LADWP Water Utility Worker Practice
Exam with Practice Questions And
Correct Answers (Ver
1. A water utility worker is preparing to repair a 12-inch ductile iron water main
break. The excavation is 8 feet deep with vertical walls and no protective system. The
soil is classified as Type B. According to OSHA 29 CFR 1926 Subpart P, which of the
following is the minimum required protective system?
A. No protective system is required for depths less than 10 feet.
B. A trench box or shoring system designed for Type B soil and a depth of at least 8 feet.
C. Sloping the trench walls to a 1:1 ratio (45 degrees).
D. A registered professional engineer must design a protective system for all excavations
deeper than 5 feet.
Answer: B. A trench box or shoring system designed for Type B soil and a depth of at
least 8 feet.
Rationale: OSHA requires a protective system for excavations 5 feet or deeper (or
shallower if unstable). For Type B soil, a trench box or shoring is required; sloping at 1:1
is insufficient (Type B requires 1:1 for short-term, but the question asks for minimum
required system, and a trench box is explicitly designed). A PE is only required for
excavations over 20 feet or when using certain methods. Option A is incorrect because 5
feet is the threshold.
Page 1 of 79
,2. A water treatment plant uses conventional coagulation, flocculation, sedimentation,
and filtration. The raw water has high turbidity and natural organic matter (NOM).
After coagulant addition, the zeta potential is measured at -5 mV. Which of the
following best describes the expected effect on filter performance and disinfection
byproduct (DBP) formation?
A. Improved filter performance and decreased DBP formation due to enhanced NOM
removal.
B. Poor filter performance and increased DBP formation due to inadequate coagulation.
C. Improved filter performance but increased DBP formation due to overdosing
coagulant.
D. No effect on filter performance or DBP formation because zeta potential is not
relevant.
Answer: A. Improved filter performance and decreased DBP formation due to
enhanced NOM removal.
Rationale: A zeta potential near zero (e.g., -5 mV) indicates optimal coagulation, where
particles are destabilized and NOM is effectively removed. This leads to better filter
performance and reduced DBP precursors. Option B would be true if zeta potential were
highly negative. Option C is incorrect because overdosing could restabilize particles.
Option D is false as zeta potential is a key indicator.
Page 2 of 79
,3. A water utility worker is calculating the volume of water in a cylindrical storage
tank with a diameter of 30 feet and a water depth of 20 feet. Using the formula V =
r²h, what is the approximate volume in gallons? (1 cubic foot = 7.48 gallons)
A. 105,000 gallons
B. 150,000 gallons
C. 211,000 gallons
D. 300,000 gallons
Answer: A. 105,000 gallons
Rationale: Radius = 15 ft, height = 20 ft. Volume = * (15)² * 20 3.1416 * 225 * 20
14,137 cubic feet. Convert to gallons: 14,137 * 7.48 105,745 gallons, closest to 105,000.
Options B, C, and D are incorrect due to calculation errors (e.g., using diameter instead of
radius or incorrect conversion).
4. A water utility worker is reviewing the latest EPA Lead and Copper Rule Revisions
(LCRR). Which of the following is a new requirement for water systems under the
LCRR?
A. Replacing all lead service lines within 5 years regardless of action level exceedance.
B. Conducting a lead service line inventory and making it publicly available.
C. Using only orthophosphate for corrosion control.
D. Testing for lead only at the treatment plant.
Answer: B. Conducting a lead service line inventory and making it publicly available.
Rationale: The LCRR (effective 2021, compliance by 2024) requires water systems to
develop a lead service line inventory and make it publicly accessible. Option A is not
required; replacement is triggered by action level exceedances or specific conditions.
Option C is too prescriptive; systems can choose corrosion control methods. Option D is
incorrect; testing is at consumer taps.
Page 3 of 79
, 5. A water utility worker is using a centrifugal pump to move water from a reservoir
to a distribution main. The pump curve indicates a shutoff head of 120 feet and a
maximum flow of 1000 gpm at 60 feet head. The system curve intersects the pump
curve at 800 gpm and 80 feet head. If the static lift increases by 10 feet, what is the
expected new operating point?
A. Flow increases, head decreases.
B. Flow decreases, head increases.
C. Flow and head both increase.
D. Flow and head remain the same.
Answer: B. Flow decreases, head increases.
