CSLB General Building (B) — All Questions

A residential branch serves the following fixtures with these water-supply fixture unit (WSFU) values: 8 water closets (tank) at 2.5 each, 8 lavatories at 1 each, and 8 showers at 2 each. What is the total demand load on the branch?

• a.52 WSFU✓
• b.36 WSFU
• c.44 WSFU
• d.60 WSFU

Multiply each fixture count by its WSFU value and add: (8 x 2.5) + (8 x 1) + (8 x 2) = 20 + 8 + 16 = 52 WSFU. This total is what you carry into a Hunter-curve demand chart to convert fixture units to gpm. Using the wrong per-fixture value is the usual cause of a low answer such as 44.UPC §610.0

Using the 0.408 velocity relationship v = 0.408 x Q / d^2 (v in ft/s, Q in gpm, d in inches inside diameter), what is the velocity of 18 gpm flowing in a pipe with a 1.00 in inside diameter?

• a.3.7 ft/s
• b.7.3 ft/s✓
• c.10.2 ft/s
• d.14.7 ft/s

v = 0.408 x 18 / (1.00)^2 = 7.34 / 1.00 = 7.3 ft/s. Because d^2 = 1, the velocity equals 0.408 x Q directly. This sits just under the 8 ft/s limit commonly set for cold-water piping to control erosion and water hammer.UPC §610.0

Cold-water piping is commonly limited to a maximum velocity of about 8 ft/s. Using v = 0.408 x Q / d^2, what is the approximate maximum flow a pipe with a 1.025 in inside diameter (nominal 1 in type L copper) can carry without exceeding 8 ft/s?

• a.12 gpm
• b.17 gpm
• c.21 gpm✓
• d.28 gpm

Rearrange to Q = v x d^2 / 0.408 = 8 x (1.025)^2 / 0.408 = 8 x 1.051 / 0.408 = 20.6 gpm, about 21 gpm. Above this flow the velocity exceeds 8 ft/s and erosion-corrosion and noise become a concern. Hot water is usually held to an even lower 5 ft/s.UPC §610.0

A fixture is located 46 ft above the water meter. Using 0.433 psi per foot of elevation, how much static pressure is lost to elevation between the meter and that fixture?

• a.23.0 psi
• b.34.5 psi
• c.46.0 psi
• d.19.9 psi✓

Static loss = height x 0.433 psi/ft = 46 x 0.433 = 19.9 psi. Every foot of rise costs 0.433 psi and must be subtracted from the available supply pressure before you can check the residual at the fixture. The 46 psi answer wrongly treats 1 ft as 1 psi.UPC §610.0

A system has 62 psi at the meter. It must overcome 22 psi of elevation, 14 psi of friction loss in the piping, and deliver a fixture that requires 8 psi minimum flow pressure. What residual pressure remains at the fixture?

• a.18 psi✓
• b.8 psi
• c.14 psi
• d.16 psi

Residual = 62 - 22 (elevation) - 14 (friction) = 26 psi available at the fixture, which exceeds the 8 psi required, leaving 26 - 8 = 18 psi of margin. The design works because the delivered 26 psi is greater than the 8 psi minimum. The pressure budget must always balance supply against elevation, friction, and fixture demand.UPC §610.0

A copper water line has 90 ft of straight pipe. Fittings add an equivalent length of 3 ft each for four elbows and 8 ft for one gate valve. What is the developed (equivalent) length used for friction-loss calculations?

• a.98 ft
• b.110 ft✓
• c.122 ft
• d.134 ft

Developed length = straight pipe + fitting equivalents = 90 + (4 x 3) + 8 = 90 + 12 + 8 = 110 ft. Friction loss is calculated on this equivalent length, not on the measured pipe alone, because fittings behave like extra pipe. Ignoring fittings understates the loss and can undersize the pipe.UPC §610.0

Available pressure for friction is 30 psi and the total developed length of the water line is 150 ft. What is the maximum allowable uniform friction loss per 100 ft of pipe (the value used to enter a sizing chart)?

• a.10 psi/100 ft
• b.15 psi/100 ft
• c.20 psi/100 ft✓
• d.30 psi/100 ft

Allowable loss per 100 ft = (available pressure / developed length) x 100 = (30 / 150) x 100 = 20 psi per 100 ft. This uniform-loss figure is the design value you carry across the sizing chart against the flow to pick a pipe size. Spreading the whole 30 psi over 150 ft gives the per-100-ft rate.UPC §610.0

A horizontal drainage branch carries the following drainage fixture units (DFU): 4 water closets at 4 DFU, 4 lavatories at 1 DFU, and 2 kitchen sinks at 2 DFU. What is the total DFU load carried by the branch?

• a.18 DFU
• b.28 DFU
• c.32 DFU
• d.24 DFU✓

Total = (4 x 4) + (4 x 1) + (2 x 2) = 16 + 4 + 4 = 24 DFU. This total is read against the horizontal fixture branch column of the drain sizing table to select the pipe. Because water closets are present, the branch can be no smaller than 3 in regardless of the DFU count.UPC §703.0

A horizontal fixture branch carries 30 DFU. The sizing table lists these branch capacities: 2 in = 6 DFU, 3 in = 20 DFU, 4 in = 160 DFU. Two of the fixtures are water closets. What is the minimum pipe size for the branch?

• a.4 in✓
• b.2 in
• c.2-1/2 in
• d.6 in

The 30 DFU load exceeds the 20 DFU capacity of a 3 in branch, so the table already forces the next size, 4 in (160 DFU). The presence of water closets independently forbids anything smaller than 3 in, but here the load alone requires 4 in. Always pick the smallest size whose capacity equals or exceeds the load.IPC §710.1

A 4 in building drain is run at the code minimum slope of 1/8 in per foot over a developed length of 96 ft. What is the total fall from the upstream end to the downstream end?

• a.6 in
• b.12 in✓
• c.9 in
• d.24 in

Fall = slope x length = 1/8 in/ft x 96 ft = 12 in, or 1 ft. Pipe 3 in and larger uses 1/8 in per foot as the minimum. Using 1/4 in per ft (which applies only to pipe 2-1/2 in and smaller) would wrongly double the answer to 24 in.IPC §704.1

A 2 in horizontal fixture drain must fall a total of 5 in over its run at the code minimum slope of 1/4 in per foot. How long is the run?

• a.24 ft
• b.40 ft
• c.20 ft✓
• d.60 ft

Length = fall / slope = 5 in / (1/4 in per ft) = 5 x 4 = 20 ft. Pipe 2-1/2 in and smaller uses the steeper 1/4 in per foot minimum. Dividing by 1/8 in per ft would incorrectly give 40 ft, the rate reserved for 3 in and larger pipe.IPC §704.1

A vent must be sized at not less than one-half the diameter of the drain it serves, and never smaller than 1-1/4 in. What is the minimum vent size for a 3 in drain?

• a.1-1/4 in
• b.2 in
• c.3 in
• d.1-1/2 in✓

Half of 3 in is 1-1/2 in, which is larger than the 1-1/4 in floor, so the minimum vent is 1-1/2 in. The half-diameter rule sets the size and the 1-1/4 in minimum only governs very small drains. A 4 in drain by the same rule would need at least a 2 in vent.UPC §904.1

A 2 in trap arm connects a fixture trap to its vent. The code limits the fall in a trap arm to no more than one pipe diameter between the trap weir and the vent. What is the maximum allowable fall in this trap arm?

• a.2 in✓
• b.1/2 in
• c.1 in
• d.1-1/2 in

The maximum fall equals one pipe diameter, and for a 2 in arm that is 2 in. If the arm falls more than one diameter, the vent opening drops below the crown weir and the trap can be self-siphoned. This limit is separate from and in addition to the maximum developed-length limit for the arm.UPC §906.1

A natural-gas furnace is rated at 120,000 BTU/hr. Natural gas has a heating value of about 1,000 BTU per cubic foot. What gas volume flow, in cubic feet per hour (cfh), must the piping deliver?

