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TCC Technical Cadet Initial Test study materials
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TCC Technical Cadet Initial Test practice questions

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Question 1Mathematics and Numerical Computation

A cadet's worksheet contains the expression 48 ÷ 6 + 3 × (5 − 2)². Applying the standard order of operations, what value should be recorded?

  1. 99
  2. 63
  3. 27
  4. 35 Correct answer.

Explanation

Order of operations requires brackets first, giving (5 − 2) = 3, then the exponent 3² = 9. Division and multiplication are then evaluated before addition: 48 ÷ 6 = 8 and 3 × 9 = 27, so the value is 8 + 27 = 35.

Question 2Mathematics and Numerical Computation

On the entry mathematics assessment a cadet answers 42 items correctly out of 60. What percentage score does this represent?

  1. 75%
  2. 72%
  3. 70% Correct answer.
  4. 68%

Explanation

A percentage is the correct count divided by the total count, multiplied by 100. Here 42 ÷ 60 = 0.70, and 0.70 × 100 gives 70%, so the cadet scored seventy per cent.

Question 3Mathematics and Numerical Computation

A machined component is specified as 2.4 metres long. What is this length when recorded in millimetres for the inspection sheet?

  1. 24 000 mm
  2. 2 400 mm Correct answer.
  3. 240 mm
  4. 0.0024 mm

Explanation

There are 1 000 millimetres in one metre, so converting metres to millimetres means multiplying by 1 000. Applying this, 2.4 × 1 000 = 2 400 millimetres, which is the value to enter on the inspection sheet.

Question 4Mathematics and Numerical Computation

A sealing compound is prepared from three powders in the ratio 5 : 3 : 2 by mass. If a cadet prepares 240 kg of the compound, what mass of the largest component is required?

  1. 80 kg
  2. 72 kg
  3. 96 kg
  4. 120 kg Correct answer.

Explanation

The ratio 5 : 3 : 2 has 5 + 3 + 2 = 10 equal parts in total, so each part has mass 240 ÷ 10 = 24 kg. The largest component takes 5 parts, giving 5 × 24 = 120 kg.

Question 5Mathematics and Numerical Computation

A cadet records five reaction-time trials. The mean of the five times is 68 units, and four of the recorded times are 62, 71, 65 and 70. What is the fifth recorded time?

  1. 66
  2. 74
  3. 70
  4. 72 Correct answer.

Explanation

The mean of five values equals their total divided by five, so the total must be 68 × 5 = 340. The four known values add to 62 + 71 + 65 + 70 = 268, leaving 340 − 268 = 72 for the fifth trial.

Question 6Mathematics and Numerical Computation

A calibration relationship used by a cadet is written as 5(2x − 3) = 4x + 9. What value of x satisfies this relationship?

  1. 2
  2. 6
  3. 3
  4. 4 Correct answer.

Explanation

Expanding the bracket gives 10x − 15 = 4x + 9. Subtracting 4x from both sides gives 6x − 15 = 9, and adding 15 gives 6x = 24, so x = 4.

Question 7Mathematics and Numerical Computation

A cylindrical standpipe has an internal radius of 5 cm and a height of 12 cm. Using π ≈ 3.14, what is its internal volume?

  1. 377 cm³
  2. 471 cm³
  3. 942 cm³ Correct answer.
  4. 1 884 cm³

Explanation

The volume of a cylinder is V = πr²h. With r = 5 cm and h = 12 cm, r² = 25, so V ≈ 3.14 × 25 × 12 = 942 cubic centimetres.

Question 8Mathematics and Numerical Computation

Two supply lines feed a reservoir. Line A alone fills it in 6 hours and line B alone fills it in 12 hours. Working together at these rates, how long will the two lines take to fill the reservoir?

  1. 4.5 hours
  2. 4 hours Correct answer.
  3. 3 hours
  4. 9 hours

Explanation

In one hour line A fills 1/6 of the reservoir and line B fills 1/12. Combined they fill 1/6 + 1/12 = 3/12 = 1/4 of the reservoir per hour, so the whole reservoir takes 4 hours.

Question 9Mathematics and Numerical Computation

A 40-litre cleaning bath contains a 25% acid solution. How much pure acid must be added to raise the concentration to 40%?