Rationale: Increasing static lift raises the system curve, moving the intersection to a lower
flow and higher head on the pump curve. This is a fundamental principle of pump-system
interaction. Options A, C, and D do not reflect the effect of increased static head.
6. A water utility worker is performing a chlorination test on a new 8-inch water main.
The main is 500 feet long and must be filled with water containing 50 mg/L free
chlorine for 24 hours. How many pounds of 65% calcium hypochlorite (Ca(OCl)) are
needed? (Volume of water = r²L; 1 gallon = 8.34 lbs; 1 ft³ = 7.48 gallons)
A. 0.5 lbs
B. 1.0 lbs
C. 1.5 lbs
D. 2.0 lbs
Answer: C. 1.5 lbs
Rationale: Volume = *(0.333 ft)²*500 174 ft³ 1300 gallons. Weight of water = 1300*8.34
10,842 lbs. Required chlorine = 50 mg/L = 50 ppm 10,842 * 50/1,000,000 = 0.542 lbs.
Since Ca(OCl) is 65% available chlorine, needed = 0.542/0.65 0.834 lbs. However, options
are rounded; the closest is 1.0 lbs, but exact calculation yields ~0.83 lbs, which is not
listed. Re-evaluating: 500 ft of 8-inch main: radius = 4 inches = 0.333 ft. Volume =
*(0.333)²*500 = 174.5 ft³. Gallons = 174.5*7.48 = 1305 gal. Weight = 1305*8.34 =
10,884 lbs. Chlorine needed = 10,884 * 50/1e6 = 0.544 lbs. Ca(OCl) needed = 0.544/0.65
= 0.837 lbs. Closest option is 1.0 lbs (B), but 0.837 is closer to 1.0 than 0.5. However, the
correct answer might be 1.0 lbs. But the options are 0.5,1.0,1.5,2.0. 0.837 rounds to 1.0. So
B is correct. But the explanation says C. This is an error. Let's correct: The correct answer
should be B. But I need to ensure the calculation is accurate. Let's recompute: 8-inch
Page 4 of 79
Exam with Practice Questions And
Correct Answers (Ver
1. A water utility worker is preparing to repair a 12-inch ductile iron water main
break. The excavation is 8 feet deep with vertical walls and no protective system. The
soil is classified as Type B. According to OSHA 29 CFR 1926 Subpart P, which of the
following is the minimum required protective system?
A. No protective system is required for depths less than 10 feet.
B. A trench box or shoring system designed for Type B soil and a depth of at least 8 feet.
C. Sloping the trench walls to a 1:1 ratio (45 degrees).
D. A registered professional engineer must design a protective system for all excavations
deeper than 5 feet.
Answer: B. A trench box or shoring system designed for Type B soil and a depth of at
least 8 feet.
Rationale: OSHA requires a protective system for excavations 5 feet or deeper (or
shallower if unstable). For Type B soil, a trench box or shoring is required; sloping at 1:1
is insufficient (Type B requires 1:1 for short-term, but the question asks for minimum
required system, and a trench box is explicitly designed). A PE is only required for
excavations over 20 feet or when using certain methods. Option A is incorrect because 5
feet is the threshold.
Page 1 of 79
,2. A water treatment plant uses conventional coagulation, flocculation, sedimentation,
and filtration. The raw water has high turbidity and natural organic matter (NOM).
After coagulant addition, the zeta potential is measured at -5 mV. Which of the
following best describes the expected effect on filter performance and disinfection
byproduct (DBP) formation?
A. Improved filter performance and decreased DBP formation due to enhanced NOM
removal.
B. Poor filter performance and increased DBP formation due to inadequate coagulation.
C. Improved filter performance but increased DBP formation due to overdosing
coagulant.
D. No effect on filter performance or DBP formation because zeta potential is not
relevant.
Answer: A. Improved filter performance and decreased DBP formation due to
enhanced NOM removal.
Rationale: A zeta potential near zero (e.g., -5 mV) indicates optimal coagulation, where
particles are destabilized and NOM is effectively removed. This leads to better filter
performance and reduced DBP precursors. Option B would be true if zeta potential were
highly negative. Option C is incorrect because overdosing could restabilize particles.
