Red Seal millwright practice exam
13 multiple-choice questions with worked solutions, grouped by the Major Work Activities on the exam. Pick an answer to check it. New questions are added each week as new videos come out.
How the real exam is weighted
The current Red Seal Industrial Mechanic (Millwright) exam has 135 questions. This is how they split across the six Major Work Activities (MWA), and how many questions this practice set has in each so far.
| Major Work Activity | Exam questions | Share | In this set | |
|---|---|---|---|---|
| A | Common occupational skills | 25 | 19 % | 3 |
| B | Rigging, hoisting/lifting and moving | 17 | 13 % | 8 |
| C | Mechanical power transmission | 32 | 24 % | 1 |
| D | Material handling / process systems | 24 | 18 % | 0 |
| E | Fluid power systems | 21 | 16 % | 1 |
| F | Maintenance, commissioning and decommissioning | 16 | 12 % | 0 |
Source: Red Seal, Industrial Mechanic (Millwright) exam information (November 2026). A new Red Seal Occupational Standard (RSOS) has been published and exams based on it are in development, with new examination weightings. Check the official page before you write.
A. Common occupational skills 25 of 135 exam questions (19 %)
An inch micrometer shows the 3 line on the sleeve, two small lines past it, and the thimble at 12. What is the reading?
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Answer: C. 0.362 in.
Sleeve 0.300 + (2 × 0.025) = 0.350; thimble 12 × 0.001 = 0.012; 0.350 + 0.012 = 0.362 in.
Each numbered line on the sleeve is 100 thousandths. So 3 is 0.300. Each small line is 25 thousandths. 2 of them is 0.050. The thimble is 1 thousandth per line, so 12 is 0.012. Add them up. 0.362 in.
A skipped the small lines. B counted only one. D counted three. The distractors are just miscounts, so slow down and count twice.
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A millwright has applied a personal lock to the disconnect for a pump motor. What must be done before the coupling guard is removed?
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Answer: B. Verify the motor cannot be started
Lock applied → verify zero energy (try to start at the local control, then return it to off) → then start work
The lock is on. But a lock on the wrong disconnect protects nobody. Before you start work, you verify that the energy is really isolated. On a motor, that usually means trying to start it at the local control, then putting the control back to off. Check for stored energy too.
A is a real step, but it comes before the lockout, not after. C doesn't isolate anything. And D is the job itself. When a question says before, or first, every option might be a real step. You're picking the one that comes next.
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Which statement about setting up a portable extension ladder is NOT correct?
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Answer: C. Set the base 1 ft out for every 2 ft up
The 4-to-1 rule: 1 ft (0.3 m) out for every 4 ft (1.2 m) up. One out for every two is far too flat.
3 of these are good practice. 1 ft out for every 4 ft up. That's the 4 : 1 rule. Extend it at least 1 m, or 3 ft, above the landing. And face the ladder when you climb.
One out for every two up puts the base much too far out, and the foot of the ladder can slip. The trap here is the word not. If you read too fast, you'll pick A, because it's the first true statement you see.
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B. Rigging, hoisting/lifting and moving 17 of 135 exam questions (13 %)
A 12,000 lb load is lifted with a two-leg bridle at 60° from horizontal. What is the tension in each leg?
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Answer: B. 6,930 lb
(12,000 ÷ 2) × 1.155 = 6,930 lb
Half the load is 6,000 lb. 60° gives a factor of 1.155. 6,000 times 1.155 is 6,930 lb.
If you picked A, you forgot the angle factor.
Learn the topic: Sling angle factor
A 3,000 kg load hangs from a two-leg bridle with a 120° included angle between the legs. What is the tension in each leg?
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Answer: C. 3,000 kg
120° included → 30° from horizontal → (3,000 ÷ 2) × 2.000 = 3,000 kg
Half of 120 is 60. 90 minus 60 is 30° from horizontal, and the factor is 2. 1,500 kg times 2 is 3,000 kg in each leg.
If you used 60° as the sling angle, you got B, and you'd have rigged a lift that's seriously overloaded.
Learn the topic: Sling angle factor
Two 10 ft slings lift a 6,000 lb load. The hook is 8 ft above the pick points. What is the tension in each leg?
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Answer: C. 3,750 lb
L ÷ H = 10 ÷ 8 = 1.250 → (6,000 ÷ 2) × 1.250 = 3,750 lb
Use L ÷ H. 10 divided by 8 gives a factor of 1.25. Half the load is 3,000 lb. 3,000 times 1.25 is 3,750 lb per leg.
If you got A, you flipped it and did H ÷ L. The factor is never less than one.
Learn the topic: Sling angle factor
A wire rope sling rated 4,000 lb vertical is used in a choker hitch with a 75° angle of choke. What is its adjusted capacity?