  1. 6 litres
  2. 15 litres
  3. 10 litres Correct answer.
  4. 9 litres

Explanation

The bath initially holds 0.25 × 40 = 10 litres of acid. Adding x litres of pure acid gives (10 + x) ÷ (40 + x) = 0.40, so 10 + x = 16 + 0.4x and 0.6x = 6, giving x = 10 litres.

Question 10Mathematics and Numerical Computation

On a calibration chart a straight line passes through the points (2, 5) and (6, 13). Assuming the same linear trend continues, what is the y-value when x = 10?

  1. 19
  2. 21 Correct answer.
  3. 17
  4. 23

Explanation

The gradient is (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2 per unit of x. From x = 6 to x = 10 the x-value rises by 4, so y rises by 2 × 4 = 8, giving y = 13 + 8 = 21.

Question 11Mathematics and Numerical Computation

A cadet records a sensor current as 0.000 045 A. Written in scientific notation, this current is:

  1. 4.5 × 10⁻⁵ A Correct answer.
  2. 0.45 × 10⁻⁴ A
  3. 45 × 10⁻⁶ A
  4. 4.5 × 10⁻⁴ A

Explanation

Scientific notation writes a number as a coefficient from 1 up to but not including 10 multiplied by a power of 10. Moving the decimal point five places to the right gives 4.5, so the power is −5.

Question 12Mathematics and Numerical Computation

A micrometer reading of 0.086 42 m is to be entered on a process sheet rounded to three significant figures. Which value should be entered?

  1. 0.08642 m
  2. 0.086 m
  3. 0.0864 m Correct answer.
  4. 0.09 m

Explanation

Leading zeros after the decimal point are not significant. The first three significant figures are 8, 6 and 4; the next digit is 2, so the third figure stays unchanged.

Question 13Mathematics and Numerical Computation

Six torque readings, in newton metres, are recorded during a fastener test: 42, 38, 45, 41, 39 and 47. What is the median reading?

  1. 41 N·m
  2. 41.5 N·m Correct answer.
  3. 43 N·m
  4. 42 N·m

Explanation

With six readings, the median is the mean of the third and fourth values after sorting. The sorted values are 38, 39, 41, 42, 45 and 47, so the median is (41 + 42)/2 = 41.5 N·m.

Question 14Mathematics and Numerical Computation

A pump delivers 3.6 cubic metres of water per hour. What is this flow rate in litres per minute?

  1. 216 L/min
  2. 60 L/min Correct answer.
  3. 6 L/min
  4. 0.06 L/min

Explanation

One cubic metre contains 1000 litres, so 3.6 m³/h is 3600 L/h. Dividing by 60 minutes per hour gives 60 L/min.

Question 15Mathematics and Numerical Computation

A 48-tooth gear drives a 16-tooth gear. If the driving gear turns at 300 revolutions per minute, what is the rotational speed of the driven gear?

  1. 900 rpm Correct answer.
  2. 300 rpm
  3. 1 200 rpm
  4. 100 rpm

Explanation

For meshing gears, speed is inversely proportional to tooth count. The ratio is 48/16 = 3, so the driven gear turns at 300 × 3 = 900 rpm.

Question 16Mathematics and Numerical Computation

A straight brace is fitted between a point 12 m up a vertical mast and a ground anchor. The brace is 13 m long. How far is the anchor from the foot of the mast, assuming the mast is perpendicular to the ground?

  1. 12.5 m
  2. 25 m
  3. 5 m Correct answer.
  4. 1 m

Explanation

The mast, ground and brace form a right triangle with the brace as the hypotenuse. By Pythagoras, distance = √(13² − 12²) = √25 = 5 m.

Question 17Mathematics and Numerical Computation

A pipe has an outside diameter of 20 mm and an inside diameter of 14 mm. Using π ≈ 3.14, what is the cross-sectional area of the pipe material?

  1. 314.00 mm²
  2. 615.44 mm²
  3. 467.86 mm²
  4. 160.14 mm² Correct answer.

Explanation

The material area is the difference between the outer and inner circle areas. Using radii 10 mm and 7 mm gives π(10² − 7²) = 3.14 × 51 = 160.14 mm².