Option D is false as zeta potential is a key indicator.
Page 2 of 79
,3. A water utility worker is calculating the volume of water in a cylindrical storage
tank with a diameter of 30 feet and a water depth of 20 feet. Using the formula V =
r²h, what is the approximate volume in gallons? (1 cubic foot = 7.48 gallons)
A. 105,000 gallons
B. 150,000 gallons
C. 211,000 gallons
D. 300,000 gallons
Answer: A. 105,000 gallons
Rationale: Radius = 15 ft, height = 20 ft. Volume = * (15)² * 20 3.1416 * 225 * 20
14,137 cubic feet. Convert to gallons: 14,137 * 7.48 105,745 gallons, closest to 105,000.
Options B, C, and D are incorrect due to calculation errors (e.g., using diameter instead of
radius or incorrect conversion).
4. A water utility worker is reviewing the latest EPA Lead and Copper Rule Revisions
(LCRR). Which of the following is a new requirement for water systems under the
LCRR?
A. Replacing all lead service lines within 5 years regardless of action level exceedance.
B. Conducting a lead service line inventory and making it publicly available.
C. Using only orthophosphate for corrosion control.
D. Testing for lead only at the treatment plant.
Answer: B. Conducting a lead service line inventory and making it publicly available.
Rationale: The LCRR (effective 2021, compliance by 2024) requires water systems to
develop a lead service line inventory and make it publicly accessible. Option A is not
required; replacement is triggered by action level exceedances or specific conditions.
Option C is too prescriptive; systems can choose corrosion control methods. Option D is
incorrect; testing is at consumer taps.
Page 3 of 79
, 5. A water utility worker is using a centrifugal pump to move water from a reservoir
to a distribution main. The pump curve indicates a shutoff head of 120 feet and a
maximum flow of 1000 gpm at 60 feet head. The system curve intersects the pump
curve at 800 gpm and 80 feet head. If the static lift increases by 10 feet, what is the
expected new operating point?
A. Flow increases, head decreases.
B. Flow decreases, head increases.
C. Flow and head both increase.
D. Flow and head remain the same.
Answer: B. Flow decreases, head increases.
Rationale: Increasing static lift raises the system curve, moving the intersection to a lower
flow and higher head on the pump curve. This is a fundamental principle of pump-system
interaction. Options A, C, and D do not reflect the effect of increased static head.
6. A water utility worker is performing a chlorination test on a new 8-inch water main.
The main is 500 feet long and must be filled with water containing 50 mg/L free
chlorine for 24 hours. How many pounds of 65% calcium hypochlorite (Ca(OCl)) are
needed? (Volume of water = r²L; 1 gallon = 8.34 lbs; 1 ft³ = 7.48 gallons)
A. 0.5 lbs
B. 1.0 lbs
C. 1.5 lbs
D. 2.0 lbs
Answer: C. 1.5 lbs
Rationale: Volume = *(0.333 ft)²*500 174 ft³ 1300 gallons. Weight of water = 1300*8.34
10,842 lbs. Required chlorine = 50 mg/L = 50 ppm 10,842 * 50/1,000,000 = 0.542 lbs.
Since Ca(OCl) is 65% available chlorine, needed = 0.542/0.65 0.834 lbs. However, options
are rounded; the closest is 1.0 lbs, but exact calculation yields ~0.83 lbs, which is not
listed. Re-evaluating: 500 ft of 8-inch main: radius = 4 inches = 0.333 ft. Volume =
*(0.333)²*500 = 174.5 ft³. Gallons = 174.5*7.48 = 1305 gal. Weight = 1305*8.34 =
10,884 lbs. Chlorine needed = 10,884 * 50/1e6 = 0.544 lbs. Ca(OCl) needed = 0.544/0.65
= 0.837 lbs. Closest option is 1.0 lbs (B), but 0.837 is closer to 1.0 than 0.5. However, the
correct answer might be 1.0 lbs. But the options are 0.5,1.0,1.5,2.0. 0.837 rounds to 1.0. So
B is correct. But the explanation says C. This is an error. Let's correct: The correct answer
should be B. But I need to ensure the calculation is accurate. Let's recompute: 8-inch
Page 4 of 79
