Ten 429A Machinist questions in the format of the Red Seal exam: four options, one correct answer, no trick questions. They are free to use, print and share. There is nothing to sign up for on this page.
01How to use these
The questions come from the TicketPrep 429A bank and are spread across the exam's work activities in the order and roughly the proportion the Red Seal Program publishes for this trade: A (performs common occupational skills), B (performs benchwork), D (machines using drill presses), E (machines using conventional lathes), F (machines using conventional milling machines), G (machines using precision grinding machines), H (machines using computer numerical control (CNC) machines). Answer each one before opening the explanation. Every explanation covers all four options, because on the real exam the wrong options are written to be plausible, and knowing why they are wrong is most of the skill.
- Pace. The exam allows four hours for 135 questions, so about 1 min 47 s per question. Try these at that pace.
- Print. The printed version shows the questions and the answer key without the explanations, so it works as a handout. Download PDF
- Mark. The pass mark is 70 percent on every Red Seal trade; seven of ten here is the same bar.
02The questions
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A punch of water-hardening tool steel must be hardened and tempered, and its finished diameter is toleranced ±0.0002 in. In what order should the work be done?
- Turn leaving stock, grind to size, then harden and temper.
- Turn leaving stock, harden and temper, then grind.
- Turn to finished size, then harden and temper.
- Turn leaving stock, temper, then harden and grind.
Show the answer
B. Correct. Remove the bulk of the metal while the steel is soft, leaving a small grinding allowance. Hardening brings distortion, scale and some decarburization, and tempering follows at once to relieve quench stress. Grinding last removes the distortion and the soft skin and holds ±0.0002 in.
Why not the others
A. Grinding to size before heat treatment wastes the precision: quenching distorts the punch and leaves scale and a decarburized skin. Grind after hardening and tempering.
C. A water-quenched part distorts and scales during hardening, so a diameter turned to finished size will not stay within ±0.0002 in. Leave grinding stock and grind after heat treatment.
D. Tempering only acts on steel that has already been hardened, where it relieves quench brittleness. Tempering first does nothing useful and leaves the hardened part untempered. The order is harden, temper, then grind.
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A new 14 tpi hacksaw blade loses several teeth within a few strokes while cutting thin-wall steel tubing. What is the cause?
- The blade has been tensioned too tightly.
- The strokes are too slow and long.
- The teeth face toward the handle.
- The pitch is too coarse for the wall.
Show the answer
D. Correct. At 14 tpi too few teeth bear on the thin wall at once, so each tooth catches the edge and snaps off. A finer blade, such as 24 or 32 tpi, keeps more teeth on the wall and cuts it smoothly.
Why not the others
A. Firm tension keeps the blade straight and is what the frame should give it. A loose blade wanders and can snap across; stripped teeth on thin walls come from too few teeth on the work.
B. Slow, full-length strokes are good hacksaw practice and spread wear over the whole blade. Stroke speed does not change how many teeth sit on the thin wall.
C. A reversed blade cuts on the return stroke and dulls quickly, but that does not make teeth snag and break on a thin wall. The tooth spacing is the problem here.
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A 32 mm hole is to be drilled in steel. The 32 mm drill's web is 4 mm thick at the point, and a pilot hole is drilled first to take the load off its chisel edge. Which pilot drill suits it best?
- 5 mm
- 22 mm
- 30 mm
- 3 mm
Show the answer
A. Correct. The chisel edge at the web does not cut well; it pushes metal aside and accounts for much of the feed thrust. A pilot just larger than the 4 mm web removes that work, so the 32 mm drill cuts with its lips, needs far less thrust and stays on location.
Why not the others
B. A 22 mm pilot is far more than the 4 mm web needs. It leaves the 32 mm drill only the outer part of each lip to cut, with little of its edge engaged to steady it, so it is prone to grab and chatter, and the pilot is itself a large hole to drill. Just over the web thickness is enough.
C. Opening a hole by only 1 mm per side is what makes a large drill grab and chatter, and the hole comes out out of round. The pilot only needs to clear the 4 mm web, so 5 mm suits.