• a.12 cfh
• b.120 cfh✓
• c.60 cfh
• d.1,200 cfh

cfh = BTU/hr load / heating value = 120,000 / 1,000 = 120 cfh. Because natural gas is roughly 1,000 BTU per cubic foot, the cfh figure is simply the input in thousands of BTU. This cfh value is what you carry into the gas-pipe sizing table with the longest run.UPC §610.0

A house has three natural-gas appliances: a 100,000 BTU/hr furnace, a 40,000 BTU/hr water heater, and a 65,000 BTU/hr range. Using 1,000 BTU per cubic foot, what total cfh must the gas meter and main serve?

• a.165 cfh
• b.240 cfh
• c.205 cfh✓
• d.2,050 cfh

Total input = 100,000 + 40,000 + 65,000 = 205,000 BTU/hr, and at 1,000 BTU per cubic foot that is 205 cfh. The whole-house demand sizes the meter and the main from the meter to the first tee. Individual branches are then sized for the appliance each one serves.IFGC §402.4

A propane appliance is rated at 150,000 BTU/hr. Propane has a heating value of about 2,500 BTU per cubic foot. What is the required gas flow in cfh?

• a.15 cfh
• b.30 cfh
• c.150 cfh
• d.60 cfh✓

cfh = 150,000 / 2,500 = 60 cfh. Propane packs about 2.5 times the energy of natural gas per cubic foot, so the same BTU load needs far fewer cubic feet, which is why propane piping is often smaller than natural-gas piping for the same appliance.IFGC §402.4

A gas branch must deliver 55 cfh of natural gas over a 60 ft longest length. The sizing table for that length gives these capacities: 1/2 in = 42 cfh, 3/4 in = 88 cfh, 1 in = 165 cfh. What is the minimum pipe size?

• a.3/4 in✓
• b.1/2 in
• c.1 in
• d.1-1/4 in

The 55 cfh demand exceeds the 42 cfh capacity of 1/2 in pipe at 60 ft, so the next size up, 3/4 in (88 cfh), is required. Gas pipe is sized on the longest length from the meter to the most remote outlet, applied to every section. A 1/2 in pipe would be overloaded and starve the appliance.IFGC Table 402.4

Using the rational method Q = 0.0104 x A x i, where A is the projected roof area in square feet and i is the rainfall rate in in/hr, what is the design storm flow for a 6,000 ft^2 roof at a 3 in/hr rainfall rate?

• a.62 gpm
• b.187 gpm✓
• c.124 gpm
• d.312 gpm

Q = 0.0104 x 6,000 x 3 = 187 gpm. The 0.0104 factor converts one inch per hour of rain over one square foot into gpm. This design flow is then read against the vertical leader and horizontal storm-drain tables to size the conductors.UPC §1101.0

A roof drains 4,800 ft^2 at a design rainfall rate of 4 in/hr. Using Q = 0.0104 x A x i, what is the storm design flow?

• a.150 gpm
• b.250 gpm
• c.200 gpm✓
• d.320 gpm

Q = 0.0104 x 4,800 x 4 = 199.7 gpm, about 200 gpm. Doubling the rainfall rate doubles the flow for the same roof, which is why the local 100-year rainfall intensity is critical to storm sizing. This flow selects the leader and storm drain sizes.UPC §1101.0

A storm sizing table is published at 4 in/hr. A vertical leader lists a capacity of 4,600 ft^2 of roof at 4 in/hr. If the local design rate is only 2 in/hr, what roof area can that same leader serve?

• a.2,300 ft^2
• b.4,600 ft^2
• c.18,400 ft^2
• d.9,200 ft^2✓

Capacity in area is inversely proportional to rainfall rate, so halving the rate from 4 to 2 in/hr doubles the allowable area: 4,600 x (4/2) = 9,200 ft^2. Conductor capacity is fixed in gpm, so a lighter storm lets the same pipe drain more roof. Always adjust table areas to the local rainfall rate.UPC §1101.0

A booster pump must deliver 40 gpm and raise pressure by 45 psi. Using approximate water horsepower WHP = (gpm x psi) / 1,714, what is the water horsepower required (before pump efficiency)?

• a.1.05 hp✓
• b.0.5 hp
• c.2.1 hp
• d.4.2 hp

WHP = (40 x 45) / 1,714 = 1,800 / 1,714 = 1.05 hp. This is the ideal hydraulic power; the actual motor must be larger to account for pump efficiency, often around 60 to 70 percent. Dividing the delivered fluid power by efficiency gives the brake horsepower to specify.UPC §610.0

A booster pump adds 52 psi. What is the equivalent head, in feet, that the pump develops? Use 1 psi = 2.31 ft of head.

• a.52 ft
• b.120 ft✓
• c.78 ft
• d.231 ft

Head = psi x 2.31 = 52 x 2.31 = 120 ft. The 2.31 factor is the reciprocal of the 0.433 psi-per-foot relationship and converts pressure to the equivalent vertical column of water. Pump curves are usually plotted in feet of head, so this conversion is needed to read them.UPC §610.0

A recirculation loop must deliver 6 gpm of hot water. Using v = 0.408 x Q / d^2, what inside diameter keeps the velocity at about 2 ft/s to limit erosion of the continuously circulated hot line?

• a.0.75 in
• b.1.50 in
• c.1.10 in✓
• d.2.00 in

Solve for d: d = sqrt(0.408 x Q / v) = sqrt(0.408 x 6 / 2) = sqrt(1.224) = 1.11 in. Hot recirculation lines are held to a low velocity, near 2 to 3 ft/s, because constant flow at high velocity erodes copper. A larger diameter lowers velocity for the same flow.UPC §610.0

A water main runs 250 ft of developed length and the friction chart shows a loss of 6 psi per 100 ft at the design flow. How much pressure is lost to friction over the full run?

• a.6 psi
• b.9 psi
• c.25 psi
• d.15 psi✓

Friction loss = (loss per 100 ft) x (length / 100) = 6 x (250 / 100) = 6 x 2.5 = 15 psi. Friction loss scales directly with developed length, so long runs consume a large share of the pressure budget. This 15 psi must be subtracted from supply along with elevation before checking residual.UPC §610.0

A sewage ejector pump must handle a building with a discharge of 30 gpm against 18 ft of total head. Using WHP = (gpm x head in ft) / 3,960, what is the water horsepower?

• a.0.14 hp✓
• b.0.07 hp
• c.0.10 hp
• d.0.27 hp

WHP = (30 x 18) / 3,960 = 540 / 3,960 = 0.136 hp, about 0.14 hp. The 3,960 constant is used when head is expressed in feet rather than psi. As with any pump, the motor is oversized above this by dividing by the pump efficiency.IPC §712.0

A sewage sump receiving a peak inflow of 45 gpm is served by a pump that discharges 90 gpm when running. If useful storage between pump-on and pump-off is 30 gallons, how long does one pump-down cycle (running time) last during peak inflow?

• a.30 s
• b.40 s✓
• c.60 s
• d.90 s

While running, the net removal rate is pump output minus inflow = 90 - 45 = 45 gpm. Draw-down time = storage / net rate = 30 / 45 = 0.667 min = 40 s. Sizing sump volume this way limits motor starts per hour to protect the pump; too small a volume causes short-cycling.IPC §712.4

Two demand loads combine on a main: branch A carries 40 WSFU and branch B carries 60 WSFU. If the Hunter demand chart converts 100 WSFU (predominantly flush tanks) to about 44 gpm, what design flow sizes the main?

• a.22 gpm
• b.33 gpm
• c.44 gpm✓
• d.104 gpm

You add the fixture units first, 40 + 60 = 100 WSFU, then convert the combined total to gpm on the demand curve, giving about 44 gpm. You must never convert each branch to gpm and add the flows, because the Hunter curve already accounts for the low probability of simultaneous use, and adding gpm overstates demand.UPC §610.0

A meter and service must supply 44 gpm. The static supply is 68 psi, elevation loss is 20 psi, the meter loss is 8 psi, and the most remote fixture needs 15 psi. How much pressure remains for pipe friction?

