Free Red Seal Welder Examination Practice Questions

Sample Practice Questions

Try these 10 sample questions from the Red Seal Welder Examination practice bank. Each includes the answer and explanation. Continue with the interactive practice above to review the full question set.

1. 1A welder is setting a portable leaning ladder against a wall, with its upper support 16 ft above level ground. Using the CCOHS 4-to-1 setup rule, how far should the base be from the wall?

   1. A.2 ft
   2. B.4 ftCorrect answer: B
   3. C.8 ft
   4. D.16 ft

   Explanation: CCOHS states, "For every 4 feet (1.2 metres) up, place the base of your ladder 1 foot (0.3 metres) from the wall or upper support it rests against." Therefore 16 ft ÷ 4 = 4 ft. Dividing by 8 gives 2 ft, dividing by 2 gives 8 ft, and using a 1-to-1 ratio gives 16 ft.

   Reference: CCOHS Ladders - Portable — How should you set up the ladder?

2. 2A welder is developing a circular cutting template with a diameter of 10 cm. Using π = 3.1416, what is the area enclosed by the template?

   1. A.78.54 cm²Correct answer: A
   2. B.31.42 cm²
   3. C.19.64 cm²
   4. D.314.16 cm²

   Explanation: The formula sheet gives the area of a circle as "π × radius squared". The radius is 10 ÷ 2 = 5 cm, so the area is 3.1416 × 5² = 78.54 cm². Using 3.1416 × 10 yields 31.42; halving the radius again yields 19.64; and using the diameter as the radius yields 314.16.

   Reference: Welder - Exam Information — Math Formulas, Area of a circle

3. 3A welder prepares gas cylinders before connecting regulators to an oxy-fuel cutting setup. According to CCOHS, which gas cylinder may be briefly cracked to clear its valve outlet?

   1. A.Acetylene
   2. B.Propane
   3. C.Hydrogen
   4. D.OxygenCorrect answer: D

   Explanation: CCOHS describes cracking as opening the cylinder valve slightly and states, "It must only be used for oxygen gas." Acetylene, propane, and hydrogen are flammable gases and are not the permitted choice.

   Reference: CCOHS Welding – Storage and Handling of Compressed Gas Cylinders — When should a cylinder be cracked?

4. 4A welder prepares a GTAW torch with the non-consumable electrode required for the process. Which material forms that electrode?

   1. A.TungstenCorrect answer: A
   2. B.Graphite
   3. C.Hafnium
   4. D.Zirconium

   Explanation: TWI describes GTAW as welding with "a non-consumable tungsten electrode". Graphite is used in carbon-arc gouging; hafnium and zirconium are electrode materials used with air plasma cutting.

   Reference: TWI Tungsten Inert Gas (TIG) Welding — Job Knowledge 6

5. 5During maintenance lockout of a hydraulic shear, an authorized worker has shut down the machine, isolated it, and relieved stored energy. The worker checks its gauge before beginning work. Which lockout step does this check perform?

   1. A.Verify isolationCorrect answer: A
   2. B.Notify affected personnel
   3. C.Identify energy sources
   4. D.Shut down the equipment

   Explanation: CCOHS lists the gauge check under "Verify Isolation" and directs workers to "Check pressure gauges to make sure hydraulic and pneumatic potential energy has been removed." Notification, energy identification, and shutdown are separate steps; a post-isolation test verifies that the energy is controlled.

   Reference: CCOHS Lockout/Tag out — 7. Verify Isolation, Testing of the equipment

6. 6A welder is laying out a circular plate template with a diameter of 8 cm. Using π = 3.1416, what is its circumference, rounded to two decimal places?

   1. A.12.57 cm
   2. B.25.13 cmCorrect answer: B
   3. C.50.27 cm
   4. D.201.06 cm

   Explanation: The official formula is "2 × π × radius". The radius is 8 ÷ 2 = 4 cm, so the circumference is 2 × 3.1416 × 4 = 25.1328 cm, or 25.13 cm. Omitting the factor of 2 gives 12.57; treating the diameter as the radius gives 50.27; and using π × diameter² gives 201.06.

   Reference: Welder - Exam Information — Math Formulas, Circumference of a circle

7. 7A conventional plasma cutting torch is configured to use air as the plasma gas. Which electrode material does TWI identify for this oxidizing gas?

   1. A.HafniumCorrect answer: A
   2. B.Tungsten
   3. C.Graphite
   4. D.Mild steel

   Explanation: For oxidizing plasma gases such as air, TWI states that "the electrode must be copper with hafnium". Tungsten is described for inert-gas plasma; graphite and mild steel are not the electrode specified for this air-plasma setup.

   Reference: TWI Plasma Arc Cutting – Process and Equipment Considerations — Gas composition

8. 8In gas metal arc welding, molten metal transfers when the wire repeatedly dips into and contacts the weld pool. Which metal transfer mode is being used?

   1. A.Spray transfer
   2. B.Globular transfer
   3. C.Short-circuit transferCorrect answer: C
   4. D.Pulsed spray transfer

   Explanation: TWI states that in short-circuiting transfer, the metal is transferred "by the wire dipping into the weld pool". Spray and pulsed transfer use droplets across the arc; globular transfer is not the repeated-contact mode described.

   Reference: TWI Metal Inert Gas Welding – Process and Applications — Metal Transfer Mode

9. 9Before mounting a clean, dry bonded abrasive wheel larger than 10 cm in diameter, a welder taps it gently with a non-metallic tool and listens to the sound. What is this inspection called?

   1. A.Liquid penetrant test
   2. B.Magnetic particle test
   3. C.Ring testCorrect answer: C
   4. D.Ultrasonic test

   Explanation: CCOHS describes the ring test: "Tap wheels gently with a non-metallic tool" and notes that an undamaged wheel gives a metallic ring. Penetrant, magnetic particle, and ultrasonic tests use different inspection principles.

   Reference: CCOHS Abrasive Wheels – Handling and Storing — What is the ring test?

10. 10A welder makes a triangular gusset template with a 12 cm base and a perpendicular height of 8 cm. What is the area of the template?

    1. A.24 cm²
    2. B.20 cm²
    3. C.48 cm²Correct answer: C
    4. D.96 cm²

    Explanation: OpenStax states, "The area of a triangle is one-half the base times the height." Thus 12 cm × 8 cm ÷ 2 = 48 cm². Multiplying without halving gives 96; halving the product twice gives 24; adding the two dimensions yields 20, rather than an area.

    Reference: Intermediate Algebra, Section 2.3 — Use Formulas to Solve Geometry Applications, triangle-area formula