D. A 3 mm pilot is smaller than the 4 mm web, so the large drill's chisel edge still has to push through metal and the thrust stays high. The pilot should be just larger than the web.
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The chart gives 25 m/min for turning this steel with an HSS tool. The bar is Ø 45 mm, and the lathe's speeds include 110, 160, 250, 360 and 560 rpm. Which speed should be set?
- 160 rpm
- 560 rpm
- 250 rpm
- 360 rpm
Show the answer
A. Correct. rpm = (1 000 × CS) ÷ (πD) = (1 000 × 25) ÷ (3.1416 × 45) = 177 rpm. The lathe has no 177 rpm step, so take the nearest speed below it, 160 rpm, which keeps the surface speed within the chart's figure.
Why not the others
B. Leaving π out gives (1 000 × 25) ÷ 45 = 556 rpm, near the 560 step. With π in the formula the answer is 177 rpm, set at the nearest lower speed, 160 rpm.
C. 250 rpm is the next step above the calculated 177 rpm and would run about 41 % faster than the chart's 25 m/min, shortening tool life. Choose the nearest lower speed, 160 rpm.
D. Using the 22.5 mm radius in place of the 45 mm diameter gives 354 rpm, which rounds to the 360 step. The formula takes the diameter: 177 rpm, set at 160 rpm.
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A taper 25 mm long runs from 40 mm to 30 mm in diameter. The compound rest's swivel scale reads the angle from the work axis. To two decimal places, at what angle is it set?
- 22.62°
- 21.80°
- 11.31°
- 78.69°
Show the answer
C. Correct. The compound is set to half the included angle: tan a = ((40 − 30) ÷ 2) ÷ 25 = 5 ÷ 25 = 0.2, so a = 11.31°. The taper's included angle is twice that, 22.62°.
Why not the others
A. This is the taper's included angle, twice the setting. A single tool point cuts one side, so the compound is set to half of it: 11.31°.
B. This uses the whole diameter difference: tan a = 10 ÷ 25. The tool cuts one side of the work, so the change in radius is used: tan a = 5 ÷ 25, a = 11.31°.
D. This is the complement, 90° − 11.31°, which would be the setting on a scale that reads from the cross-slide. This scale reads from the work axis, so it is set to 11.31°.
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When gang milling with cutters of different diameters on one arbor, which cutter's diameter is used to calculate the spindle speed?
- The average of all the cutters'.
- The widest cutter's.
- The largest cutter's.
- The smallest cutter's.
Show the answer
C. Correct. All the cutters turn at one rpm, and the largest has the highest surface speed. Setting the rpm from its diameter keeps it at or below the chart cutting speed; the smaller cutters simply run a little below theirs.
Why not the others
A. An average still runs the largest cutter faster than its chart speed. The rpm is set for the largest cutter so that no cutter goes over it.
B. Width affects the load and the feed, not the surface speed of the teeth. Surface speed depends on diameter, so the largest diameter sets the rpm.
D. The smallest cutter gives the highest rpm for the cutting speed, and that rpm would drive the largest cutter's teeth well over their chart speed, overheating and dulling them.
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A 6 mm drill is making a hole 45 mm deep in mild steel on a vertical mill. At about 20 mm deep, chips stop coming up the flutes and the drill starts to squeal. What should the machinist do?
- Flood more cutting fluid and keep feeding steadily.
- Raise the spindle speed so the flutes throw the chips clear.
- Raise the feed so the drill pushes the chips out faster.
- Withdraw the drill at intervals to clear out the chips.
Show the answer
D. Correct. As the hole deepens, chips pack in the flutes and stop flowing out; the squeal is the drill binding in them. Pecking, withdrawing the drill fully at intervals to clear the chips and let fluid in, lets it reach 45 mm without breaking.
Why not the others
A. Fluid cannot get past chips already packed in the flutes, and feeding on drives the drill deeper into them until it seizes or snaps. The drill must come out at intervals to clear the chips.
B. Higher speed does not lift chips out of a deep hole; it adds heat to a drill that is already binding in packed chips. Withdraw the drill at intervals to clear them.
