤ Cc ≤ 3, alongside the fines and
gradation requirements in D2487.
5. A soil plots above the A-line on the USCS plasticity chart and has liquid limit below 50. Which
fine-grained group family is generally indicated, subject to organic checks?
A. Silt of high plasticity, symbol MH.
B. Clay of high plasticity, symbol CH.
C. Well-graded sand, symbol SW.
D. Clay of low plasticity, symbol CL.
Answer: D. Clay of low plasticity, symbol CL.
Rationale: For inorganic fine-grained soil, a point above the A-line indicates clay-like
plasticity, while LL below 50 indicates low plasticity (CL). Classification still requires the
prescribed laboratory data and consideration of organic behavior (D2487, D4318).
6. For a liquid limit of 42, the approximate A-line plasticity index is 0.73(LL − 20). Which value
is closest?
A. 30.7.
B. 42.0.
C. 45.3.
D. 16.1.
Answer: D. 16.1.
Rationale: PI_A = 0.73(42 − 20) = 0.73 × 22 = 16.06, approximately 16.1. A measured PI above
or below this line helps distinguish clay-like from silt-like plasticity in USCS, but does not
replace the complete D2487 classification.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 3
, ASTM SOILS TECHNICIAN | PRACTICE EXAM
7. A soil has LL = 48 and PL = 21. What is its plasticity index?
A. 21.
B. 2.3.
C. 27.
D. 69.
Answer: C. 27.
Rationale: Plasticity index is PI = LL − PL = 48 − 21 = 27 percentage points (D4318). PI
describes the water-content range over which the fine-grained soil behaves plastically; it is not
a direct measure of strength by itself.
8. In the multipoint Casagrande liquid-limit procedure, what water content is reported as the
liquid limit?
A. The water content at which the sample first passes the No. 200 sieve.
B. The water content corresponding to 25 blows on the flow curve.
C. The water content at which a 3.2-mm thread first forms.
D. The average water content from any two arbitrary blow counts.
Answer: B. The water content corresponding to 25 blows on the flow curve.
Rationale: ASTM D4318 obtains the liquid limit from the flow relationship at 25 blows for the
specified groove-closure procedure. The plastic-limit thread procedure is separate and is not
used to define LL.
9. During the plastic-limit test, a soil thread crumbles as it is rolled to about 3.2 mm diameter.
What does the technician determine?
A. The water content of the crumbled soil as the plastic limit, using the required replicate
procedure.
B. The specific gravity of soil solids.
C. The liquid limit from the number of rolls.
D. The shrinkage limit from the diameter alone.
Answer: A. The water content of the crumbled soil as the plastic limit, using the required
replicate procedure.
Rationale: D4318 defines the plastic limit from the water content at which the soil thread
crumbles at the specified small diameter, following the method's preparation and replicate
steps. PI is subsequently calculated as LL minus PL.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 4
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ASTM SOILS
TECHNICIAN PRACTICE EXAM
2026/2027 EDITION
150 Original Questions | Integrated Answers & Detailed Rationales
Classification • Sampling • Compaction • Strength • Field Testing
SOIL MECHANICS • LABORATORY & FIELD METHODS
• • • • • • • •
◆
• • • • • • • •
◆
• • • • • • • •
• • • • • • • • • 01 •
• • • • • • • •
◆
• • • • • • • •
• • • • • • • •
,ASTM SOILS TECHNICIAN | PRACTICE EXAM
Section 1: Brief Introduction
This 150-question ASTM soils technician practice exam reviews soil classification, sampling,
laboratory testing, field methods, compaction control, and geotechnical interpretation. It
includes calculations and scenarios covering moisture, specific gravity, gradation, Atterberg
limits, permeability, consolidation, shear strength, CBR, equipment safety, quality control, and
reporting.
Section 2: The Complete Exam
Select the single best response. The keyed answer and a concise rationale follow each item. Questions mix standards
selection, procedure, calculation, data interpretation, safety, and applied soil mechanics.
1. A soil specimen has 62% by dry mass passing the No. 200 sieve. Under the USCS, which broad
group is indicated before plasticity-based symbol assignment?
A. Clean gravel, because the No. 4 sieve controls the fine/coarse split.
B. Fine-grained soil, because at least half of the material passes the No. 200 sieve.
C. Organic soil, because fines content alone identifies organic matter.
D. Coarse-grained soil, because most particles are visible without magnification.
Answer: B. Fine-grained soil, because at least half of the material passes the No. 200 sieve.
Rationale: ASTM D2487 divides soils by whether 50% or more passes the 75-µm No. 200 sieve:
those at or above that threshold are fine-grained. Liquid limit, plasticity index, and organic
behavior then determine the group symbol.
2. A coarse-grained soil has most of its coarse fraction passing the No. 4 sieve but retained
above the No. 200 sieve. Which primary fraction is indicated?
A. Clay, because particle shape is not measured by sieving.
B. Sand, because more than half of the coarse fraction passes the No. 4 sieve.
C. Gravel, because every particle retained on No. 200 is gravel.
D. Silt, because No. 4 is a fine-grained sieve.
Answer: B. Sand, because more than half of the coarse fraction passes the No. 4 sieve.