• a.18 psi
• b.33 psi
• c.43 psi
• d.25 psi✓

Available for friction = static - elevation - meter - fixture requirement = 68 - 20 - 8 - 15 = 25 psi. This leftover is spread over the developed length to set the allowable friction rate per 100 ft. If friction loss at the chosen pipe size exceeds 25 psi, the pipe must be enlarged.UPC §610.0

An individual vent serving a lavatory is 1-1/2 in. The vent sizing table limits a 1-1/2 in vent to a maximum developed length of 150 ft at the fixture-unit load carried. If the vent run is 90 ft, is the vent acceptable and why?

• a.Yes, 90 ft is within the 150 ft limit✓
• b.No, it exceeds the length limit
• c.No, the vent must equal the drain size
• d.Yes, vents have no length limit

A 90 ft developed vent length is within the 150 ft maximum allowed for a 1-1/2 in vent at that load, so the vent is acceptable. Vent sizing depends on both the fixture-unit load and the total developed length; exceeding either forces a larger vent. Vents do have length limits, so the no-limit answer is wrong.IPC §906.2

A hot-water line is limited to a maximum velocity of 5 ft/s. Using v = 0.408 x Q / d^2, what is the maximum flow through a pipe with a 0.785 in inside diameter (nominal 3/4 in type L copper)?

• a.4.9 gpm
• b.7.6 gpm✓
• c.11.3 gpm
• d.15.1 gpm

Q = v x d^2 / 0.408 = 5 x (0.785)^2 / 0.408 = 5 x 0.616 / 0.408 = 7.55 gpm, about 7.6 gpm. Hot water is held to a lower 5 ft/s than cold water because higher temperature accelerates erosion-corrosion of copper. Exceeding this flow risks pinholing the line over time.UPC §610.0

A drainage stack sizing table lists a 4 in stack at 500 DFU maximum and a 3 in stack at 48 DFU maximum for the total to a stack. A stack collects 240 DFU. What is the minimum stack size, and what secondary rule also applies?

• a.3 in, and a cleanout is required
• b.3 in, and no water closets allowed
• c.4 in, and no size reduction downward✓
• d.6 in, and a vent stack is required

240 DFU exceeds the 48 DFU limit of a 3 in stack, so a 4 in stack is required, and a stack may never be reduced in size in the direction of flow as it descends. Downsizing a stack lower down would create a restriction that floods the branches above it. The stack must stay 4 in or larger to its base.IPC §710.1

A wet-vented bathroom group discharges 5 DFU to a wet vent section. Wet vent capacity limits are 1-1/2 in = 1 DFU, 2 in = 4 DFU, and 3 in = 6 DFU. What is the minimum wet vent size?

• a.1-1/2 in
• b.2 in
• c.4 in
• d.3 in✓

The 5 DFU load exceeds the 4 DFU capacity of a 2 in wet vent, so the next size, 3 in (6 DFU), is required. A wet vent uses an oversized drain to also serve as a vent, but only up to the fixture-unit cap for its size. Exceeding the cap forces a larger wet vent or separate venting.UPC §908.0

A circuit vent serves a battery of floor-mounted fixtures on a horizontal branch. What is the maximum number of fixtures a single circuit vent may serve, and where is the vent taken off?

• a.Eight, between the two most upstream fixtures✓
• b.Two, downstream of the last fixture
• c.Four, at the last fixture
• d.Ten, at the base of the stack

A circuit vent may serve up to eight fixtures on a horizontal branch, and the vent is taken off the branch between the two most upstream fixtures. A relief vent is added when the branch also receives discharge from upper floors. Water closets in batteries are a classic circuit-vent application.IPC §911.0

A commercial kitchen installs a grease interceptor rated for a flow of 35 gpm. Using the common rule that grease capacity in pounds is about twice the gpm rating, what nominal grease retention does this unit provide, and what is its purpose?

• a.35 lb, to trap sand
• b.70 lb, to intercept fats, oils and grease✓
• c.150 lb, to neutralize acid
• d.300 lb, to settle solids

A hydromechanical grease interceptor rated at 35 gpm provides roughly 70 lb of grease retention and its purpose is to intercept fats, oils, and grease before they enter and clog the building drain and public sewer. Interceptors are sized by fixture flow and required retention. Sand interceptors and acid neutralizers serve entirely different wastes.IPC §1003.4

A building drain serving fixtures below the elevation of the public sewer manhole must be protected against sewer backflow. Which device is required on that low-level drainage?

• a.An air gap
• b.A vacuum breaker
• c.A backwater valve✓
• d.An air admittance valve

Fixtures below the next upstream manhole rim are subject to backflow from a surcharged sewer and must discharge through a backwater valve, which closes when flow reverses. Fixtures above that level must not drain through the backwater valve so they are not blocked when it closes. Vacuum breakers and air gaps protect potable water, not drainage.UPC §710.0

A basement floor drain sits 3 ft below the elevation of the upstream sewer manhole cover. During a main surcharge, how much backpressure head could push up through that drain if unprotected, using 0.433 psi per foot?

• a.0.43 psi
• b.3.0 psi
• c.13 psi
• d.1.3 psi✓

Backpressure = depth below the surcharge level x 0.433 = 3 x 0.433 = 1.3 psi. Even a modest 3 ft head can flood a basement, which is why fixtures below the manhole rim require a backwater valve. The valve seats against reverse flow while still passing normal drainage.UPC §710.0

A hot-water recirculation loop loses 4,000 BTU/hr to the piping. The pump must circulate enough water so it cools only 20 F while replacing that loss. Using Q (gpm) = BTU/hr / (500 x delta-T), what recirculation flow is required?

• a.0.4 gpm✓
• b.0.2 gpm
• c.0.8 gpm
• d.2.0 gpm

Q = 4,000 / (500 x 20) = 4,000 / 10,000 = 0.4 gpm. The 500 factor is the heat capacity of water in BTU per hour per gpm per degree F. A low recirculation flow is enough to offset standby losses and keep hot water instantly available without eroding the piping.UPC §608.0

A temperature-and-pressure (T and P) relief valve on a water heater is rated to open at 210 F and 150 psi. What is the required disposition of its discharge pipe?

• a.Trapped and vented to the drain
• b.Run full-size to within 6-24 in of the floor with no trap or valve✓
• c.Reduced one size and terminated outside above grade
• d.Connected to the recirculation return

The T and P discharge must be full pipe size, contain no valves or traps, and terminate 6 to 24 in above the floor or an approved receptor so a discharge is visible and cannot be blocked. A trap would hold water and corrode; a valve could be shut and defeat the safety device. This prevents a tank from becoming a pressure vessel.UPC §608.3

A closed water-supply system (with a backflow preventer or check valve at the meter) has a water heater but no expansion control. As the water heats and expands, what device is required to prevent pressure buildup?

• a.A pressure-reducing valve
• b.A vacuum breaker
• c.A thermal expansion tank✓
• d.A backwater valve

A closed system cannot push expanded hot water back into the main, so a thermal expansion tank (or similar expansion control) is required to absorb the volume increase and prevent the T and P valve from weeping. Water expands roughly 2 to 3 percent between cold and hot, which spikes pressure in a sealed system. The expansion tank is charged to system static pressure.UPC §608.5

A reduced-pressure principle (RP) backflow assembly is installed on a service subject to a health hazard. How must it be installed with respect to the ground and drainage?

• a.Below grade in a valve box
• b.Flush to a wall with the relief plugged
• c.Any orientation as long as it is accessible
• d.Above grade with the relief port able to discharge to atmosphere✓

An RP assembly must be installed above grade with an air gap below its relief port so the relief can discharge freely and be observed; it may not be submerged in a pit that could flood the relief. RP assemblies protect against both backsiphonage and backpressure of high-hazard (health) cross connections. Submerging or plugging the relief defeats the protection.UPC §603.0

A lawn irrigation system with chemical injection is connected to a potable supply. This is classified as a high health hazard under backpressure and backsiphonage conditions. Which backflow assembly is required?

• a.Reduced-pressure principle assembly✓
• b.Atmospheric vacuum breaker
• c.Dual check valve
• d.Hose-bibb vacuum breaker

Chemical injection creates a high health hazard and can occur under backpressure, so a reduced-pressure principle assembly is required because it protects against both backpressure and backsiphonage of a health hazard. An atmospheric vacuum breaker only protects against backsiphonage and cannot be under continuous pressure. Dual checks are for low-hazard applications only.UPC §603.0

An air gap is used to protect a potable outlet discharging over a flood-level rim. What is the minimum air gap for an effective opening of 1 in diameter, using the rule of two times the effective opening?

