ASBOG FG Practice Questions (2026) — Free Sample Questions with Answers

Answer: D

12 ft: This is the clay-to-sand contact, a change within the soil overburden.

14 ft: This is the water level encountered during drilling, not a lithologic contact.

21 ft: This is the top of weathered rock, but the highly weathered shale still behaves like soil (samples break down to clayey gravel), so it is not competent rock.

26 ft: Correct. Auger refusal at 26 ft and slightly weathered to fresh cored shale with 95% recovery and RQD 70–85% mark the top of competent rock.

Distinguish weathered rock that samples like soil from competent rock indicated by auger refusal, good core recovery and fair-to-good RQD.

Answer: A

Deposition of A, B, and C; faulting on F; intrusion of E; erosion; deposition of D: Correct. Superposition puts A, B, C first. Fault F offsets A–C, and dike E crosses F without being offset, so the dike is younger (cross-cutting relationships). Both F and E are truncated at the erosion surface beneath D, so erosion and D come last.

Deposition of A, B, and C; faulting on F; erosion; deposition of D; intrusion of E: E stops at the base of D, so the dike is older than D and the erosion surface.

Deposition of A, B, and C; erosion; deposition of D; faulting on F; intrusion of E: F does not offset D or the erosion surface, so faulting came before erosion and deposition of D.

Deposition of A, B, and C; intrusion of E; faulting on F; erosion; deposition of D: If E predated faulting, the dike would be offset where it crosses F; it is continuous.

Use superposition for the layers, then cross-cutting relationships: a feature is younger than anything it cuts and older than any surface that truncates it.

Answer: A

metamorphic grade increases toward the east: Correct. In the Barrovian sequence chlorite → biotite → garnet → staurolite marks increasing temperature, so grade rises eastward.

the zones reflect changes in protolith, not grade: In rocks of similar pelitic composition, successive index minerals record grade, not protolith.

staurolite formed at lower temperature than biotite: Staurolite is a higher-grade index mineral than biotite.

metamorphic grade increases toward the west: Reverses the index-mineral sequence; chlorite (west) is the lowest grade.

Barrovian index minerals in pelites appear in order of increasing grade: chlorite, biotite, garnet, staurolite, kyanite, sillimanite.

Answer: A

Laterally adjacent environments shifted landward during a transgression.: Correct. By Walther's law, conformable vertically stacked facies were laterally adjacent. Marsh (most landward), then lagoon, then barrier (most seaward) stacking upward means seaward facies moved over landward ones: a transgression.

The three environments existed at the same place at the same time.: Vertically stacked facies at one place formed one after another, not simultaneously.

Each contact marks a large time gap between unrelated environments.: Walther's law applies to conformable successions with no major gaps.

Laterally adjacent environments shifted seaward during a regression.: A regression would stack landward facies (marsh) over seaward ones (barrier), the reverse of this order.

Walther's law: in a conformable succession, facies that are vertically adjacent were deposited in laterally adjacent environments; their stacking order shows the direction of shoreline shift.

Answer: B

Glacial till; eolian dune sand: Reverses the samples: the steep (well-sorted) curve is Sample 1, not Sample 2.

Eolian dune sand; glacial till: Correct. Sample 1's steep curve shows a well-sorted fine sand (median about 0.2 mm, almost no gravel or fines), typical of wind-blown dune sand. Sample 2's flat curve spans gravel to mud (38% finer than 0.063 mm), the very poor sorting of till.

Loess; debris-flow deposit: Loess is silt; Sample 1 has only 1% finer than 0.063 mm, so it is sand, not silt.

Fluvial gravel bar; offshore mud: Sample 1 contains no gravel (100% finer than 1 mm), and Sample 2 is a gravel–sand–mud mixture, not mud.

A steep cumulative curve means good sorting (eolian, beach); a flat curve spanning many size classes means poor sorting (till, debris flow).

Answer: C

Toward the northwest: Reversed sense: the blunt end faces the direction the ice came from, so the ice moved away from the northwest, not toward it.

