On a Hjulström diagram, a stream bed contains clay (0.002 mm), silt (0.02 mm), fine sand (0.2 mm), and pebbles (20 mm). As flow velocity gradually increases from near zero, which size fraction is entrained FIRST?

Fine sand, because its erosion velocity is lowest

Silt, because it settles the most slowly

Clay, because its particles are the smallest

Pebbles, because they project farthest into flow

In a region of active crustal shortening, borehole breakouts and focal mechanisms indicate σ₁ is horizontal and trends E–W, σ₂ is horizontal and trends N–S, and σ₃ is vertical. According to Anderson's theory, the expected faults are:

thrust faults striking N–S and dipping about 30°

normal faults striking N–S and dipping about 60°

thrust faults striking E–W and dipping about 60°

strike-slip faults striking about 30° from E–W

A Shelby-tube sample of a marine clay has a natural water content of 38%, a liquid limit of 45, and a plastic limit of 22. What is the liquidity index, and what does it indicate about the in-situ consistency?

0.70; the clay lies within its plastic range

1.43; the clay is wetter than its liquid limit

0.30; the clay is stiff, near its plastic limit

0.36; the clay is near its shrinkage limit

A geologist is retained to characterize subsurface conditions for a proposed hillside residential subdivision in an area she has not worked in before. What should she do FIRST?

Review published maps, prior reports, borings and historical air photos

Run a geophysical survey to map depth to bedrock across the whole property

Prepare a cost estimate for slope stabilization based on regional experience

Mobilize a drill rig to obtain continuous core at the planned building pads

Downgradient of a petroleum hydrocarbon release, dissolved arsenic in groundwater has risen from below detection to 40 µg/L, although the fuel contains no arsenic. Dissolved oxygen is near zero and dissolved iron is elevated. What is the most likely cause of the arsenic increase?

Reductive dissolution of iron oxyhydroxides that had adsorbed arsenic

Precipitation of arsenic sulfides as sulfate in the plume is reduced

Desorption of arsenic caused by acid generated as the fuel degrades

Oxidation of arsenopyrite caused by oxygen supplied by the plume

Air sparging is proposed for dissolved benzene in a sand aquifer. Boring logs show a laterally continuous silt layer 1 m above the water table and a basement 20 m away. What is the PRIMARY design concern?

Sparged vapors may spread under the silt toward the basement

The silt layer will stop air from entering the saturated zone

Benzene is too soluble to partition into the injected air stream

Injected air will raise dissolved iron and clog the sand aquifer

Staged construction of an embankment over soft, low-permeability clay requires close tracking of construction-induced excess pore pressures so the next lift can be timed. Which instrumentation is MOST appropriate?

Vibrating-wire piezometers installed by the fully grouted method

Monitoring wells screened across the clay and underlying sand

Open-standpipe piezometers read weekly by water-level tape

In-place inclinometers to infer pore pressure from displacement

ASBOG Geology Exam

Free ASBOG Practice Questions

10 sample questions for the ASBOG Geology Exam, with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Domain A · General and Field Geology, Domain B · Mineralogy, Petrology, Geochemistry, Domain D · Geomorphology and Surficial Processes, Domain C · Sedimentology, Stratigraphy, Paleontology, Domain E · Structure, Tectonics, Seismology, Domain F · Hydrogeology, Domain G · Engineering Geology, Domain H · Economic Geology and Energy.