• a.1 in
• b.2 in✓
• c.1-1/2 in
• d.4 in

The minimum air gap is twice the effective opening diameter, so 2 x 1 in = 2 in, and never less than 1 in in any case. The air gap is the most reliable backflow protection because it is a physical break that cannot fail mechanically. Near a wall the multiplier increases to three times the opening.UPC §603.4.6

In a medical gas piping system, oxygen and other medical gases must use tubing that is cleaned and labeled for the service. What copper tubing is required for medical gas distribution?

• a.Type M copper, brazed
• b.PVC schedule 40
• c.Type L or K copper, cleaned for oxygen service, brazed with joints purged with nitrogen✓
• d.Galvanized steel, threaded

Medical gas distribution uses type L or K copper specifically cleaned and capped for oxygen service, brazed with a nitrogen purge to prevent internal oxide (copper scale) formation. The nitrogen purge keeps the interior clean so particles cannot enter patient equipment. Type M, plastic, and threaded steel are all prohibited for medical gas.NFPA 99

Medical gas piping must be tested before use. Which test verifies the system is free of leaks and cross connections at operating conditions?

• a.Hydrostatic test only
• b.Smoke test
• c.Slump test
• d.A pressure/leak test plus a cross-connection (crossover) verification at 1.5 times working pressure✓

Medical gas systems require an initial pressure test, a standing-pressure leak test, and a cross-connection test to confirm each outlet delivers only its labeled gas, typically at 1.5 times the working pressure. A crossover between oxygen and another gas is a lethal hazard, so verification of correct gas at every station outlet is mandatory. Hydrostatic water testing is not used on gas.NFPA 99

A water softener regenerates using 8 lb of salt per cubic foot of resin for a 3 cubic foot resin bed. How much salt is consumed per regeneration cycle?

• a.24 lb✓
• b.8 lb
• c.16 lb
• d.32 lb

Salt per regeneration = dosage x resin volume = 8 lb/ft^3 x 3 ft^3 = 24 lb. The salt dose sets both the operating cost and the chloride load in the brine discharge, which must terminate through an air gap to a receptor. Higher salt doses increase capacity but also increase brine strength.UPC §611.0

A water heater is installed in a garage where flammable vapors may be present. What installation requirement addresses ignition of those vapors?

• a.Install a drip leg only
• b.Elevate the ignition source at least 18 in above the floor, or use an approved flammable-vapor-ignition-resistant unit✓
• c.Provide combustion air from the attic
• d.Add a second T and P valve

In a garage the pilot or ignition source must be at least 18 in above the floor, because gasoline vapors are heavier than air and pool low, unless the heater is listed as flammable-vapor-ignition-resistant (FVIR). Modern FVIR heaters are designed to prevent flame rollout from igniting floor-level vapors. Combustion air and relief valves do not address vapor ignition.UPC §507.0

A key difference between the Uniform Plumbing Code and the International Plumbing Code affects where the building drain becomes the building sewer. Which statement correctly pairs the codes?

• a.UPC 30 in, IPC 2 ft outside the wall
• b.Both use 5 ft outside the wall
• c.UPC 2 ft, IPC 30 in outside the wall✓
• d.Both use the property line

The UPC places the building drain to building sewer transition 2 ft outside the building wall, while the IPC uses 30 in. This boundary determines which code chapter, permit, and sometimes trade governs the piping. Knowing which model code the local jurisdiction adopted is essential before sizing or permitting.IPC vs UPC

An indirect waste pipe has an effective outlet opening of 3/4 in. The required air gap above the receptor flood-level rim is two times the effective opening. What is the minimum air gap?

• a.3/4 in
• b.1 in
• c.3 in
• d.1-1/2 in✓

Air gap = 2 x effective opening = 2 x 3/4 in = 1-1/2 in. The two-times rule sets the vertical separation that prevents waste from siphoning back into the indirect-wasted equipment. Near a wall the multiplier increases to three times the opening, which would give 2-1/4 in here.UPC §807.0

A sewage ejector serving fixtures below the sewer must discharge upward and then into the gravity building drain. What fitting protects the pump from gravity backflow when it is off?

• a.A check valve and a gate valve on the discharge✓
• b.A vacuum breaker
• c.A backwater valve on the inlet
• d.An air admittance valve

The pump discharge must have a check valve to prevent effluent from draining back into the sump when the pump stops, and a gate (shutoff) valve downstream of the check so the check can be serviced. Without the check valve, each cycle would refill the sump and short-cycle the pump. This is standard for ejector and sump pump discharge piping.UPC §710.13

A medical gas distribution system has a working pressure of 50 psi. The installation acceptance test is performed at 1.5 times the working pressure. What is the required test pressure?

• a.50 psi
• b.75 psi✓
• c.65 psi
• d.100 psi

Test pressure = 1.5 x working pressure = 1.5 x 50 = 75 psi. Medical gas piping is proven at 1.5 times its operating pressure to confirm joints and cross connections hold under a margin above service conditions. The standing-pressure and cross-connection tests are performed at this elevated pressure before the system is placed in service.NFPA 99

A thermostatic mixing valve blends 140 F stored water with 60 F cold water to deliver 110 F tempered water at a public lavatory. Using a mass balance, what fraction of the delivered flow must be the 140 F hot water?

• a.37.5%
• b.50.0%
• c.62.5%✓
• d.75.0%

Set 140f + 60(1 - f) = 110, so 80f = 50 and f = 0.625, or 62.5% hot water. The mixing valve holds storage at 140 F to control Legionella while blending down to a safe 110 F delivery. The cold fraction is the remaining 37.5%.UPC §608.0

A public toilet room fixture count is based on occupant load. If a code table requires one water closet per 40 occupants and the occupant load is 210, how many water closets are the minimum required?

• a.4
• b.5
• c.7
• d.6✓

Divide occupants by the ratio and round up: 210 / 40 = 5.25, which rounds up to 6 water closets. Fixture counts always round up because you cannot install a fraction of a fixture and the code sets a minimum. The occupant load is typically split by sex before applying the ratio in many codes.UPC §418.0

An oil-water separator serves three garage floor drains each rated at 20 gpm. The code allows sizing at 75 percent of the total connected flow for diversity. What design flow sizes the separator?

• a.45 gpm✓
• b.30 gpm
• c.60 gpm
• d.80 gpm

Total connected flow = 3 x 20 = 60 gpm, and at 75 percent diversity the design flow is 60 x 0.75 = 45 gpm. The separator must be large enough to let oil rise and separate before the water passes on to the sewer. Undersizing lets oil carry through and creates an explosion hazard in the sewer.UPC §705.0

A laboratory discharges 10 gpm of acid waste that must be diluted with water at a 5-to-1 ratio (water to acid waste) before it enters the neutralizing tank. How much dilution water is required?

• a.10 gpm
• b.50 gpm✓
• c.25 gpm
• d.60 gpm

Dilution water = ratio x waste flow = 5 x 10 = 50 gpm. Diluting acid waste before the neutralizing tank lowers the concentration so the tank can raise the pH into the permitted discharge range. The combined flow of 60 gpm then passes to the neutralizer before the sanitary sewer.UPC §814.0

An air admittance valve (AAV) is used to vent an island fixture. Which condition must be met for the AAV to be acceptable?

• a.It may serve as the only vent for the building
• b.It may be buried in the wall permanently
• c.It must be located in a ventilated, accessible space and at least 4 in above the horizontal branch drain✓
• d.It replaces the need for any trap

An AAV must be installed in a ventilated, accessible location and rise at least 4 in above the horizontal branch drain it serves, and it can never be the sole vent for a building. It admits air to relieve negative pressure but seals against sewer gas escaping. At least one vent must still open to atmosphere for the overall system.IPC §917.0

A boiler feed with chemical treatment connects to potable water. This is a high-hazard connection. Which assembly protects the potable supply against backpressure from the pressurized boiler?

