Secondary 5 Design And Applied Technology Practice Paper: Design Process And User Needs, Materials And Mechanics, Structures And Loads, Mechanisms And Transmission, Electronics And Control, Sustainable Design And Evaluation (100 Marks)

Secondary 5 Design and Applied Technology Practice Paper: Design Process and User Needs, Materials and Mechanics, Structures and Loads, Mechanisms and Transmission, Electronics and Control, Sustainable Design and Evaluation (100 marks)

2026-11-14

Design and Applied Technology – Secondary 5 – Practice Paper

Secondary 5 DAT practice paper with 24 parts and worked answers on design cycle durations, survey shares, ergonomic dimensions, prototype cost, stress, density,

Paper details　Secondary 5 Design and Applied Technology practice paper: 24 parts with worked answers on stress, density, beams, gears, pulley systems, circuits and life-cycle

Section A: Design Process and User Needs

Show the working and give the rounding asked.

#	Question	Answer and marking notes
1	A product development project lasts 40 working days. Research takes 25 per cent, ideation 30 per cent, prototyping 30 per cent and testing 15 per cent. Find the days for each stage and check the total.	Answer: research 40 x 25 per cent = 10 days, ideation 40 x 30 per cent = 12 days, prototyping 12 days, testing 40 x 15 per cent = 6 days; total 10 + 12 + 12 + 6 = 40 days.
2	A survey of 250 users finds 52 per cent prefer model A and 28 per cent prefer model B, the rest prefer model C. Find the number of users for each model and the percentage for model C.	Answer: model A 250 x 52 per cent = 130 users, model B 250 x 28 per cent = 70 users, model C 250 – 130 – 70 = 50 users, which is 50 / 250 x 100 per cent = 20 per cent.
3	A user is 172 cm tall. Ergonomic guidance puts the bench height at 0.43 of stature and the seat height at 0.25 of stature. Find both dimensions to 1 decimal place.	Answer: bench 172 x 0.43 = 73.96, so 74.0 cm; seat 172 x 0.25 = 43.0 cm; the clearance between them is 74.0 – 43.0 = 31.0 cm.
4	A prototype costs 180 dollars in material, 6 hours of machining at 95 dollars per hour and 250 dollars of testing. Development cost of 4800 dollars is spread over 8 units. Find the direct cost per unit, the development share per unit and the total cost per unit.	Answer: machining 6 x 95 = 570 dollars, direct cost 180 + 570 + 250 = 1000 dollars, development share 4800 / 8 = 600 dollars, total 1000 + 600 = 1600 dollars per unit.

Section B: Materials and Mechanics

Write the formula, substitute the values and give the units.

#	Question	Answer and marking notes
5	A tie rod carries a tensile force of 2400 newtons and its cross-section is 30 mm by 40 mm. Find the cross-sectional area and the tensile stress.	Answer: area 30 x 40 = 1200 square millimetres; stress = force / area = 2400 / 1200 = 2.0 newtons per square millimetre, that is 2.0 MPa.
6	A metal sample has a mass of 486 grams and a volume of 180 cubic centimetres. Find its density, then the volume of a 1200 gram part made of the same material to 1 decimal place.	Answer: density 486 / 180 = 2.7 grams per cubic centimetre; part volume 1200 / 2.7 = 444.4 cubic centimetres.
7	A rod of original length 800 mm stretches to 800.96 mm. The Young modulus of the material is 200000 MPa. Find the strain and the stress.	Answer: extension 800.96 – 800 = 0.96 mm; strain 0.96 / 800 = 0.0012; stress = 200000 x 0.0012 = 240 MPa.
8	A material has an ultimate strength of 480 MPa. Find the factor of safety at a working stress of 120 MPa, then at 160 MPa, and comment on the risk.	Answer: factor of safety 480 / 120 = 4.0, then 480 / 160 = 3.0; the margin falls from 4 to 3, so tolerance of load error and fatigue is reduced and inspection must be tighter.

Section C: Structures and Loads

State the point about which moments are taken and give units.

#	Question	Answer and marking notes
9	A simply supported beam spans 6 m and carries a point load of 900 newtons at 2 m from the left support. Find both reactions and check the total.	Answer: taking moments about the right support, left reaction = 900 x (6 – 2) / 6 = 600 newtons and right reaction = 900 – 600 = 300 newtons; the sum 600 + 300 = 900 newtons matches the load.
10	A force of 250 newtons acts at 0.6 m from a pivot. Find the moment, then the force needed at 0.9 m for the same moment to 1 decimal place.	Answer: moment 250 x 0.6 = 150 newton metres; at 0.9 m the force is 150 / 0.9 = 166.7 newtons, so a longer arm needs a smaller force.
11	A triangulated truss spans 4.8 m with a height of 1.8 m; a diagonal runs from mid-span to one end. Find the diagonal length to 2 decimal places and its angle to the horizontal to 1 decimal place.	Answer: horizontal projection 4.8 / 2 = 2.4 m; length = square root of (2.4 squared + 1.8 squared) = square root of 9.00 = 3.00 m; angle = inverse tangent of 1.8 / 2.4 = 36.9 degrees.
12	Two columns are both 40 mm wide but 200 mm and 120 mm deep. The second moment of area varies as the cube of depth. Find the ratio of bending resistance to 2 decimal places and state the design implication.	Answer: ratio = 200 cubed / 120 cubed = 8000000 / 1728000 = 4.63; raising the depth from 120 to 200 mm multiplies bending resistance by 4.63, so deepening is far more effective than thickening, though lateral stability must still be checked.

