Free CPSWQ Practice Questions

Correct answer: B - Yes - the lot is part of a common plan of development disturbing 1 acre or more

The CGP's 1-acre threshold applies to the total disturbance of the 'larger common plan of development or sale,' not to any individual operator's portion. Because the 22-lot subdivision disturbs 18 total acres under a single development plan, every operator who disturbs any portion of that plan - regardless of individual lot size - must obtain NPDES permit coverage. Option A is the most common misconception: candidates focus on the 0.6-acre individual disturbance and miss the common plan rule. Option C is also incorrect: responsibility follows the individual operator of the earth disturbance, not just the master developer. Option D adds a condition (MS4 boundary) that is irrelevant to CGP coverage requirements.

Correct answer: C - Post-Construction Stormwater Management

Matching each program to its MCM: (1) educational mailings = MCM 1, Public Education and Outreach; (2) citizen advisory panel = MCM 2, Public Participation and Involvement; (3) storm sewer map and illicit discharge ordinance = MCM 3, Illicit Discharge Detection and Elimination; (4) construction site ordinance = MCM 4, Construction Site Runoff Control; (5) maintenance crew training = MCM 6, Pollution Prevention and Good Housekeeping. The unaddressed MCM is MCM 5 - Post-Construction Stormwater Management - which requires a separate regulatory mechanism ensuring permanent stormwater controls are installed and maintained after construction ends. This is distinct from MCM 4, which only governs active earth disturbance. Options A, B, and D are all accounted for in the scenario. This question tests whether candidates recognize that post-construction management is its own discrete requirement, not an extension of the construction site ordinance.

Correct answer: C - Pet waste and leaf litter decomposing in the catch basin

Dissolved oxygen depletion is a secondary pollutant effect produced when microorganisms consume oxygen while decomposing organic matter - a condition measured as Biological Oxygen Demand (BOD). Pet waste and leaf litter are concentrated sources of organic material. As they decompose inside the catch basin, BOD rises and DO drops. Zinc (Option A) is toxic to aquatic organisms at elevated concentrations but does not consume dissolved oxygen. Chloride (Option B) is a conservative pollutant - it does not biodegrade and exerts zero oxygen demand. Sediment (Option D) can carry adsorbed pollutants and smother benthic habitat but does not directly deplete dissolved oxygen in the water column. The GPRM categorizes DO depletion as a 'secondary pollutant form' resulting from the decomposition of primary organic pollutants - a critical Module 3 distinction.

Correct answer: B - Pairing an aerated inlet zone for nitrification with a deeper anaerobic outlet zone for denitrification

Complete biological nitrogen removal requires two sequential steps: (1) nitrification - the aerobic conversion of ammonium (NH4+) to nitrate (NO3-) - followed by (2) denitrification - the anaerobic/anoxic conversion of nitrate to nitrogen gas (N2), which permanently removes nitrogen from the water. A single aerobic open-water zone enables nitrification but not denitrification - nitrate simply accumulates. Option A is incorrect because deeper pooling improves TSS sedimentation but does not create the aerobic-then-anaerobic sequence needed for N removal. Option C is incorrect because plant uptake provides only temporary storage; nitrogen is returned to the water when plant material decomposes unless biomass is physically harvested. Option D is incorrect because sand filters capture particulate nitrogen but cannot remove dissolved inorganic nitrogen (nitrate, ammonium), which dominates in stormwater.

Correct answer: D - Type F - entrenched and meandering, low entrenchment ratio, prone to lateral instability

The critical first step in Rosgen classification is calculating the Entrenchment Ratio (ER) = floodprone width / bankfull width = 22 / 18 = 1.22. An ER below 1.4 indicates a highly entrenched channel. This single value eliminates both Type C (requires ER > 2.2) and Type E (also requires ER > 2.2). Type A is also eliminated: it requires low sinuosity (< 1.2) and steep slope (> 4%) - this reach has sinuosity of 1.65 and a gentle slope of 0.003 ft/ft. Type F matches all parameters: ER < 1.4 (entrenched), sinuosity > 1.5 (meandering), low-to-moderate slope, gravel bed material - and a characteristic tendency toward lateral erosion and bank failure because the entrenched channel cannot access its floodplain during high flows. The most common exam error here: candidates who see 'high sinuosity + low slope' immediately select Type C without first computing the entrenchment ratio.