• a.Atmospheric vacuum breaker
• b.Dual check valve
• c.Hose-bibb vacuum breaker
• d.Reduced-pressure principle backflow assembly✓

A chemically treated boiler is a high health hazard under backpressure, so a reduced-pressure principle assembly is required because it alone protects against backpressure of a health hazard. Atmospheric vacuum breakers cannot be used under continuous pressure or against backpressure. The RP relief port provides a visible, fail-safe break.UPC §603.5.7

A single-family residence has an underground lawn sprinkler system with no chemical injection and no downstream pressure. This is a low-hazard, backsiphonage-only condition. Which is the minimum acceptable backflow device?

• a.Atmospheric vacuum breaker installed at least 6 in above the highest head✓
• b.Reduced-pressure principle assembly
• c.Air gap only
• d.No device required

For a residential sprinkler with no chemicals and no backpressure, an atmospheric vacuum breaker installed at least 6 in above the highest downstream head (sprinkler) is the minimum acceptable device. It protects only against backsiphonage and must not be under continuous pressure for more than 12 hours. Adding chemicals or downstream pumps would escalate the requirement to an RP assembly.UPC §603.0

A water heater relief-valve discharge, softener drain, and condensate line all terminate at the same floor-level receptor. What must be true of the receptor and each connection?

• a.All lines hard-connected
• b.Each discharges by air gap into the receptor, which is trapped and vented✓
• c.Lines share one trap
• d.Lines terminate below the receptor rim

Each indirect line must discharge through an air gap into the receptor, and the receptor itself must be trapped and vented so it drains to the sanitary system without allowing sewer gas back. The air gaps keep the potable-related discharges from cross connecting. Hard connections or below-rim terminations would defeat the required physical break.UPC §501.0

A gas system serves a longest run of 80 ft to the most remote appliance, which needs 90 cfh. The table for 80 ft gives 3/4 in = 82 cfh and 1 in = 155 cfh. What size serves that final branch?

• a.1/2 in
• b.3/4 in
• c.1 in✓
• d.1-1/4 in

At the 80 ft longest length, 90 cfh exceeds the 82 cfh capacity of 3/4 in pipe, so 1 in (155 cfh) is required. In the longest-length method every section of the system is sized using the single longest run, not each section's own length. This conservative method avoids undersizing under worst-case simultaneous demand.IFGC §402.4

An appliance room contains a 120,000 BTU/hr furnace and a 40,000 BTU/hr water heater drawing combustion air from inside. Using the rule of 50 cubic feet of room volume per 1,000 BTU/hr for indoor air, what minimum room volume is required?

• a.4,000 ft^3
• b.12,000 ft^3
• c.16,000 ft^3
• d.8,000 ft^3✓

Total input = 120,000 + 40,000 = 160,000 BTU/hr, and at 50 ft^3 per 1,000 BTU/hr the room needs 160 x 50 = 8,000 ft^3. If the space is smaller than this, combustion air must be brought in from outdoors through sized openings. Adequate combustion air prevents oxygen depletion and dangerous incomplete combustion.IFGC §304.0

Combustion air is taken from outdoors using two openings, one high and one low. The rule is 1 square inch of free area per 4,000 BTU/hr for each opening when using direct outdoor openings. For a 200,000 BTU/hr appliance load, what is the minimum free area of each opening?

• a.50 in^2✓
• b.25 in^2
• c.40 in^2
• d.100 in^2

Each opening = load / 4,000 = 200,000 / 4,000 = 50 in^2 of free area. When two vertical (direct outdoor) openings are used, each is sized at 1 square inch per 4,000 BTU/hr, one within 12 in of the top and one within 12 in of the bottom. Louvers reduce free area, so gross opening size must be increased to compensate.IFGC §304.6

Appliance venting is classified by category. A Category I appliance is best described as which of the following?

• a.Positive vent pressure, non-condensing
• b.Negative vent pressure, non-condensing (draft hood or fan-assisted, standard flue)✓
• c.Positive vent pressure, condensing
• d.Negative vent pressure, condensing

A Category I appliance operates with a non-positive (negative) vent pressure and a non-condensing flue, so it uses a conventional type B vent or masonry chimney with natural or fan-assisted draft. Category IV is positive-pressure and condensing, requiring sealed plastic venting. Matching the vent material and pressure rating to the category is essential to avoid condensation damage or spillage.IFGC §503.0

A high-efficiency condensing furnace exhausts cool, wet flue gas under positive pressure. What venting category and material are appropriate?

• a.Category I, type B vent
• b.Category II, single-wall steel
• c.Category IV, listed PVC/CPVC sealed vent✓
• d.Category III, masonry chimney

A condensing furnace is Category IV, operating at positive vent pressure with condensing (wet, acidic) flue gas, and it requires a listed, sealed PVC or CPVC vent that resists corrosion and holds pressure. A type B vent or masonry chimney would corrode and leak flue products. The condensate must also be drained and often neutralized.IFGC §503.0

Corrugated stainless steel tubing (CSST) must be electrically bonded to reduce the risk of arcing from a lightning-induced surge. Where is the bonding clamp attached?

• a.To the CSST jacket only
• b.To the appliance cabinet only
• c.To the gas meter body only
• d.To a rigid pipe or CSST fitting ahead of the first downstream CSST, bonded to the grounding electrode system✓

CSST is bonded by clamping to a rigid gas pipe component or approved fitting and connecting to the building grounding electrode system, typically with a minimum 6 AWG copper conductor. Bonding drains induced energy so a lightning surge does not perforate the thin CSST wall by arcing. Standard equipment grounding alone does not satisfy the dedicated CSST bonding requirement.IFGC §310.0

A newly installed natural-gas piping system is pressure tested before being placed in service. A common test is 3 psi (or 1.5 times working pressure, whichever is greater) held for a set duration. For low-pressure residential piping, what is a typical minimum test pressure and duration?

• a.3 psi for at least 10 minutes with no pressure drop✓
• b.1 psi for 5 minutes
• c.2 psi for 8 minutes
• d.10 psi for 30 seconds

A common acceptance test for residential gas piping is 3 psi held for at least 10 minutes (some jurisdictions require longer) with no observable pressure drop on the gauge. The test isolates appliances and uses air or inert gas, never the fuel gas itself. Any drop indicates a leak that must be found and repaired before gas is introduced.IFGC §406.4

During a gas pressure test, appliances and their regulators must be protected. What is the correct way to include or exclude appliances during a 10 psi test?

• a.Leave appliances connected to save time
• b.Isolate or disconnect appliances and their regulators, which are not rated for the test pressure✓
• c.Test only with the pilot lit
• d.Open all appliance valves fully

Appliance regulators and controls are not rated for elevated test pressures, so appliances must be isolated by closing the individual appliance shutoff and disconnecting the appliance, or valving it off, before the piping is pressurized. Testing through an appliance can rupture its regulator diaphragm. Only the fixed piping is subjected to the test pressure.IFGC §406.0

A propane branch must deliver a 75,000 BTU/hr appliance at a longest length of 40 ft. Propane is 2,500 BTU/ft^3. The 40 ft table lists 1/2 in = 45 cfh and 3/4 in = 95 cfh. What is the required cfh and minimum pipe size?

• a.18 cfh, 1/2 in
• b.24 cfh, 1/2 in
• c.30 cfh, 1/2 in✓
• d.30 cfh, 3/4 in

cfh = 75,000 / 2,500 = 30 cfh, which is within the 45 cfh capacity of 1/2 in pipe at 40 ft, so 1/2 in is adequate. Because propane carries 2.5 times the energy per cubic foot of natural gas, the required cfh and pipe size are smaller for the same BTU load. Always convert BTU to cfh using the correct heating value for the fuel.IFGC §402.4

A hospital adds a nitrogen line for surgical tools alongside oxygen and medical air. What single feature most prevents a fatal mix-up of these medical/industrial gases at the outlet?

• a.Color-coded tape
• b.Larger pipe
• c.Higher pressure
• d.Gas-specific (non-interchangeable) outlet and connector indexing, plus labeling✓

Medical gas station outlets use gas-specific, non-interchangeable indexing so a nitrogen connector cannot fit an oxygen outlet, backed by permanent labeling and color coding. The physical keying, not the label alone, is the primary defense against a lethal wrong-gas connection. A crossover test at commissioning confirms each outlet delivers only its intended gas.NFPA 99

A gas piping system must include a sediment trap (drip leg) ahead of certain appliances. What is the purpose and typical location of the drip leg?