Question 1 · Domain A · General and Field Geology
On a 1:62,500 topographic quadrangle, a geologist measures 4.8 in between two outcrops along a straight traverse. What is the ground distance between the outcrops?
- A.About 4.73 mi
- B.About 56.8 mi
- C.About 0.39 mi
- D.About 4.80 mi
Show answer and explanation
Answer: A. About 4.73 mi
Ground distance = 4.8 in × 62,500 = 300,000 in; 300,000 in ÷ 63,360 in/mi = 4.73 mi . Choosing about 4.80 mi treats 1:62,500 as exactly one inch to the mile, a common shortcut that is about 1% off. About 56.8 mi comes from dividing inches by 5,280 (feet per mile) without first converting to feet, and about 0.39 mi divides by 12 twice.
Question 2 · Domain A · General and Field Geology
A suite of porphyritic andesites and dacites contains hornblende phenocrysts, follows a calc-alkaline trend, and shows enrichment in Ba, Rb and Sr with a pronounced negative Nb–Ta anomaly on a spider diagram. In what tectonic setting did the magmas most likely form?
- A.Above a subduction zone
- B.In a young continental rift
- C.Over an intraplate hotspot
- D.At a mid-ocean ridge
Show answer and explanation
Answer: A. Above a subduction zone
Calc-alkaline, hydrous (hornblende-bearing) intermediate magmas enriched in fluid-mobile large-ion lithophile elements (Ba, Rb, Sr) and depleted in Nb–Ta are the geochemical signature of melting above a subduction zone , where slab fluids flux the mantle wedge. Mid-ocean ridge basalts are tholeiitic and depleted in incompatible elements. Hotspot and rift magmas are typically alkaline or tholeiitic and lack the Nb–Ta trough.
Question 3 · Domain D · Geomorphology and Surficial Processes
On a Hjulström diagram, a stream bed contains clay (0.002 mm), silt (0.02 mm), fine sand (0.2 mm), and pebbles (20 mm). As flow velocity gradually increases from near zero, which size fraction is entrained FIRST?
- A.Fine sand, because its erosion velocity is lowest
- B.Silt, because it settles the most slowly
- C.Clay, because its particles are the smallest
- D.Pebbles, because they project farthest into flow
Show answer and explanation
Answer: A. Fine sand, because its erosion velocity is lowest
The erosion-velocity curve reaches a minimum near 0.1–0.5 mm, so fine sand is entrained at the lowest velocity. Clay and silt require higher velocities because cohesion and smooth bed surfaces (particles within the viscous sublayer) resist entrainment, a classic misconception being that the smallest particles move first. Slow settling explains why silt and clay stay in suspension once entrained, not how easily they are eroded.
Question 4 · Domain E · Structure, Tectonics, Seismology
In a region of active crustal shortening, borehole breakouts and focal mechanisms indicate σ₁ is horizontal and trends E–W, σ₂ is horizontal and trends N–S, and σ₃ is vertical. According to Anderson's theory, the expected faults are:
- A.thrust faults striking N–S and dipping about 30°
- B.normal faults striking N–S and dipping about 60°
- C.thrust faults striking E–W and dipping about 60°
- D.strike-slip faults striking about 30° from E–W
Show answer and explanation
Answer: A. thrust faults striking N–S and dipping about 30°
With σ₃ vertical, Anderson's theory predicts reverse/thrust faults ; they contain σ₂ (so strike N–S) and form at about 30° to σ₁, giving dips near 30°. Normal faults require σ₁ vertical, strike-slip faults require σ₂ vertical, and a thrust striking E–W would be parallel to σ₁ rather than perpendicular to it.
Question 5 · Domain G · Engineering Geology
A Shelby-tube sample of a marine clay has a natural water content of 38%, a liquid limit of 45, and a plastic limit of 22. What is the liquidity index, and what does it indicate about the in-situ consistency?
- A.0.70; the clay lies within its plastic range
- B.1.43; the clay is wetter than its liquid limit
- C.0.30; the clay is stiff, near its plastic limit
- D.0.36; the clay is near its shrinkage limit
Show answer and explanation
Answer: A. 0.70; the clay lies within its plastic range
LI = (w − PL)/(LL − PL) = (38 − 22)/(45 − 22) = 16/23 = 0.70, which falls between 0 and 1, so the clay is in the plastic range and relatively soft. "1.43; the clay is wetter than its liquid limit" inverts the ratio, and "0.30; the clay is stiff, near its plastic limit" is actually the consistency index, (LL − w)/PI. "0.36" results from dividing by LL instead of PI.
Question 6 · Domain A · General and Field Geology
A geologist is retained to characterize subsurface conditions for a proposed hillside residential subdivision in an area she has not worked in before. What should she do FIRST?
- A.Review published maps, prior reports, borings and historical air photos
- B.Run a geophysical survey to map depth to bedrock across the whole property
- C.Prepare a cost estimate for slope stabilization based on regional experience
