Tradesman Exam Questions and correct answers
rated A+
1. In the context of advanced electrical theory, a three-phase induction motor operating at 60 Hz
has a synchronous speed of 1200 rpm. If the rotor speed is measured as 1140 rpm under full load,
what is the slip percentage, and how does this affect rotor frequency?
A. Slip = 5%, rotor frequency = 3 Hz
B. Slip = 5%, rotor frequency = 60 Hz
C. Slip = 10%, rotor frequency = 6 Hz
D. Slip = 10%, rotor frequency = 60 Hz
Answer: A
Rationale: Slip is calculated as (Ns - Nr)/Ns = (1200-1140)/1200 = 0.05 = 5%. Rotor frequency = slip ×
stator frequency = 0.05 × 60 = 3 Hz. Option B incorrectly uses stator frequency; C and D have wrong
slip values.
2. A structural steel beam is subjected to a bending moment that causes a maximum stress of 36
ksi. The yield strength of the steel is 50 ksi. Using the maximum distortion energy theory (von
Mises criterion), what is the factor of safety if the beam is in pure bending?
A. 1.39
B. 1.39 but for plane stress with zero shear, von Mises stress equals normal stress, so factor of safety = 50/36 =
1.39
C. 1.39, but the correct calculation is yield strength divided by von Mises stress, which equals 50/36 = 1.39
D. 1.39
Answer: B
Rationale: In pure bending, the stress state is uniaxial (one normal stress, zero shear). The von Mises
stress for uniaxial stress equals the normal stress itself (36 ksi). Factor of safety = yield strength / von
Mises stress = 50/36 = 1.39. Options A, C, and D are numerically correct but lack the reasoning that
von Mises stress equals normal stress in this case. Option B is the most complete explanation.
3. A plumbing system includes a backflow preventer installed on a potable water line serving a
chemical mixing tank. The tank operates at a pressure of 80 psi, while the water supply pressure is
60 psi. Which type of backflow preventer is most appropriate, and what is the primary reason?
A. Reduced pressure zone (RPZ) assembly, because it provides the highest level of protection against
backpressure backflow
B. Double check valve assembly (DCVA), because it is adequate for non-health hazard applications
C. Pressure vacuum breaker (PVB), because it prevents backsiphonage
D. Atmospheric vacuum breaker (AVB), because it is the simplest and most cost-effective
Answer: A
Rationale: The chemical tank creates a backpressure condition (80 psi > 60 psi), which requires a device
that can handle both backpressure and backsiphonage. An RPZ assembly is the only type that provides
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, protection against backpressure backflow and is suitable for health hazard (chemical) applications.
DCVA is for non-health hazard; PVB and AVB cannot handle backpressure.
4. In HVAC design, a building's cooling load is 500,000 BTU/hr. The design supply air temperature
is 55°F, and the return air temperature is 75°F. What is the required airflow in CFM? (Use
standard air density of 0.075 lb/ft³ and specific heat of air = 0.24 BTU/lb-°F)
A. Approximately 20,000 CFM
B. Approximately 25,000 CFM
C. Approximately 30,000 CFM
D. Approximately 35,000 CFM
Answer: B
Rationale: Cooling load Q = 1.08 × CFM × ”T (sensible heat equation). ”T = 75 - 55 = 20°F. So CFM =
Q / (1.08 × T) = 500,000 / (1.08 × 20) = 500,.6 23,148 CFM. The closest option is 25,000 CFM.
Option A is too low; C and D are too high.
5. A concrete mix design requires a water-to-cement ratio of 0.45 by weight. The coarse aggregate
has a specific gravity of 2.65 and a dry-rodded unit weight of 100 lb/ft³. If the total volume of
concrete is 1 cubic yard, and the cement content is 600 lb, what is the approximate weight of coarse
aggregate required, assuming the mix is not air-entrained and using the absolute volume method?
(Assume fine aggregate is the remainder)
A. 1600 lb
B. 1800 lb
C. 2000 lb
D. 2200 lb
Answer: C
Rationale: Total volume = 27 ft³. Cement volume = 600 / (3.15 × 62.4) = 3.05 ft³ (assuming specific
gravity of cement = 3.15). Water volume = (0.45 × 600) / 62.4 = 4.33 ft³. Coarse aggregate volume =
(dry-rodded weight / specific gravity × 62.4) but we need weight. Using ACI method: coarse aggregate
volume 0.6 × 27 = 16.2 ft³ (for typical fineness modulus). Weight = 16.2 × 100 = 1620 lb? Actually,
dry-rodded unit weight is 100 lb/ft³, so weight = volume × 100. But the volume of coarse aggregate in
the mix is not the dry-rodded volume; it's the solid volume. The correct approach: absolute volume of
coarse aggregate = (weight) / (2.65 × 62.4). The sum of absolute volumes of cement, water, air, coarse,
fine = 27 ft³. Assume 2% air = 0.54 ft³. Then volume of coarse + fine = 27 - 3.05 - 4.33 - 0.54 = 19.08
ft³. Typically coarse aggregate volume is about 60% of that = 11.45 ft³. Then weight = 11.45 × 2.65 ×
62.4 = 1895 lb. Closest to 2000 lb. However, the problem expects use of dry-rodded weight: typical
coarse aggregate weight per cubic yard is about 2000 lb. So answer C is correct.
6. A tradesman is installing a 200-amp, 120/240-volt, single-phase service. The service entrance
conductors are to be sized per the National Electrical Code (NEC). The calculated load is 180
amps. The conductors will be run in a raceway with two other current-carrying conductors, and
the ambient temperature is 86°F. What is the minimum required ampacity for the ungrounded
conductors? (Use 75°C rated conductors, and refer to NEC 310.15(B)(16) and Table
310.15(B)(2)(a))
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rated A+
1. In the context of advanced electrical theory, a three-phase induction motor operating at 60 Hz
has a synchronous speed of 1200 rpm. If the rotor speed is measured as 1140 rpm under full load,
what is the slip percentage, and how does this affect rotor frequency?
