Secondary 6 Physical Education Practice Paper: Fitness And Training Principles, Energy Systems And Nutrition, Exercise Physiology And Cardiovascular Response, Sport Psychology And Motivation, Sports Injury And First Aid, Sport In Society And Fair Play

Secondary 6 Physical Education Practice Paper: Fitness and Training Principles, Energy Systems and Nutrition, Exercise Physiology and Cardiovascular Response, Sport Psychology and Motivation, Sports Injury and First Aid, Sport in Society and Fair Play

2026-11-14

Physical Education – Secondary 6 – Practice Paper

S6 PE paper, 24 parts: 1600 m run, target heart-rate zone, cardiac output, motivation survey, injury rate, icing schedule, event surplus; figures computed in code.

Paper details　S6 Physical Education practice paper, 24 parts and 100 marks, covering training principles, energy systems, cardiovascular response, motivation, injury care and fair play; every figure is computed in code and asserted.

Section A: Fitness and Training Principles (16 marks)

Source A: A Secondary 6 student runs 1,600 m in 8 minutes 40 seconds; the target is 8 minutes flat.
Source B: The student is 16 years old; maximum heart rate is estimated as 220 minus age, and training intensity is set at 65 to 80 per cent of that maximum.
Source C: Weekly mileage over four weeks was 12, 14, 16 and 18 km; the student trains four times a week for 45 minutes per session.

#	Question	Answer and marking notes
1	(a) From Source A, find how many seconds faster the target is and the percentage improvement (one decimal place).	Answer: the target is 40 seconds faster (520 minus 480); the improvement is 40/520 x 100 = 7.7 per cent. Marking: 2 marks for the method, 1 mark for the correct value.
2	(b) From Source B, calculate the maximum heart rate and the target rates at 65 per cent and 80 per cent of it (nearest whole number).	Answer: maximum heart rate is 220 minus 16 = 204 beats per minute; 65 per cent gives 204 x 0.65 = about 133 and 80 per cent about 163 beats per minute. Marking: 2 marks for the maximum, 1 mark for each target rate.
3	(c) From Source C, calculate the total mileage over four weeks and the average weekly increase.	Answer: total mileage is 12 + 14 + 16 + 18 = 60 km; mileage rises from 12 to 18 km, so the average weekly increase is (18 – 12)/3 = 2.0 km. Marking: 2 marks for the total, 2 marks for the average increase.
4	(d) Using Sources B and C, explain how the student applies the principle of progressive overload, and calculate the total training minutes per week.	Answer: four sessions of 45 minutes give 180 minutes a week; mileage climbs week by week from 12 to 18 km so the load rises gradually, while intensity stays within 65 to 80 per cent of maximum heart rate. Marking: 2 marks for the minutes, 2 marks for the progression, 1 mark for intensity control.

Section B: Energy Systems and Nutrition (16 marks)

Source A: A 60 kg athlete takes in 2,800 kcal a day with 1.6 g of protein per kilogram of body mass.
Source B: A 100 m sprint takes about 10 seconds and a marathon about 150 minutes; the sprint relies on the ATP-PC system and the marathon on the aerobic system.
Source C: One meal contains 60 g of carbohydrate, 25 g of protein and 15 g of fat; carbohydrate and protein give about 4 kcal per gram and fat about 9 kcal per gram.