C. More feed makes thicker chips that pack the flutes faster and raises the torque on a drill that is already binding. The chips must be cleared by withdrawing the drill at intervals.
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While surface grinding soft, annealed low-carbon steel with a fine, dense wheel, the operator sees the finish turn scratchy and bright specks of metal embedded in the wheel face. What is the cause?
- The wheel face is glazed.
- The wheel grade is too soft.
- The wheel is out of balance.
- The wheel face is loaded.
Show the answer
D. Correct. Loading is metal packed into the pores of the wheel face, seen as bright specks or smears. Soft, ductile steel clogs a fine, dense wheel and the loaded spots rub and scratch. Dress the face open, and use a coarser, more open wheel with plenty of coolant.
Why not the others
A. A glazed face is evenly shiny because the grains have worn flat, with no metal lodged in it, and it tends to burn rather than scratch. Embedded metal specks mean loading.
B. A soft wheel sheds grains readily and keeps renewing its face, so it is less prone to load, not more. Loading on soft steel points to a wheel too fine or too dense for the material.
C. Imbalance shows as evenly spaced chatter marks and vibration, not as metal lodged in the face. The bright specks point to loading.
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Tool 5 has just been loaded with T05 M06, and its measured length is stored in offset number 5. Which block applies the length offset as the tool comes down to its clearance height?
- G00 H05 Z25.
- G00 G43 D05 Z25.
- G00 G41 D05 Z25.
- G00 G43 H05 Z25.
Show the answer
D. Correct. G43 switches on tool length compensation using the value in H05, where tool 5's length is stored, so the tool tip, not the spindle gauge line, stops at Z25. above work zero. An H number copied from another tool's block is a common cause of crashes, so check that it matches the tool in the spindle.
Why not the others
A. On most controls an H number on its own does not switch on length compensation; it needs G43 in the block. Without it the control positions as if the tool had no length, so where lengths are measured from the spindle gauge line the tip comes down a full tool length below Z25.
B. G43 is the right code, but it takes the length from the register named by an H word. D is the address cutter radius compensation reads, so this block does not call up tool 5's length: the length offset must be G43 H05.
C. G41 with D05 is cutter radius compensation in the XY plane; it does nothing for tool length. Length compensation is G43 with an H number.
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Four holes drilled from the work offset all measure 0.05 mm too far in the plus-X direction from datum A, and their spacing is correct. What is adjusted?
- Each hole's X coordinate in the program, by 0.05 mm.
- The drill's radius wear offset, reduced by 0.05 mm.
- The X work offset, shifted 0.05 mm toward minus.
- The drill's length offset, raised by 0.05 mm.
Show the answer
C. Correct. When every feature is displaced by the same amount and the spacing is right, the datum is off, not the tool or the program. Shifting the X work offset 0.05 mm in the minus direction moves the whole pattern back onto datum A.
Why not the others
A. Editing every coordinate puts a set-up error into the proven program and leaves the real fault, the work offset, in place for every other feature.
B. Drilling runs on the spindle centreline without cutter compensation, so a radius offset has no effect on hole positions.
D. A length offset moves the tool in Z only; it cannot shift a hole pattern sideways.
03Answer key
| Question | Answer | Work activity |
|---|---|---|
| Q1 | B | Performs common occupational skills |
| Q2 | D | Performs benchwork |
| Q3 | A | Machines using drill presses |
| Q4 | A | Machines using conventional lathes |
| Q5 | C | Machines using conventional lathes |
| Q6 | C | Machines using conventional milling machines |
| Q7 | D | Machines using conventional milling machines |
| Q8 | D | Machines using precision grinding machines |
| Q9 | D | Machines using computer numerical control (CNC) machines |
| Q10 | C | Machines using computer numerical control (CNC) machines |
04More free questions
For the 429A exam itself, the 429A exam guide covers the work activities, their weights and how the exam is scored, and the 429A Machinist practice exam page describes the full TicketPrep 429A question bank. The same free format is available for the other 12 trades, plus the free guides on the pass mark, question counts, exam day and how to study.
How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it.
05Questions by exam block
Working on one part of the 429A exam? Each block has its own page with five more free questions, the tasks the block covers and their weights.