- D.Mobilize a drill rig to obtain continuous core at the planned building pads
Show answer and explanation
Answer: A. Review published maps, prior reports, borings and historical air photos
A sound investigation begins with a review of published geologic maps, prior reports, borings and historical aerial photographs , which identifies landslides, faults and problem units and guides where field exploration is needed. Drilling or geophysics before this review risks putting exploration in the wrong places. Estimating stabilization costs before any site data exist assumes a conclusion the investigation is meant to test.
Question 7 · Domain F · Hydrogeology
Downgradient of a petroleum hydrocarbon release, dissolved arsenic in groundwater has risen from below detection to 40 µg/L, although the fuel contains no arsenic. Dissolved oxygen is near zero and dissolved iron is elevated. What is the most likely cause of the arsenic increase?
- A.Precipitation of arsenic sulfides as sulfate in the plume is reduced
- B.Desorption of arsenic caused by acid generated as the fuel degrades
- C.Oxidation of arsenopyrite caused by oxygen supplied by the plume
- D.Reductive dissolution of iron oxyhydroxides that had adsorbed arsenic
Show answer and explanation
Answer: D. Reductive dissolution of iron oxyhydroxides that had adsorbed arsenic
Biodegradation of hydrocarbons consumes oxygen and drives the aquifer toward iron-reducing conditions, and reductive dissolution of iron oxyhydroxides releases the arsenic they had adsorbed, consistent with high dissolved iron. The plume removes oxygen rather than supplying it, so sulfide oxidation is unlikely. Sulfide precipitation would remove arsenic from solution, and biodegradation does not usually acidify buffered aquifers enough to cause desorption.
Question 8 · Domain A · General and Field Geology
A geologist is evaluating whether wastewater injection could reactivate a cohesionless basement fault. On the fault, the total normal stress is 60 MPa, the shear stress is 25 MPa, the friction coefficient is 0.6, and the existing (hydrostatic) pore pressure is 15 MPa. By how much must pore pressure increase to reach failure?
- A.About 18.3 MPa
- B.About 35.0 MPa
- C.About 3.3 MPa
- D.About 26.7 MPa
Show answer and explanation
Answer: C. About 3.3 MPa
Slip occurs when τ = μ(σₙ − P), so 25 = 0.6(60 − P) → 60 − P = 41.7 → P = 18.3 MPa; the required increase above hydrostatic is 18.3 − 15 = 3.3 MPa . "About 18.3 MPa" is the total failure pore pressure, not the increase, and "About 26.7 MPa" subtracts the hydrostatic pressure from τ/μ (41.7 − 15), confusing effective stress with pore pressure.
Question 9 · Domain F · Hydrogeology
Air sparging is proposed for dissolved benzene in a sand aquifer. Boring logs show a laterally continuous silt layer 1 m above the water table and a basement 20 m away. What is the PRIMARY design concern?
- A.The silt layer will stop air from entering the saturated zone
- B.Benzene is too soluble to partition into the injected air stream
- C.Injected air will raise dissolved iron and clog the sand aquifer
- D.Sparged vapors may spread under the silt toward the basement
Show answer and explanation
Answer: D. Sparged vapors may spread under the silt toward the basement
Air injected below the water table strips VOCs into the vadose zone; a low-permeability layer above the water table can trap the vapors and divert them laterally, creating a vapor-intrusion risk to nearby structures. Sparging must therefore be paired with a properly designed SVE system and soil-gas monitoring near the basement. Benzene is volatile enough (high Henry's constant) to be stripped effectively, and a silt layer above the water table does not prevent air entry below it.
Question 10 · Domain A · General and Field Geology
Staged construction of an embankment over soft, low-permeability clay requires close tracking of construction-induced excess pore pressures so the next lift can be timed. Which instrumentation is MOST appropriate?
- A.Monitoring wells screened across the clay and underlying sand
- B.Open-standpipe piezometers read weekly by water-level tape
- C.In-place inclinometers to infer pore pressure from displacement
- D.Vibrating-wire piezometers installed by the fully grouted method
Show answer and explanation
Answer: D. Vibrating-wire piezometers installed by the fully grouted method
Vibrating-wire piezometers measure pressure with negligible volume change, so they respond rapidly in low-permeability clay, and the fully grouted installation is accepted practice. "Open-standpipe piezometers" have a long hydrodynamic time lag in clay because water must flow in to register a change. Long-screened wells cross-connect the clay with the sand and record a mixed head, and inclinometers measure deformation, not pore pressure.

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