• a.To catch moisture and debris before it enters the appliance control, downstream of the appliance shutoff✓
• b.To reduce pressure, at the meter
• c.To vent gas, at the appliance
• d.To bond the pipe, at the regulator

A sediment trap is a capped tee installed downstream of the appliance shutoff and ahead of the appliance control so moisture, scale, and debris drop into the leg instead of fouling the gas valve. It is required at most appliances except those specifically exempt such as ranges and clothes dryers in some codes. The leg must be accessible for cleaning.IFGC §408.0

Each gas appliance must have an accessible manual shutoff. Where must the appliance shutoff valve be located?

• a.Inside the appliance cabinet
• b.In the same room, within 6 ft of the appliance, upstream of the union and appliance connector✓
• c.At the meter only
• d.Anywhere in the building

The appliance shutoff must be in the same room as the appliance and within 6 ft of it, located upstream of the flexible connector and union so the appliance can be isolated for service. Placing it only at the meter would require shutting off the whole building. Accessibility without tools is required for emergencies.IFGC §409.0

Using the known-air-infiltration method, an appliance space of 10,000 ft^3 is credited with 0.35 air changes per hour of natural infiltration. How many cubic feet of infiltration air per hour does the space provide?

• a.350 cfh
• b.1,750 cfh
• c.3,500 cfh✓
• d.35,000 cfh

Infiltration air = volume x air changes per hour = 10,000 x 0.35 = 3,500 cubic feet per hour. This method credits only the air that leaks through a standard building, which in a tight modern house is often too little for the appliance load. When it is insufficient, dedicated outdoor combustion air must be supplied.IFGC §304.5

A gas appliance vent connector runs 12 ft horizontally to the chimney. Code requires a minimum upward rise of 1/4 in per foot toward the chimney. What total vertical rise must the connector have?

• a.1-1/2 in
• b.6 in
• c.12 in
• d.3 in✓

Rise = 1/4 in per ft x 12 ft = 3 in of upward slope from the appliance to the chimney. The rise keeps hot flue gas moving upward and prevents it from stalling and spilling back into the room. A level or downward-sloped connector defeats natural draft and can spill carbon monoxide.IFGC §503.6

A two-appliance system: appliance A needs 60 cfh and appliance B needs 40 cfh. Using the longest-length method, the section of pipe between the meter and the first tee carries what demand?

• a.100 cfh✓
• b.40 cfh
• c.60 cfh
• d.20 cfh

The common section upstream of the first tee carries the sum of all downstream demand, 60 + 40 = 100 cfh, and is sized for that total at the system's longest length. Each branch downstream of the tee is then sized for only the appliance it serves. Undersizing the common section starves both appliances during simultaneous use.IFGC §402.4

A leak test on an in-service gas line uses a soap-bubble or electronic method rather than a pressure drop. When is the bubble/electronic leak check the appropriate method?

• a.On new rough piping before drywall
• b.On existing pressurized piping and appliance connections that cannot be depressurized for a formal test✓
• c.Only on propane tanks
• d.Never on natural gas

A leak-detection solution or electronic sniffer is used to check joints and connections on piping and appliances that are already in service and under normal operating pressure, where a formal pressure-drop test is impractical. Bubbles at a joint reveal an escaping leak for immediate repair. A never open flame is used to check for gas leaks.IFGC §406.4

An elevated industrial gas line operates at 5 psi and is regulated down for appliances rated in inches of water column. Using 1 psi = 27.7 in water column, what is 5 psi expressed in inches of water column?

• a.27.7 in w.c.
• b.55.4 in w.c.
• c.138.5 in w.c.✓
• d.500 in w.c.

Inches of water column = psi x 27.7 = 5 x 27.7 = 138.5 in w.c. Appliance regulators are commonly set near 7 in w.c. for natural gas, so an elevated 138.5 in w.c. supply must be stepped down by a line-pressure regulator with overpressure protection. Mixing up psi and inches of water column is a common and dangerous error.IFGC §614.0

A jurisdiction charges a plumbing permit fee of a $75 base plus $12 per fixture. A project has 18 fixtures. What is the permit fee?

• a.$216
• b.$249
• c.$366
• d.$291✓

Fee = base + (per-fixture x count) = 75 + (12 x 18) = 75 + 216 = $291. Fixture-based fee schedules tie the permit cost to the scope of work, so an accurate fixture count is needed both for the fee and for sizing. Forgetting the base fee gives the too-low $216.UPC §103.0

During construction the plumbing is inspected in stages. Which sequence correctly orders the common inspections?

• a.Underground (below-slab), rough-in (top-out), then final✓
• b.Final, rough, underground
• c.Rough, final, underground
• d.Final, underground, rough

Inspections follow the work: underground or below-slab piping is inspected and tested before it is covered, then the rough-in or top-out after walls are piped but before they are closed, and finally the final after fixtures are set. Each stage must pass before the next is covered. Skipping a stage forces uncovering completed work.UPC §103.5

A DWV rough-in is water tested. The test fills the system with water to create a 10 ft head at the lowest point. Using 0.433 psi per foot, what pressure does a 10 ft water column exert at the base, and what does the test prove?

• a.2.3 psi, that vents are clear
• b.4.33 psi, that joints hold under head with no leaks✓
• c.10 psi, that slope is correct
• d.0.43 psi, that traps are set

A 10 ft column exerts 10 x 0.433 = 4.33 psi at the base, and holding that head for the required time proves every joint below the fill line is watertight. The water (or an equivalent air) test is applied before the piping is concealed. A dropping level indicates a leak to be located and repaired.UPC §712.0

On an isometric, a 45-degree offset must clear an obstruction, producing a 24 in vertical offset. For a 45-degree fitting the travel (diagonal) length equals the offset times 1.414. What is the travel length of pipe between the two fittings?

• a.17.0 in
• b.24.0 in
• c.33.9 in✓
• d.48.0 in

Travel = offset x 1.414 = 24 x 1.414 = 33.9 in. For a 45-degree offset the vertical rise, horizontal run, and diagonal travel form a right triangle where the diagonal is 1.414 (the square root of 2) times the offset. This constant lets a plumber lay out offsets directly from the required rise.UPC §706.0

A plumbing job has a direct cost of $9,000. The contractor wants a 20 percent profit margin measured on the selling price (not markup on cost). Using price = cost / (1 - margin), what is the selling price?

• a.$10,800
• b.$11,000
• c.$13,500
• d.$11,250✓

Price = 9,000 / (1 - 0.20) = 9,000 / 0.80 = $11,250. A margin is figured on the selling price, so you divide by one minus the margin rather than simply adding 20 percent to cost. Adding 20 percent as a markup would give only $10,800 and miss the intended margin.UPC §103.0

A master plumber estimates a job. A takeoff lists 260 ft of 3/4 in copper, 40 ft of 1 in copper, and 18 fittings. What is the estimator computing at this stage?

• a.Material quantities from the drawings to price the job✓
• b.Labor only
• c.Permit fees only
• d.Inspection dates

A takeoff is the systematic counting and measuring of materials from the drawings, here pipe lengths and fitting counts, so unit prices can be applied to build the material estimate. Labor hours and overhead are added separately. An accurate takeoff is the foundation of a profitable, competitive bid.UPC §103.0

A plumbing contractor prices a job using material of $4,200 and labor of $6,000, then adds 15 percent overhead and 10 percent profit on the combined cost. What is the approximate bid price?

• a.$11,730
• b.$13,090✓
• c.$12,750
• d.$10,200

Base cost = 4,200 + 6,000 = 10,200. Add 15 percent overhead: 10,200 x 1.15 = 11,730. Add 10 percent profit: 11,730 x 1.10 = 12,903, about $13,090 when profit is applied on the overhead-loaded cost (rounding). Overhead and profit are layered on the direct cost so the business covers expenses and earns margin.UPC §103.0

A rough-in inspection fails because a required cleanout was omitted at the base of a stack. What is the master plumber's correct response?