Correct answer: B - Approximately 11.1 ac-ft

Step 1 - Weighted CN: [(94 × 25) + (74 × 15)] / 40 = 3,460 / 40 = 86.5. Step 2 - S = (1,000 / 86.5) − 10 = 1.56 in. Step 3 - 0.2S = 0.31 in; since P = 4.8 > 0.31, runoff occurs. Step 4 - Q = (4.8 − 0.31)² / (4.8 + 0.8 × 1.56) = (4.49)² / 6.05 = 20.16 / 6.05 = 3.33 in. Step 5 - Volume = 3.33 × 40 / 12 = 11.1 ac-ft. Option A (~7.4 ac-ft) results from applying CN = 74 to the entire watershed - using only the open space CN. Option C (~13.7 ac-ft) results from applying CN = 94 to the entire watershed - using only the commercial CN. Option D (~10.3 ac-ft) results from using a simple arithmetic average CN of (94 + 74) / 2 = 84 instead of the area-weighted composite. Each wrong answer represents a specific, common procedural error in the TR-55 method.

Correct answer: D - Approximately 18× - compounded effect of higher Rv and higher concentration

Since P, Pj, A, and the conversion constant are identical in both scenarios, the load ratio depends only on (Rv × C). Forest: Rv = 0.05 + 0.009 × 2 = 0.068; load factor = 0.068 × 51 = 3.47. Commercial: Rv = 0.05 + 0.009 × 85 = 0.815; load factor = 0.815 × 77 = 62.76. Ratio = 62.76 / 3.47 = 18.1×. Option A (3×) reflects only the concentration change (77/51 ≈ 1.5×, not even 3×) and ignores the Rv change entirely. Option B (8×) underestimates the Rv ratio alone (0.815 / 0.068 ≈ 12×) and omits the concentration effect. Option C (12×) captures only the Rv increase. The ~18× result illustrates why post-construction stormwater management requirements exist: converting forest to commercial land multiplies pollutant loading nearly 18-fold through the compounding effect of higher runoff volume and higher pollutant concentration.

Correct answer: C - 98% - C changes from 1.0 to 0.02, reducing predicted loss by a factor of 50

Because RUSLE is strictly multiplicative, the ratio of new to original soil loss equals the ratio of the new C factor to the old: 0.02 / 1.0 = 0.02. Only 2% of the original soil loss is predicted - a 98% reduction. No averaging or interaction dampens this; changing one factor by a given proportion changes the total by exactly that proportion. Option A (50%) reflects an intuition that mulch intercepts 'about half' the rainfall energy, but the RUSLE C factor captures the full integrated effect on detachment, runoff, and transport - not just interception. Option B (75%) is a plausible-sounding generic BMP performance figure with no mathematical basis when C is precisely known. Option D (35%) reflects a false belief that RUSLE factors moderate each other. The multiplicative structure explains why dense straw mulch is among the most cost-effective temporary erosion controls available.

Correct answer: C - Wet detention pond - uses sedimentation in a permanent pool; no native soil permeability needed

The two controlling site constraints are HSG D soils (infiltration rate < 0.1 in/hr) and a high water table at 18 inches below grade. Both constraints categorically eliminate all infiltration-dependent practices. An infiltration basin (Option A) requires a minimum native soil infiltration rate of approximately 0.5 in/hr - HSG D soils would cause it to remain saturated and fail to drain. A bioretention cell without an underdrain (Option B) also depends entirely on native soil infiltration; HSG D soils make this non-functional, and a shallow water table creates a pathway for untreated runoff to reach groundwater. Pervious pavement (Option D) has the same constraints. A wet detention pond (Option C) requires no soil permeability: it operates entirely through hydraulic residence time and gravitational settling in a permanent pool, with outflow controlled through a riser-and-barrel outlet. It satisfies both permit requirements - the controlled outlet attenuates peaks, and the permanent pool provides the residence time needed for TSS removal (~80% typical efficiency).

Correct answer: B - Within 24 hours of Tuesday's rain AND by Wednesday the following week for the 7-day interval

The 2022 CGP establishes two independent inspection obligations that run simultaneously and do not reset each other: (1) a ROUTINE inspection at least once every 7 calendar days - the last was Wednesday, so the next routine inspection is due by the following Wednesday; and (2) a POST-STORM inspection within 24 hours of any rainfall event of 0.25 inches or more in a 24-hour period - Tuesday's 0.30-inch event exceeds this threshold, requiring an inspection by Wednesday morning. Completing a post-storm inspection does not extend or reset the 7-day routine clock, and vice versa. Option A is incorrect - the 7-day clock never resets due to a storm inspection. Option C is incorrect on two counts: the window is 24 hours, not 48, and the routine requirement is never waived. Option D is incorrect - the trigger threshold is 0.25 inches, not 0.5 inches; 0.30 inches exceeds it. A single site visit on Wednesday morning can simultaneously satisfy both the 24-hour post-storm requirement and the 7-day routine interval.