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Answer: D. 2,220 lb
4,000 × 0.75 (choker) × 0.74 (60°–89° choke) = 2,220 lb
Start with the choker rating. 75 % of 4,000 is 3,000 lb. 75° falls in the 60 to 89 row, so take 74 % of that. That's 2,220 lb.
C is the trap. That's applying the choke reduction but forgetting the choker rating itself.
Learn the topic: Sling angle factor
A 2,400 kg pump is lifted with a two-leg bridle at 45° from horizontal. What is the minimum rating required for each leg?
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Answer: B. 1,697 kg
(2,400 ÷ 2) × 1.414 = 1,697 kg
Half of 2,400 is 1,200 kg. 45° gives a factor of 1.414. 1,200 times 1.414 is 1,697 kg.
On the job, you'd go up to the next sling size. On the exam, they want the number. D is what you get if you forget to divide by the number of legs.
Learn the topic: Sling angle factor
A 4,000 lb load is lifted with a two-leg bridle. Each leg is 30° from vertical. What is the tension in each leg?
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Answer: B. 2,310 lb
30° from vertical → 60° from horizontal → (4,000 ÷ 2) × 1.155 = 2,310 lb
Convert first. 30 from vertical is 60 from horizontal. Half the load is 2,000 lb. 2,000 times 1.155 is 2,310 lb.
If you used 30° as the sling angle, you doubled it and got D. Same numbers, one word different, and the answer changes completely.
Learn the topic: Sling angle factor
A 20,000 lb load is lifted with a four-leg bridle at 60° from horizontal. Using the two-leg rule, what tension must each leg be rated for?
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Answer: C. 11,550 lb
Two legs carry: (20,000 ÷ 2) × 1.155 = 11,550 lb
With the two leg rule, you plan as if only two legs carry the load. Half of 20,000 is 10,000 lb. Times 1.155 gives 11,550 lb.
If you picked A, you divided by 4. That's the trap the question is built around.
Learn the topic: Sling angle factor
An 8,000 lb gearbox is lifted with a two-leg bridle. Each leg is 60° from vertical. What is the tension in each leg?
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Answer: C. 8,000 lb
60° from vertical → 90° − 60° = 30° from horizontal → (8,000 ÷ 2) × 2.000 = 8,000 lb
Convert first. 90 minus 60 is 30° from horizontal. The factor for 30 is 2. Half the load is 4,000 lb. 4,000 times 2 is 8,000 lb in each leg.
Now look at the wrong ones. A forgot the angle factor. B used 60 as if it were from horizontal. D forgot to divide by 2 legs. Every wrong answer is a real mistake. And 30° from horizontal is the practical minimum, so on the job this rig needs a second look.
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C. Mechanical power transmission 32 of 135 exam questions (24 %)
A deep groove ball bearing is stamped 6208. What is its bore diameter?
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Answer: B. 40 mm
Last two digits = bore code 08; codes 04 to 96: bore = code × 5 → 08 × 5 = 40 mm
The last two digits are the bore code. Here that's 08. For codes from 04 up to 96, you multiply the code by 5. 08 times 5 is 40 mm. The first digits tell you the bearing type and series, not the size of the bore.
A read the code as the size. C read the type and series digits as the size. D read the last three digits. Below 04, the codes are fixed values, so learn those four: 10, 12, 15 and 17 mm.
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E. Fluid power systems 21 of 135 exam questions (16 %)
A hydraulic cylinder has a 4 in bore. The relief valve is set at 1,500 psi. Can it push a 20,000 lbf load?
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Answer: C. No: it can push only about 18,850 lbf
A = π × 4² ÷ 4 = 12.566 in²; F = P × A = 1,500 × 12.566 ≈ 18,850 lbf < 20,000 lbf → it stalls at relief
Force is pressure times area. First, the area of the piston. π times the bore squared, divided by 4. 4² is 16. Times π, divided by 4, is 12.566 in². Now the force. 1,500 lb on every square inch, times 12.566, is about 18,850 lb. That's the most it can push before the relief opens. 20,000 is more than that.
The cylinder stalls and the oil goes over the relief. A squared the bore but forgot to divide by 4. B used the bore as if it were the radius. D multiplied by the bore, not the area. Area is where most hydraulics marks are lost. Work it out first, every time.
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How to use this practice exam
Answer each question before you open the solution. When you check, go through every option: why the right answer is right, and which mistake leads to each wrong one. For the full picture of how the exam works, read the Red Seal millwright exam guide.
Get the full program
CrewForge is building lessons, question banks and timed practice exams for every Red Seal millwright topic. Level 1 opens for early access in Winter 2027.
Exam preparation, not a procedure. On the job, check the tag, the manufacturer's data and your lift or work plan.