Rationale: In USCS, gravel versus sand is assigned by whether the coarse fraction is
predominantly retained on or passes the 4.75-mm No. 4 sieve. The No. 200 sieve distinguishes
fines from the coarse fraction (D2487, D6913/D6913M).
3. A gravelly soil contains 8% fines and meets the USCS dual-symbol fines range. Which
classification approach is appropriate?
A. Classify it as CH because all soils with 8% fines are high-plasticity clay.
B. Use only an AASHTO group index in place of a USCS symbol.
C. Ignore the fines and report only GW.
D. Use a dual symbol that reflects both the gravel gradation and the fines type, such as GW-GM
or GW-GC when supported by test results.
Answer: D. Use a dual symbol that reflects both the gravel gradation and the fines type,
such as GW-GM or GW-GC when supported by test results.
Rationale: For a coarse-grained soil with about 5–12% fines, D2487 uses a dual symbol to
communicate gradation and fines characteristics. The second symbol depends on plasticity
testing, not visual appearance alone.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 2
,ASTM SOILS TECHNICIAN | PRACTICE EXAM
4. A clean sand has D10 = 0.20 mm, D30 = 0.60 mm, and D60 = 1.60 mm. What do its gradation
coefficients indicate?
A. Cu = 4 and Cc = 0.75, automatically classifying the sample as SW.
B. Cu = 0.125 and Cc = 8, indicating poorly graded gravel.
C. Cu = 8 and Cc = 1.125, satisfying the usual USCS well-graded sand coefficient criteria if other
requirements are met.
D. Cu = 8 and Cc = 4.8, which fails the curvature criterion.
Answer: C. Cu = 8 and Cc = 1.125, satisfying the usual USCS well-graded sand coefficient
criteria if other requirements are met.
Rationale: Cu = D60/D10 = 1.60/0.20 = 8; Cc = D30²/(D10D60) = 0.36/0.32 = 1.125. For clean
sand, the usual SW coefficient checks are Cu ≥ 6 and 1 ≤ Cc ≤ 3, alongside the fines and
gradation requirements in D2487.
5. A soil plots above the A-line on the USCS plasticity chart and has liquid limit below 50. Which
fine-grained group family is generally indicated, subject to organic checks?
A. Silt of high plasticity, symbol MH.
B. Clay of high plasticity, symbol CH.
C. Well-graded sand, symbol SW.
D. Clay of low plasticity, symbol CL.
Answer: D. Clay of low plasticity, symbol CL.
Rationale: For inorganic fine-grained soil, a point above the A-line indicates clay-like
plasticity, while LL below 50 indicates low plasticity (CL). Classification still requires the
prescribed laboratory data and consideration of organic behavior (D2487, D4318).
6. For a liquid limit of 42, the approximate A-line plasticity index is 0.73(LL − 20). Which value
is closest?
A. 30.7.
B. 42.0.
C. 45.3.
D. 16.1.
Answer: D. 16.1.
Rationale: PI_A = 0.73(42 − 20) = 0.73 × 22 = 16.06, approximately 16.1. A measured PI above
or below this line helps distinguish clay-like from silt-like plasticity in USCS, but does not
replace the complete D2487 classification.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 3
, ASTM SOILS TECHNICIAN | PRACTICE EXAM
7. A soil has LL = 48 and PL = 21. What is its plasticity index?
A. 21.
B. 2.3.
C. 27.
D. 69.
Answer: C. 27.
Rationale: Plasticity index is PI = LL − PL = 48 − 21 = 27 percentage points (D4318). PI
describes the water-content range over which the fine-grained soil behaves plastically; it is not
a direct measure of strength by itself.
8. In the multipoint Casagrande liquid-limit procedure, what water content is reported as the
liquid limit?
A. The water content at which the sample first passes the No. 200 sieve.
B. The water content corresponding to 25 blows on the flow curve.
C. The water content at which a 3.2-mm thread first forms.
D. The average water content from any two arbitrary blow counts.
Answer: B. The water content corresponding to 25 blows on the flow curve.
Rationale: ASTM D4318 obtains the liquid limit from the flow relationship at 25 blows for the
specified groove-closure procedure. The plastic-limit thread procedure is separate and is not
used to define LL.
9. During the plastic-limit test, a soil thread crumbles as it is rolled to about 3.2 mm diameter.
What does the technician determine?
A. The water content of the crumbled soil as the plastic limit, using the required replicate
procedure.
B. The specific gravity of soil solids.
C. The liquid limit from the number of rolls.
D. The shrinkage limit from the diameter alone.
Answer: A. The water content of the crumbled soil as the plastic limit, using the required
replicate procedure.
Rationale: D4318 defines the plastic limit from the water content at which the soil thread
crumbles at the specified small diameter, following the method's preparation and replicate
steps. PI is subsequently calculated as LL minus PL.
ASTM SOILS TECHNICIAN • 2026/2027 | Page 4