Question 18Mathematics and Numerical Computation

For an electrical heater, power is related to voltage and resistance by P = V² / R. A heater rated at 1200 W is connected to a 240 V supply. What is its resistance?

  1. 20 Ω
  2. 0.2 Ω
  3. 4800 Ω
  4. 48 Ω Correct answer.

Explanation

Rearranging P = V²/R gives R = V²/P. Substituting V = 240 V and P = 1200 W gives R = 57,600/1200 = 48 Ω.

Question 19Mathematics and Numerical Computation

A production run has a defect rate of 3%. A cadet inspects three components selected independently. What is the probability that exactly one of the three is defective, to three significant figures?

  1. 0.0847 Correct answer.
  2. 0.0282
  3. 0.9127
  4. 0.00262

Explanation

The binomial probability for exactly one defect is C(3,1)(0.03)(0.97)². This equals 3 × 0.03 × 0.9409 = 0.0847 to three significant figures.

Question 20Mathematics and Numerical Computation

A thermal sensor's indicated temperature falls by one half every 30 minutes after a furnace is switched off. At switch-off the indicated temperature is 640°C. What is the indicated temperature 2.5 hours later?

  1. 80°C
  2. 40°C
  3. 20°C Correct answer.
  4. 10°C

Explanation

A fall by half every 30 minutes means the number of half-lives is 150/30 = 5. The reading is 640 × (1/2)^5 = 640/32 = 20°C.

Question 21Mathematics and Numerical Computation

A cadet checks a coolant reservoir that is 2/5 full. A further 1/4 of the reservoir's capacity is then added. What fraction of the reservoir's capacity is now filled?

  1. 13/20 Correct answer.
  2. 1/3
  3. 7/20
  4. 11/20

Explanation

Convert both fractions to twentieths: 2/5 = 8/20 and 1/4 = 5/20. Adding these gives 8/20 + 5/20 = 13/20 of the reservoir's capacity. The remaining 7/20 is the empty portion, not the filled portion.

Question 22Mathematics and Numerical Computation

A fastener is tightened from 80 N·m to 92 N·m. What is the percentage increase in torque?

  1. 18%
  2. 15% Correct answer.
  3. 12%
  4. 13%

Explanation

The increase is 92 − 80 = 12 N·m. Percentage increase is calculated from the original value: (12 / 80) × 100 = 15%.

Question 23Mathematics and Numerical Computation

A cadet must cut a 5.25 m cable into 7 equal lengths. What is the length of each piece?

  1. 0.65 m
  2. 0.80 m
  3. 0.75 m Correct answer.
  4. 0.70 m

Explanation

Dividing the total length by the number of pieces gives 5.25 ÷ 7 = 0.75 m. Checking the result, 0.75 m × 7 = 5.25 m.

Question 24Mathematics and Numerical Computation

For a temperature limit, the inequality 3x − 7 < 2x + 5 must hold. Which range of x satisfies it?

  1. x < 2
  2. x > 12
  3. x > 2
  4. x < 12 Correct answer.

Explanation

Subtract 2x from both sides to obtain x − 7 < 5. Adding 7 to both sides gives x < 12, so the allowable range is all values below 12.

Question 25Mathematics and Numerical Computation

On a drawing with scale 1:250, a pipe run measures 48 mm. What is the actual length in metres?

  1. 9.6 m
  2. 6 m
  3. 12 m Correct answer.
  4. 15 m

Explanation

A scale of 1:250 means the actual length is 250 times the drawing length. Thus 48 mm × 250 = 12,000 mm, and 12,000 mm ÷ 1,000 = 12 m.

Question 26Mathematics and Numerical Computation

A trapezoidal support plate has parallel sides of 18 cm and 12 cm, and a perpendicular height of 10 cm. What is its area?

  1. 240 cm²
  2. 120 cm²
  3. 180 cm²
  4. 150 cm² Correct answer.

Explanation

The area of a trapezium is the average of the parallel sides multiplied by the perpendicular height. Therefore area = ((18 + 12) / 2) × 10 = 15 × 10 = 150 cm².

Question 27Mathematics and Numerical Computation

The velocity of a test carriage is modelled by v = u + at. Which expression gives t when v, u and a are known and a ≠ 0?