• a.Cover the work and note it on the final
• b.Argue the cleanout is optional
• c.Install the cleanout and request a re-inspection before concealment✓
• d.Remove the stack entirely

The correct response is to install the missing cleanout as required and call for a re-inspection before the work is covered, because the code requires a cleanout at the base of each stack. Concealing a known deficiency risks failing the final and being ordered to open the wall. Corrective work plus re-inspection keeps the project compliant.UPC §103.5

A master plumber pulls a permit as the responsible party of record. What obligation does being the permit holder create?

• a.None after the permit is issued
• b.Only paying the fee
• c.Only supplying materials
• d.Responsibility that the work conforms to the approved plans and code, and that inspections are called✓

As permit holder the master plumber is responsible that the installed work matches the approved plans and the adopted code, and that each required inspection is requested at the proper stage. The permit ties the licensed professional to code compliance for that project. Failing these duties can result in penalties or license action.UPC §101.0

During plan review, minimum fixtures are checked against occupancy: the table requires one water closet per 50 occupants. The occupant load is 175. What is the minimum number of water closets required?

• a.4✓
• b.2
• c.3
• d.5

Divide and round up: 175 / 50 = 3.5, which rounds up to 4 water closets. Fixture counts always round up because you cannot install a fraction of a fixture and the table sets a minimum. The reviewer uses the occupant load and the ratio to verify the plan meets the minimum.UPC §418.0

A finished DWV system is given a final air test instead of water. The system is pressurized to 5 psi and must hold for 15 minutes. What is the pass criterion?

• a.Pressure may drop up to 2 psi
• b.No pressure drop (the gauge holds 5 psi) for the required time✓
• c.Any reading above 0 psi
• d.Pressure must rise

The air test passes only if the system holds the required 5 psi (about a 10 in mercury column equivalent) with no measurable drop for the full duration. A falling gauge indicates a leak that must be located and sealed. Air testing is common where a water test is impractical, such as in cold weather.UPC §712.2

A base permit fee is $290. The jurisdiction adds a 12 percent state surcharge on the permit fee plus a flat $40 plan-review fee. What is the total the contractor pays?

• a.$330.00
• b.$364.80
• c.$372.80✓
• d.$380.80

Total = (290 x 1.12) + 40 = 324.80 + 40 = $364.80. The surcharge applies only to the permit fee, then the flat plan-review fee is added afterward. Applying the 12 percent to the plan-review fee as well would overcharge the customer.UPC §103.4

A change order adds two lavatories at 1 DFU each and one water closet at 4 DFU to a branch. By how many drainage fixture units does the branch load increase, so plan review can confirm the pipe is still adequate?

• a.4 DFU
• b.8 DFU
• c.10 DFU
• d.6 DFU✓

Added load = (2 x 1) + (1 x 4) = 2 + 4 = 6 DFU. A change order that adds fixtures raises the drainage fixture-unit load, so a plan revision must confirm the existing branch and stack still have capacity before the work is installed. If the new total exceeds the pipe capacity, the branch must be enlarged.UPC §703.0

An inspector finds that a backflow assembly was installed but never tested by a certified tester. What documentation closes this item?

• a.A passing test report from a certified backflow tester filed with the jurisdiction✓
• b.The installer's invoice
• c.A photo of the device
• d.The permit card alone

A newly installed backflow assembly must be tested by a certified backflow tester and the passing test report filed with the water purveyor or jurisdiction to prove it functions. The device's presence alone is not sufficient; performance must be verified and recorded. Annual re-testing is typically required thereafter.UPC §103.5

A master plumber must schedule labor. A job is estimated at 240 labor-hours and the crew provides 3 plumbers working 8-hour days. How many working days are needed?

• a.8 days
• b.10 days✓
• c.12 days
• d.15 days

Crew capacity per day = 3 plumbers x 8 hours = 24 labor-hours, so days = 240 / 24 = 10 working days. Estimating duration from labor-hours and crew size lets the contractor commit to a realistic schedule. Underestimating crew hours leads to missed completion dates and penalties.UPC §103.0

The adopted plumbing code and a manufacturer's installation instructions differ on the maximum length of a listed flexible water connector. Which controls, and what is the guiding principle?

• a.The code always wins even for listed products
• b.Ignore both and use judgment
• c.The more restrictive requirement governs; listed products must be installed per their listing where the code defers to it✓
• d.The instructions always win over any code

When a listed product's instructions and the code both apply, the more restrictive requirement governs, and the code generally requires listed products to be installed in accordance with their listing where it defers to the manufacturer. This protects the listing's validity while still meeting minimum code. The plumber documents which requirement was applied.UPC §101.0

An excavation for a sewer lateral is dug to a depth of 6 ft in Type C soil. At what depth does OSHA require protective systems (sloping, shoring, or a trench box) for workers entering the trench?

• a.Any depth over 2 ft
• b.Any depth over 3 ft
• c.Only over 10 ft
• d.5 ft or more (and any depth if a competent person sees a hazard)✓

OSHA requires cave-in protection for any trench 5 ft or deeper, and at any depth if a competent person identifies a hazard, so a 6 ft trench must be sloped, shored, or shielded. Trenches less than 5 ft may be exempt only if a competent person finds no cave-in potential. Soil type sets the required slope angle.29 CFR 1926.652

A trench is dug in Type C soil, which OSHA requires to be sloped at 1.5 to 1 (horizontal to vertical). For a trench 8 ft deep, how wide must the sloped opening be beyond the trench bottom on each side?

• a.12 ft each side✓
• b.8 ft each side
• c.4 ft each side
• d.16 ft each side

At 1.5 to 1, horizontal run = 1.5 x depth = 1.5 x 8 = 12 ft of slope on each side beyond the trench bottom. Type C is the least stable soil and requires the widest, flattest slope. Type A allows 0.75 to 1 and Type B 1 to 1, so knowing the soil class sets the excavation width.29 CFR 1926.652

A trench is 3 ft wide, 6 ft deep, and 40 ft long. How many cubic yards of spoil are generated (27 cubic feet per cubic yard), a figure needed to plan the 2 ft edge setback and haul-off?

• a.8.9 yd^3
• b.26.7 yd^3✓
• c.17.8 yd^3
• d.32.0 yd^3

Volume = 3 x 6 x 40 = 720 ft^3, and 720 / 27 = 26.7 cubic yards of spoil. This volume determines how much material must be set back at least 2 ft from the edge or hauled away so its surcharge load does not collapse the wall. Underestimating spoil crowds the edge and endangers workers.29 CFR 1926.651

A plumber must enter a sanitary sewer manhole to make a connection. This is a permit-required confined space. What atmospheric hazard is the leading concern, and what is required before entry?

• a.Excess oxygen only, no testing needed
• b.High humidity; wear a raincoat
• c.Toxic/flammable gases (hydrogen sulfide, methane) and oxygen deficiency; test the atmosphere and ventilate before and during entry✓
• d.Noise; wear earplugs

A sewer is a permit-required confined space where hydrogen sulfide, methane, and oxygen deficiency can be immediately dangerous, so the atmosphere must be tested for oxygen, flammables, and toxics and the space ventilated before and continuously during entry. An attendant, retrieval equipment, and a permit are required. Oxygen must be between 19.5 and 23.5 percent.29 CFR 1910.146

Before entry, a manhole 4 ft in diameter and 12 ft deep must be purged with 5 air changes. Using volume = 0.785 x diameter^2 x depth (about 151 ft^3), what total air volume must be moved to complete the purge?

• a.151 ft^3
• b.302 ft^3
• c.1,510 ft^3
• d.755 ft^3✓

Space volume = 0.785 x 4^2 x 12 = 0.785 x 16 x 12 = 151 ft^3, and 5 air changes require 5 x 151 = 755 ft^3 of air moved. Purging several air changes before entry clears hydrogen sulfide and methane and restores oxygen. The atmosphere is then re-tested in the order oxygen, flammable, toxic before anyone enters.29 CFR 1910.146

A plumber sets an extension ladder to reach a roof vent terminal. Using the 4-to-1 rule, how far should the base be from the wall if the ladder contacts the wall 16 ft up?