A. Slip = 5%, rotor frequency = 3 Hz
B. Slip = 5%, rotor frequency = 60 Hz
C. Slip = 10%, rotor frequency = 6 Hz
D. Slip = 10%, rotor frequency = 60 Hz
Answer: A
Rationale: Slip is calculated as (Ns - Nr)/Ns = (1200-1140)/1200 = 0.05 = 5%. Rotor frequency = slip ×
stator frequency = 0.05 × 60 = 3 Hz. Option B incorrectly uses stator frequency; C and D have wrong
slip values.
2. A structural steel beam is subjected to a bending moment that causes a maximum stress of 36
ksi. The yield strength of the steel is 50 ksi. Using the maximum distortion energy theory (von
Mises criterion), what is the factor of safety if the beam is in pure bending?
A. 1.39
B. 1.39 but for plane stress with zero shear, von Mises stress equals normal stress, so factor of safety = 50/36 =
1.39
C. 1.39, but the correct calculation is yield strength divided by von Mises stress, which equals 50/36 = 1.39
D. 1.39
Answer: B
Rationale: In pure bending, the stress state is uniaxial (one normal stress, zero shear). The von Mises
stress for uniaxial stress equals the normal stress itself (36 ksi). Factor of safety = yield strength / von
Mises stress = 50/36 = 1.39. Options A, C, and D are numerically correct but lack the reasoning that
von Mises stress equals normal stress in this case. Option B is the most complete explanation.
3. A plumbing system includes a backflow preventer installed on a potable water line serving a
chemical mixing tank. The tank operates at a pressure of 80 psi, while the water supply pressure is
60 psi. Which type of backflow preventer is most appropriate, and what is the primary reason?
A. Reduced pressure zone (RPZ) assembly, because it provides the highest level of protection against
backpressure backflow
B. Double check valve assembly (DCVA), because it is adequate for non-health hazard applications
C. Pressure vacuum breaker (PVB), because it prevents backsiphonage
D. Atmospheric vacuum breaker (AVB), because it is the simplest and most cost-effective
Answer: A
Rationale: The chemical tank creates a backpressure condition (80 psi > 60 psi), which requires a device
that can handle both backpressure and backsiphonage. An RPZ assembly is the only type that provides
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, protection against backpressure backflow and is suitable for health hazard (chemical) applications.
DCVA is for non-health hazard; PVB and AVB cannot handle backpressure.
4. In HVAC design, a building's cooling load is 500,000 BTU/hr. The design supply air temperature
is 55°F, and the return air temperature is 75°F. What is the required airflow in CFM? (Use
standard air density of 0.075 lb/ft³ and specific heat of air = 0.24 BTU/lb-°F)
A. Approximately 20,000 CFM
B. Approximately 25,000 CFM
C. Approximately 30,000 CFM
D. Approximately 35,000 CFM
Answer: B
Rationale: Cooling load Q = 1.08 × CFM × ”T (sensible heat equation). ”T = 75 - 55 = 20°F. So CFM =
Q / (1.08 × T) = 500,000 / (1.08 × 20) = 500,.6 23,148 CFM. The closest option is 25,000 CFM.
Option A is too low; C and D are too high.
5. A concrete mix design requires a water-to-cement ratio of 0.45 by weight. The coarse aggregate
has a specific gravity of 2.65 and a dry-rodded unit weight of 100 lb/ft³. If the total volume of
concrete is 1 cubic yard, and the cement content is 600 lb, what is the approximate weight of coarse
aggregate required, assuming the mix is not air-entrained and using the absolute volume method?
(Assume fine aggregate is the remainder)
A. 1600 lb
B. 1800 lb
C. 2000 lb
D. 2200 lb
Answer: C
Rationale: Total volume = 27 ft³. Cement volume = 600 / (3.15 × 62.4) = 3.05 ft³ (assuming specific
gravity of cement = 3.15). Water volume = (0.45 × 600) / 62.4 = 4.33 ft³. Coarse aggregate volume =
(dry-rodded weight / specific gravity × 62.4) but we need weight. Using ACI method: coarse aggregate
volume 0.6 × 27 = 16.2 ft³ (for typical fineness modulus). Weight = 16.2 × 100 = 1620 lb? Actually,
dry-rodded unit weight is 100 lb/ft³, so weight = volume × 100. But the volume of coarse aggregate in
the mix is not the dry-rodded volume; it's the solid volume. The correct approach: absolute volume of
coarse aggregate = (weight) / (2.65 × 62.4). The sum of absolute volumes of cement, water, air, coarse,
fine = 27 ft³. Assume 2% air = 0.54 ft³. Then volume of coarse + fine = 27 - 3.05 - 4.33 - 0.54 = 19.08
ft³. Typically coarse aggregate volume is about 60% of that = 11.45 ft³. Then weight = 11.45 × 2.65 ×
62.4 = 1895 lb. Closest to 2000 lb. However, the problem expects use of dry-rodded weight: typical
coarse aggregate weight per cubic yard is about 2000 lb. So answer C is correct.
6. A tradesman is installing a 200-amp, 120/240-volt, single-phase service. The service entrance
conductors are to be sized per the National Electrical Code (NEC). The calculated load is 180
amps. The conductors will be run in a raceway with two other current-carrying conductors, and
the ambient temperature is 86°F. What is the minimum required ampacity for the ungrounded
conductors? (Use 75°C rated conductors, and refer to NEC 310.15(B)(16) and Table
310.15(B)(2)(a))
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