  1. (v − u) / a Correct answer.
  2. (u − v) / a
  3. v/u − a
  4. (v + u) / a

Explanation

Subtract u from both sides of v = u + at to get v − u = at. Dividing both sides by a gives t = (v − u) / a, provided a is not zero.

Question 28Mathematics and Numerical Computation

A metal sample has a mass of 2.7 kg and a volume of 0.003 m³. What is its density in kg/m³?

  1. 1 100 kg/m³
  2. 810 kg/m³
  3. 900 kg/m³ Correct answer.
  4. 9 000 kg/m³

Explanation

Density is mass divided by volume: 2.7 ÷ 0.003 = 900. The units are kg/m³, so the density is 900 kg/m³.

Question 29Mathematics and Numerical Computation

In a fastener batch, 4 bolts and 3 nuts have a combined mass of 47 g. In the same batch, 2 bolts and 5 nuts have a combined mass of 41 g. What is the mass of one bolt?

  1. 5 g
  2. 6 g
  3. 11 g
  4. 8 g Correct answer.

Explanation

Let b be the bolt mass and n the nut mass. The equations are 4b + 3n = 47 and 2b + 5n = 41. Doubling the second gives 4b + 10n = 82; subtracting the first gives 7n = 35, so n = 5. Then 4b + 15 = 47, so 4b = 32 and b = 8 g.

Question 30Mathematics and Numerical Computation

A rectangular tank measuring 1.5 m by 0.8 m by 0.6 m is filled by a pump delivering 12 litres per minute. How long will it take to fill the empty tank?

  1. 72 min
  2. 45 min
  3. 90 min
  4. 60 min Correct answer.

Explanation

The tank volume is 1.5 × 0.8 × 0.6 = 0.72 m³. Since 1 m³ = 1,000 litres, this is 720 litres. At 12 litres per minute, the time is 720 ÷ 12 = 60 minutes.

Question 31Mathematics and Numerical Computation

A cadet records the masses of five identical washers as 2.4 g, 2.5 g, 2.3 g, 2.6 g and 2.2 g. What is the mean mass of this sample?

  1. 2.4 g Correct answer.
  2. 2.3 g
  3. 2.5 g
  4. 2.6 g

Explanation

The mean is the sum of the readings divided by the number of readings. The sum is 12.0 g, and 12.0 g ÷ 5 = 2.4 g.

Question 32Mathematics and Numerical Computation

A torque specification is 45 N·m. A cadet needs the value in newton-centimetres (N·cm). Using 1 m = 100 cm, what is the specification?

  1. 4500 N·cm Correct answer.
  2. 450 N·cm
  3. 0.45 N·cm
  4. 45 N·cm

Explanation

Torque in N·m is converted to N·cm by multiplying by 100 because 1 m equals 100 cm. Therefore 45 × 100 = 4500 N·cm.

Question 33Mathematics and Numerical Computation

A coolant concentrate is mixed with water in the ratio 1:9 by volume. To make 20 litres of the mixture, how many litres of concentrate are required?

  1. 10 L
  2. 1.9 L
  3. 1.8 L
  4. 2.0 L Correct answer.

Explanation

The ratio has 1 + 9 = 10 parts. The concentrate is 1 part of the 20-litre total, so 20 L ÷ 10 = 2.0 L.

Question 34Mathematics and Numerical Computation

A rectangular access panel measures 0.45 m by 0.30 m. What is its area in square centimetres?

  1. 13.5 cm²
  2. 135 cm²
  3. 13 500 cm²
  4. 1350 cm² Correct answer.

Explanation

Convert both dimensions to centimetres first: 0.45 m = 45 cm and 0.30 m = 30 cm. The area is 45 cm × 30 cm = 1350 cm².

Question 35Mathematics and Numerical Computation

A hydraulic press applies a force of 2400 N over a piston area of 0.08 m². What pressure, in pascals, is produced?

  1. 192 Pa
  2. 30 000 Pa Correct answer.
  3. 300 Pa
  4. 19 200 Pa

Explanation

Pressure is force divided by area: P = F / A. Substituting gives 2400 N ÷ 0.08 m² = 30 000 Pa.