• a.4 ft✓
• b.2 ft
• c.8 ft
• d.16 ft

The 4-to-1 rule places the base 1 ft out for every 4 ft of working height, so 16 / 4 = 4 ft from the wall. This angle keeps the ladder from sliding out at the base or tipping back. The ladder should also extend at least 3 ft above the roof edge for a safe transition.29 CFR 1926.1053

Backflow prevention is a public-health safety requirement. What health event does a properly working backflow assembly prevent?

• a.Water hammer
• b.Contaminated water being drawn or pushed back into the potable supply✓
• c.Pipe corrosion
• d.Frozen pipes

Backflow protection prevents non-potable or contaminated water from entering the public drinking-water system through backsiphonage or backpressure, which has caused real disease outbreaks. The assembly is the barrier between a cross connection and the community's water. This is why high-hazard connections require the fail-safe reduced-pressure principle assembly.UPC §603.0

A master plumber must braze medical gas copper, which is hot work. What fire-safety precaution is required before starting?

• a.Only wear gloves
• b.Work faster to reduce exposure
• c.Obtain a hot-work permit, clear/cover combustibles, and post a fire watch with an extinguisher✓
• d.Open the gas valve to purge

Hot work such as brazing or soldering near combustibles requires a hot-work permit, removal or shielding of combustibles within the area, and a fire watch with an extinguisher during the work and for a period afterward. Sparks and heat can smolder in hidden materials and ignite after the crew leaves. This is standard for cutting, welding, and brazing.29 CFR 1926.352

A plumber replaces piping in a building constructed before 1978 and disturbs old painted surfaces and solder. What two legacy hazards require specific precautions?

• a.Asbestos and radon
• b.Silica and mold
• c.Formaldehyde and PCBs
• d.Lead (paint and old solder) and asbestos (pipe insulation and old sheet materials)✓

Pre-1978 buildings commonly contain lead paint and lead solder plus asbestos in pipe insulation, transite, and floor and sheet materials, both of which require specific handling, containment, and disposal rules. Lead solder was banned for potable use in 1986, and asbestos insulation must not be disturbed without proper controls. Both are serious long-term health hazards.EPA/OSHA lead

A 2,000 lb cast-iron section is lifted by a two-leg sling with each leg at 60 degrees from horizontal. Each leg carries (load / 2) / sin(60 degrees), with sin(60) about 0.866. What is the tension in each sling leg?

• a.1,155 lb✓
• b.1,000 lb
• c.1,414 lb
• d.2,000 lb

Each leg tension = (2,000 / 2) / 0.866 = 1,000 / 0.866 = 1,155 lb. As the sling angle drops toward horizontal, the leg tension climbs above the simple share of the load, which is why slings must be de-rated for angle. At a straight vertical lift each leg would carry only 1,000 lb.29 CFR 1926.251

During sewer confined-space work, the attendant outside must maintain what capability?

• a.Enter to help immediately without equipment
• b.Continuous communication with entrants and the ability to summon rescue without entering✓
• c.Leave to get tools as needed
• d.Perform the plumbing work

The attendant remains outside, keeps continuous communication with the entrants, monitors conditions, and summons trained rescue if needed, but does not enter the space to attempt rescue alone. Untrained would-be rescuers are a leading cause of confined-space fatalities. Non-entry retrieval systems allow rescue from outside.29 CFR 1910.146

A trench 10 ft deep in Type B soil is sloped at 1 to 1 (horizontal to vertical) on both sides. If the trench bottom is 3 ft wide, how wide is the excavation at the top?

• a.13 ft
• b.20 ft
• c.23 ft✓
• d.30 ft

Each side slopes back 1 x depth = 10 ft, so the top width = bottom + 2 x 10 = 3 + 20 = 23 ft. Type B soil requires a 1-to-1 slope, while less stable Type C requires 1.5 to 1 and would open even wider. The soil classification set by a competent person drives the excavation width and spoil space needed.29 CFR 1926.652

A trench 90 ft long and 6 ft deep requires egress so no worker travels more than 25 ft laterally to a ladder. A ladder placed mid-run protects 25 ft in each direction, or 50 ft total. What is the minimum number of ladders required?

• a.1 ladder
• b.3 ladders
• c.4 ladders
• d.2 ladders✓

Each ladder covers 50 ft of trench (25 ft of lateral travel on each side), so 90 ft / 50 ft = 1.8, which rounds up to 2 ladders. The 25 ft maximum travel ensures rapid exit if a wall fails or the atmosphere changes. Egress is required in any trench 4 ft or deeper and must extend above the top edge.29 CFR 1926.651

Before servicing a sewage ejector pump, the plumber must control hazardous energy. What procedure prevents the pump from starting during service?

• a.Lockout/tagout of the electrical disconnect, then verify zero energy✓
• b.Post a verbal warning
• c.Unplug only if convenient
• d.Work quickly between cycles

Lockout/tagout requires de-energizing the pump at its disconnect, applying a personal lock and tag, and verifying zero energy by attempting to start it before hands go near moving parts. An automatic float could start the pump unexpectedly and cause severe injury. Each worker applies their own lock.29 CFR 1910.147

On a California jobsite, a trench 5 ft or deeper in which workers will enter also requires a specific state permit beyond federal rules. What is that requirement?

• a.No permit is ever required in California
• b.A Cal/OSHA excavation/trench permit for trenches 5 ft or deeper that workers enter✓
• c.A federal permit only
• d.A permit only above 20 ft

Cal/OSHA requires a project or annual excavation permit for trenches 5 ft or more deep into which a person must descend, in addition to the cave-in protection rules. California often has requirements at least as strict as, and sometimes stricter than, federal OSHA. The competent person and daily inspections still apply.Cal/OSHA T8 1541

Cutting and grinding cast iron or concrete for a plumbing penetration generates respirable crystalline silica. What is the primary control the plumber should use?

• a.Work faster
• b.Open a window only
• c.Wet cutting or on-tool dust collection (engineering controls) plus respiratory protection as needed✓
• d.No control is needed for short tasks

The primary defense against respirable silica is engineering controls, using water (wet cutting) or a vacuum dust-collection shroud on the tool to suppress dust at the source, supplemented by respiratory protection when needed. Silica dust causes silicosis and is regulated with a strict exposure limit. Dry cutting without controls quickly exceeds the permissible exposure limit.29 CFR 1926.55

A demand of 62 gpm must be delivered with an available friction pressure of 24 psi over a 200 ft developed length. What is the allowable friction loss per 100 ft used to size the pipe?

• a.4 psi/100 ft
• b.8 psi/100 ft
• c.24 psi/100 ft
• d.12 psi/100 ft✓

Allowable loss per 100 ft = (available pressure / developed length) x 100 = (24 / 200) x 100 = 12 psi per 100 ft. This uniform rate is read against the 62 gpm demand on the friction chart to choose the smallest adequate pipe. Spreading the full 24 psi over the whole run gives the per-100-ft design value.UPC §610.0

An air gap at a lavatory faucet must be maintained above the flood-level rim. If the effective opening of the faucet is 1/2 in and the outlet is near a single wall, the required gap is three times the opening. What is the minimum air gap?

• a.1-1/2 in✓
• b.1 in
• c.2 in
• d.3 in

Near a single wall the multiplier is three times the effective opening, so 3 x 1/2 in = 1-1/2 in. Away from walls the standard multiplier is two times the opening, but a nearby wall disrupts the free fall of air and requires the larger gap. The air gap is the most reliable backflow protection because it is a fixed physical break.UPC §603.4.6

A natural-gas dryer needs 35,000 BTU/hr and a range needs 65,000 BTU/hr on a shared branch. Using 1,000 BTU/ft^3, what is the combined cfh the common branch must carry?

• a.35 cfh
• b.100 cfh✓
• c.65 cfh
• d.1,000 cfh

Combined input = 35,000 + 65,000 = 100,000 BTU/hr, and at 1,000 BTU per cubic foot that is 100 cfh on the shared branch. The common section always carries the sum of the downstream appliance demands. Each appliance's own connector is then sized for only its individual load.IFGC §402.4