Toward the northeast: Perpendicular to the long axes; drumlins are elongated parallel to flow, not across it.

Toward the southeast: Correct. These are drumlins. Their long axes parallel ice flow, the blunt stoss end faces up-ice (northwest) and the tapered lee end points down-ice (southeast).

Toward the southwest: Also perpendicular to the long axes; this reading treats the hills as transverse ridges.

Drumlins are elongated parallel to ice flow, with the steep blunt (stoss) end up-ice and the gentle tapered tail down-ice.

Answer: C

Normal faulting: Normal faulting shows the reverse pattern: a white (dilatational) center with shaded sides.

Strike-slip faulting: Strike-slip mechanisms show four alternating quadrants with both nodal planes near vertical, drawn as straight lines through the center.

Thrust faulting: Correct. Compressional first motions at the center mean the tension (T) axis is near vertical and the pressure (P) axis horizontal (E–W): thrust faulting on a N–S plane dipping about 45°.

Oblique-normal faulting: Oblique-normal slip has an off-center pattern with a white region near the center; this mechanism is pure dip-slip with a shaded center.

Beach balls: shaded center = thrust/reverse, white center = normal, four quadrants with straight nodal lines = strike-slip.

Answer: C

1. Offset: 45 m × 1,000 mm/m = 45,000 mm
2. Slip rate = offset ÷ age = 45,000 mm ÷ 15,000 yr
3. Slip rate = 3.0 mm/yr

0.003 mm/yr: This is the rate in meters per year (45 m ÷ 15,000 yr) mislabeled as millimeters.

0.3 mm/yr: Converted meters to centimeters (× 100) and labeled the result millimeters.

3.0 mm/yr: Correct. 45 m = 45,000 mm; 45,000 mm ÷ 15,000 yr = 3.0 mm/yr.

30 mm/yr: Divided by 1,500 years: dropped a zero from the age.

Slip rate = displacement ÷ age of the offset feature; keep units consistent (mm/yr is standard).

Answer: D

Structural dome: A dome has the oldest rocks in the center and dips pointing outward, the reverse of this map.

Monocline: A monocline is a single step in otherwise flat strata and gives straight, not closed, outcrop belts.

Doubly plunging anticline: A doubly plunging anticline also gives a closed elongate pattern, but with the oldest rocks in the core and outward dips.

Structural basin: Correct. Outcrop belts form closed, concentric ellipses with the youngest unit (Permian Pw) in the center, and all dips point inward toward the center.

On flat ground, closed outcrop patterns with youngest beds in the center and inward dips mean a basin; oldest in the center and outward dips mean a dome.

Answer: D

1. 1. Hydraulic gradient: i = Δh/L = (846.0 ft − 840.0 ft)/1,200 ft = 0.005 (dimensionless).
2. 2. Darcy flux: q = K·i = 50 ft/day × 0.005 = 0.25 ft/day.
3. 3. Seepage velocity: v = q/nₑ = 0.25 ft/day ÷ 0.25 = 1.0 ft/day.

0.06 ft/day: Multiplies the Darcy flux by effective porosity (0.25 × 0.25) instead of dividing by it.

0.25 ft/day: This is the Darcy flux q = Ki, a volumetric flux per unit cross-sectional area, not the velocity of the water through the pores.

0.71 ft/day: Divides by total porosity (0.35) instead of effective porosity; dead-end and isolated pores do not carry flow.

1.0 ft/day: Correct: i = 6.0/1,200 = 0.005; q = Ki = 0.25 ft/day; v = q/nₑ = 0.25/0.25 = 1.0 ft/day.

Darcy flux q = Ki; the average linear (seepage) velocity is v = Ki/nₑ, always faster than q because water moves only through the effective pores.

Answer: C

Well 1: Well 1 is in the sandstone outcrop (recharge area), where the aquifer is unconfined and the water level lies below land surface.