Question 36Mathematics and Numerical Computation

A test carriage travels 180 m in 12 s at constant speed. What is its speed in kilometres per hour?

  1. 54 km/h Correct answer.
  2. 45 km/h
  3. 216 km/h
  4. 15 km/h

Explanation

Speed is distance divided by time: 180 m ÷ 12 s = 15 m/s. Converting metres per second to kilometres per hour multiplies by 3.6, giving 15 × 3.6 = 54 km/h.

Question 37Mathematics and Numerical Computation

The resistance of a sensor is modelled by R = R₀(1 + αT), where R₀ = 100 Ω, α = 0.004 per °C and T = 50 °C. What is R?

  1. 102 Ω
  2. 200 Ω
  3. 120 Ω Correct answer.
  4. 140 Ω

Explanation

Substitute the given values: R = 100 × (1 + 0.004 × 50). The product 0.004 × 50 equals 0.2, so R = 100 × 1.2 = 120 Ω.

Question 38Mathematics and Numerical Computation

A component's mass is specified as 250 g ± 2%. What is the maximum acceptable mass?

  1. 252.5 g
  2. 245 g
  3. 250 g
  4. 255 g Correct answer.

Explanation

A tolerance of ±2% of 250 g is 0.02 × 250 = 5 g. The maximum acceptable mass is the nominal value plus the tolerance: 250 + 5 = 255 g.

Question 39Mathematics and Numerical Computation

A rectangular support bracket has a length that is 4 cm greater than its width. Its area is 96 cm². What is the width of the bracket?

  1. 6 cm
  2. 12 cm
  3. 8 cm Correct answer.
  4. 10 cm

Explanation

Let the width be w cm; then the length is w + 4 cm. The area equation is w(w + 4) = 96, which gives w² + 4w - 96 = 0. Solving the quadratic gives the positive width w = 8 cm.

Question 40Mathematics and Numerical Computation

A cylindrical tank has a diameter of 1.2 m and a height of 2.5 m. Using π ≈ 3.14, what is its volume to the nearest 0.01 m³?

  1. 11.30 m³
  2. 1.13 m³
  3. 2.83 m³ Correct answer.
  4. 2.26 m³

Explanation

The radius is half the diameter, so r = 0.6 m. Volume is πr²h = 3.14 × 0.6² × 2.5 = 3.14 × 0.9 = 2.826 m³, which rounds to 2.83 m³.

Question 41Physics and Applied Mechanics

A cadet vehicle starts from rest and accelerates uniformly at 2.0 m/s² for 5.0 s along a straight test track. What distance does it cover during this interval?

  1. 25 m Correct answer.
  2. 50 m
  3. 5.0 m
  4. 10 m

Explanation

For uniform acceleration from rest the displacement is s = ½at². Substituting a = 2.0 m/s² and t = 5.0 s gives s = 0.5 × 2.0 × 25 = 25 m. The product at = 10 m/s is a velocity, not a distance, so that shortcut cannot be used here.

Question 42Physics and Applied Mechanics

A 1200 kg vehicle travelling on a level road is brought down in speed by a constant braking force of 4800 N. What is the magnitude of its deceleration?

  1. 9.8 m/s²
  2. 5.8 m/s²
  3. 0.25 m/s²
  4. 4.0 m/s² Correct answer.

Explanation

Newton's second law for a body of constant mass gives a = F/m. Dividing the net braking force 4800 N by the mass 1200 kg yields 4.0 m/s². Because the force opposes motion, this value is a deceleration acting along the direction of travel.

Question 43Physics and Applied Mechanics

A 4.0 m plank is balanced on a pivot at its centre. A 30 kg load is placed 1.5 m to the left of the pivot. How far to the right of the pivot must a 45 kg load be placed to keep the plank horizontal?

  1. 1.0 m Correct answer.
  2. 0.67 m
  3. 2.0 m
  4. 1.5 m

Explanation

For rotational balance the clockwise and anticlockwise moments about the pivot must be equal, so m₁d₁ = m₂d₂. Here 30 × 1.5 = 45 N·m-equivalent on the left, requiring 45 × d = 45 on the right, which gives d = 1.0 m. The heavier load therefore sits closer to the pivot than the lighter one.