Section D: Mechanisms and Transmission

Give efficiency as a percentage to the nearest whole number.

#	Question	Answer and marking notes
13	A driving gear with 20 teeth runs at 1200 revolutions per minute and drives a 60 tooth gear. Find the driven speed and the torque multiplication.	Answer: speed 1200 x 20 / 60 = 400 revolutions per minute; torque ratio 60 / 20 = 3, so torque rises threefold while speed falls to one third.
14	A pulley system has 4 rope strands supporting a load of 600 newtons. Find the ideal effort, then the actual effort if the efficiency is 80 per cent.	Answer: ideal effort 600 / 4 = 150 newtons; actual effort 600 / (4 x 0.80) = 187.5 newtons; the 37.5 newton difference is taken up by friction.
15	A lever carries a load of 450 newtons on a load arm of 0.2 m, with an effort arm of 0.9 m. Find the effort needed and the mechanical advantage.	Answer: load moment 450 x 0.2 = 90 newton metres; effort 90 / 0.9 = 100 newtons; mechanical advantage 0.9 / 0.2 = 4.5.
16	A lead screw has a lead of 5 mm and turns at 120 revolutions per minute. Find the feed rate, then the revolutions and time in seconds to move a table 250 mm.	Answer: feed rate 5 x 120 = 600 mm per minute; revolutions 250 / 5 = 50; time 50 / (120 / 60) = 25 seconds.

Section E: Electronics and Control

Give currents in milliamperes and state the unit of power.

#	Question	Answer and marking notes
17	A 12 volt supply is connected across a 240 ohm resistor. Find the current in milliamperes.	Answer: current = voltage / resistance = 12 / 240 = 0.05 amperes = 50 milliamperes.
18	Resistors of 220 ohm and 330 ohm are in series across 12 volts. Find the total resistance, the current to 1 decimal place and the voltage across the 330 ohm resistor to 1 decimal place.	Answer: total resistance 220 + 330 = 550 ohm; current 12 / 550 = 21.8 milliamperes; the drop across 330 ohm is 21.8 milliamperes x 330 ohm = 7.2 volts, which is 7.2 / 12 = 60 per cent of the supply.
19	Resistors of 300 ohm and 600 ohm are in parallel across 12 volts. Find the equivalent resistance and the total current in milliamperes.	Answer: equivalent resistance = (300 x 600) / (300 + 600) = 200 ohm; total current 12 / 200 = 0.06 amperes = 60 milliamperes, the sum of 40 and 20 milliamperes in the two branches.
20	A light emitting diode has a forward drop of 2 volts and a rated current of 20 milliamperes, driven from a 5 volt supply. Find the series resistor, its dissipation and a suitable power rating.	Answer: the resistor must drop 5 – 2 = 3 volts, so resistance = 3 / 0.020 = 150 ohm; dissipation = 0.020 squared x 150 = 0.06 watts, so a 0.25 watt part gives more than four times the margin.

Section F: Sustainable Design and Evaluation

Give money in dollars and emissions in kilograms of carbon dioxide equivalent.

#	Question	Answer and marking notes
21	A product costs 1200 dollars, uses 180 kilowatt hours a year at 1.2 dollars per kilowatt hour, lasts 8 years and costs 150 dollars to dispose of. Find the life-cycle cost.	Answer: energy cost 180 x 1.2 x 8 = 1728 dollars; life-cycle cost 1200 + 1728 + 150 = 3078 dollars; energy is 1728 / 3078 x 100 per cent = 56.1 per cent of the total.
22	Virgin material emits 4.2 kilograms of carbon dioxide equivalent per kilogram and recycled material 1.5 kilograms. With 60 per cent recycled content, find the average emission and the reduction to 1 decimal place.	Answer: average = 0.6 x 1.5 + 0.4 x 4.2 = 0.90 + 1.68 = 2.58 kilograms; reduction = (4.2 – 2.58) / 4.2 x 100 per cent = 38.6 per cent.
23	A grade A model uses 210 kilowatt hours a year and a grade B model 320 kilowatt hours, at 1.2 dollars per kilowatt hour. Find the electricity cost difference over 10 years.	Answer: the yearly difference is 320 – 210 = 110 kilowatt hours; over 10 years 110 x 1.2 x 10 = 1320 dollars, that is 132 dollars a year.
24	Annual output is 12000 units, each carrying 2.58 kilograms of carbon dioxide equivalent. Find the annual total in tonnes to 1 decimal place, then the reduction in kilograms if a design of 2.2 kilograms per unit is adopted.	Answer: annual total 12000 x 2.58 = 30960 kilograms = 31.0 tonnes; at 2.2 kilograms it is 12000 x 2.2 = 26400 kilograms, a cut of 30960 – 26400 = 4560 kilograms, or 14.7 per cent.

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