Well 2: Well 2 is artesian (confined), but it sits on a hill where the potentiometric surface is below land surface, so water rises in the casing without flowing out.

Well 3: Correct: Well 3 is screened in the confined sandstone in the valley, where the potentiometric surface (dashed line) lies above land surface, so the well flows.

Well 4: Well 4 is screened in the overlying siltstone, above the sandstone; its level reflects the shallow unit, not the sandstone potentiometric surface.

A well in a confined aquifer is artesian; it flows only where the potentiometric surface stands above land surface.

Answer: A

inelastic compaction of clay interbeds as effective stress rises: Correct: lowering heads reduces pore pressure and raises effective stress; slow-draining clay aquitards compact permanently, so the land surface settles.

collapse of abandoned underground mine workings: Mine collapse produces local pits and troughs over workings, not basin-wide subsidence.

hydrocompaction of dry sediment on first wetting: Hydrocompaction happens when dry, low-density deposits are wetted, the opposite of declining water levels.

dissolution of buried evaporite beds: Evaporite dissolution causes local sinkholes or collapse, not basin-wide subsidence tied to water-level decline.

Aquifer-system compaction: head decline → higher effective stress → largely irreversible compaction of fine-grained interbeds → subsidence and fissures.

Answer: B

1. Area = 160 acres × 43,560 ft²/acre = 6,969,600 ft²
2. Volume = 6,969,600 ft² × 6.5 ft = 45,302,400 ft³
3. Unit weight of coal = 1.30 × 62.4 lb/ft³ = 81.12 lb/ft³
4. In-place weight = 45,302,400 ft³ × 81.12 lb/ft³ = 3,674,930,688 lb ÷ 2,000 lb/ton = 1,837,465 tons
5. Recoverable = 1,837,465 tons × 0.75 ≈ 1,378,000 tons ≈ 1.38 million tons

1.06 million tons: Uses the unit weight of water (62.4 lb/ft³) and forgets to multiply by the specific gravity of 1.30.

1.38 million tons: Correct. 45,302,400 ft³ × 81.12 lb/ft³ ÷ 2,000 = 1,837,465 tons in place × 0.75 = about 1.38 million tons recoverable.

1.84 million tons: This is the in-place tonnage; it ignores the 75% mining recovery.

2.45 million tons: Divides by the recovery factor instead of multiplying (1,837,465 ÷ 0.75).

Coal tonnage = area × thickness × (specific gravity × 62.4 lb/ft³) ÷ 2,000, then × recovery for recoverable reserves.

Answer: A

A low-density body, such as salt, centered near 1,200 m: Correct. The symmetric negative anomaly reaches its minimum (about −14 mGal) at 1,200 m, indicating a mass deficit such as a salt dome, which is less dense than the compacted sediments around it.

Two dense bodies centered near 0 m and 2,400 m: The profile ends are the flat background level, not separate highs; the only anomaly is the symmetric low in the middle.

A regional gradient from crustal thinning: A regional effect would be a broad, one-directional slope; this anomaly is local, symmetric and closes on both sides.

A dense mafic body centered near 1,200 m: A denser-than-host body produces a gravity high; this profile is a low, so the sign is reversed.

Gravity lows mark mass deficits (salt, sediment-filled grabens) and highs mark mass excesses; the anomaly peak locates the body.

Answer: D

Well-sorted, uncemented quartz sandstone: Large, well-connected pores give a clean, well-sorted sandstone both high porosity and high permeability.

Dolostone with connected vugs: Connected vugs raise permeability along with porosity; this is a good carbonate reservoir.

Fractured granite: Crystalline rock has very low porosity (a few percent at most); its fractures give it permeability, the reverse of the pattern asked for.

Unfractured marine shale: Correct. Clay-rich mudrocks can hold 20–40% water-filled porosity, but their pore throats are so small that permeability is extremely low, which is why shales act as seals and aquitards.

Porosity is storage and permeability is connectivity; fine-grained rocks combine high porosity with low permeability because of tiny pore throats.