Question 44Physics and Applied Mechanics

A 1500 kg vehicle increases its speed from 10 m/s to 20 m/s on a level road. What is the increase in its kinetic energy?

  1. 75 kJ
  2. 225 kJ Correct answer.
  3. 150 kJ
  4. 450 kJ

Explanation

The change in kinetic energy is ½m(v² − u²), because kinetic energy depends on the square of speed. Substituting 0.5 × 1500 × (400 − 100) gives 225 000 J, or 225 kJ. The speed difference alone (10 m/s) must not be used in place of the difference of squares.

Question 45Physics and Applied Mechanics

A pressure sensor is lowered to a depth of 20 m in seawater of density 1025 kg/m³. Taking g = 9.81 m/s², what gauge pressure due to the water column acts on the sensor?

  1. about 100 kPa
  2. about 2.0 MPa
  3. about 200 kPa Correct answer.
  4. about 20 kPa

Explanation

Hydrostatic gauge pressure in a fluid column is p = ρgh, independent of the shape of the container. Substituting 1025 × 9.81 × 20 gives roughly 2.0 × 10⁵ Pa, which is about 200 kPa. This is the pressure contribution of the water alone, additional to atmospheric pressure at the surface.

Question 46Physics and Applied Mechanics

A 0.20 kg ball strikes a vertical wall horizontally at 15 m/s and rebounds along the same line at 10 m/s. What is the magnitude of the change in the ball's momentum?

  1. 2.0 kg·m/s
  2. 1.0 kg·m/s
  3. 3.0 kg·m/s
  4. 5.0 kg·m/s Correct answer.

Explanation

Momentum is a vector, so the reversal of direction must be carried in the sign: Δp = m(v_final − v_initial) = 0.20 × (10 − (−15)). This gives 5.0 kg·m/s. Adding the two speeds rather than subtracting them is essential whenever a body rebounds along its incoming line.

Question 47Physics and Applied Mechanics

A 40 kg crate is pushed across a level floor at constant velocity by a horizontal force of 120 N. Taking g = 9.81 m/s², what is the coefficient of kinetic friction between the crate and the floor?

  1. 0.31 Correct answer.
  2. 3.0
  3. 0.50
  4. 0.12

Explanation

At constant velocity the horizontal push exactly balances kinetic friction, and on a level floor the normal force equals the weight. Thus μ = F/(mg) = 120/(40 × 9.81) ≈ 0.31. The coefficient is a dimensionless ratio of two forces, so it cannot exceed the value expected from the balance of forces here.

Question 48Physics and Applied Mechanics

A steel tie rod of cross-sectional area 4.0 × 10⁻⁴ m² carries a tensile force of 20 kN. What is the tensile stress in the rod?

  1. 8.0 MPa
  2. 5.0 MPa
  3. 50 MPa Correct answer.
  4. 200 MPa

Explanation

Direct tensile stress is defined as force divided by cross-sectional area, σ = F/A. Substituting 20 000 N and 4.0 × 10⁻⁴ m² gives 5.0 × 10⁷ Pa, which is 50 MPa. Stress always has the units of pressure, so force multiplied by area is dimensionally wrong.

Question 49Physics and Applied Mechanics

A projectile is launched over level ground at 40 m/s at 30° above the horizontal. Neglecting air resistance and taking g = 9.81 m/s², what horizontal range does it achieve?

  1. 82 m
  2. 141 m Correct answer.
  3. 282 m
  4. 163 m

Explanation

For level ground the range is R = v² sin(2θ)/g. With v = 40 m/s and 2θ = 60°, R = 1600 × 0.866/9.81 ≈ 141 m. The flight time and maximum height can be found separately, but range follows directly from this single expression.

Question 50Physics and Applied Mechanics

A simple pendulum has a length of 1.0 m and swings through a small angle. Taking g = 9.81 m/s², what is its period of oscillation?

  1. 6.3 s
  2. 1.0 s
  3. 0.32 s
  4. 2.0 s Correct answer.

Explanation

For small oscillations the period of a simple pendulum is T = 2π√(L/g), independent of the mass of the bob. Here √(1.0/9.81) ≈ 0.319 s, and multiplying by 2π gives about 2.0 s. The small-angle condition is what makes this simple harmonic approximation valid.