#	Question	Answer and marking notes
5	(a) From Source A, calculate the daily protein need in grams and the percentage of daily energy it supplies (one decimal place).	Answer: protein need is 60 x 1.6 = 96 g; that supplies 384 kcal, which is 384/2800 x 100 = 13.7 per cent. Marking: 1 mark for grams, 1 for kcal, 1 for the percentage.
6	(b) From Source C, calculate the energy of the meal and the share supplied by carbohydrate (one decimal place).	Answer: carbohydrate 60 x 4 = 240 kcal; protein 25 x 4 = 100 kcal; fat 15 x 9 = 135 kcal; the meal totals 475 kcal and carbohydrate is 240/475 x 100 = 50.5 per cent. Marking: 1 mark for each energy value, 1 for the percentage.
7	(c) From Source B, calculate how many times longer the marathon takes than the 100 m sprint.	Answer: 150 minutes is 9000 seconds, so the ratio is 9000/10 = 900 times. Marking: 2 marks for the unit conversion, 2 for the ratio.
8	(d) Using Source B, explain why the two events rely on different energy systems.	Answer: the sprint is maximal and lasts about 10 seconds, too short for oxygen delivery, so the ATP-PC system supplies energy at once; the marathon is lower in intensity but lasts 150 minutes, so carbohydrate and fat are broken down aerobically. Marking: 2 marks for sprint intensity and duration, 2 for the marathon, 1 for oxygen supply.

Section C: Exercise Physiology and Cardiovascular Response (17 marks)

Source A: At rest the heart rate is 60 beats per minute with a stroke volume of 70 mL; during exercise the heart rate is 150 beats per minute with a stroke volume of 110 mL.
Source B: Cardiac output equals heart rate multiplied by stroke volume.
Source C: Five minutes after exercise stops, the heart rate falls from 150 to 90 beats per minute.

#	Question	Answer and marking notes
9	(a) From Sources A and B, calculate the cardiac output at rest in litres per minute.	Answer: 60 x 70 = 4200 mL per minute, about 4.2 litres per minute. Marking: 2 marks for the formula, 1 for the unit conversion.
10	(b) From Sources A and B, calculate the cardiac output during exercise in litres per minute and how many times the resting value it is (one decimal place).	Answer: 150 x 110 = 16500 mL per minute, that is 16.5 litres per minute, which is 16.5/4.2 = 3.9 times the resting value. Marking: 2 marks for cardiac output, 2 for the ratio, 1 for units.
11	(c) From Source C, calculate the fall in heart rate during recovery in beats and as a percentage (one decimal place).	Answer: the rate falls by 150 – 90 = 60 beats per minute, which is 60/150 x 100 = 40.0 per cent. Marking: 2 marks for the fall, 3 for the percentage.
12	(d) Explain two mechanisms that raise cardiac output during exercise.	Answer: sympathetic activity raises the heart rate from 60 to 150 beats per minute, and greater venous return plus stronger contraction raises stroke volume from 70 to 110 mL; together they lift cardiac output from about 4.2 to 16.5 litres per minute for oxygen delivery and carbon dioxide removal. Marking: 2 marks for the heart-rate mechanism, 2 for stroke volume, 1 for linking data.

Section D: Sport Psychology and Motivation (17 marks)

Source A: A survey of 200 students found 90 with intrinsic motivation, 70 with extrinsic motivation and 40 with no motivation.
Source B: The short-term goal is a 3 cm gain in the sit-and-reach in four weeks; the long-term goal is 6 cm in twelve weeks.
Source C: A pre-competition anxiety score of 22 out of 40 falls to 14 afterwards.

#	Question	Answer and marking notes
13	(a) From Source A, calculate the percentage of students in each of the three motivation groups.	Answer: intrinsic 90/200 = 45.0 per cent, extrinsic 35.0 per cent and amotivated 20.0 per cent, which sums to 100. Marking: 1 mark each.
14	(b) From Source B, calculate the average weekly progress of the short-term and long-term goals in cm per week (two decimal places) and how many times larger the long-term goal is.	Answer: short term is 3/4 = 0.75 cm per week and long term 6/12 = 0.5 cm per week; the long-term goal is 6/3 = 2 times larger. Marking: 1 mark per rate, 2 for the multiple, 1 for units.
15	(c) From Source C, calculate the percentage fall in anxiety from before to after the competition (one decimal place).	Answer: (22 – 14)/22 x 100 = 36.4 per cent. Marking: 3 marks for the method, 2 for the answer.
16	(d) Give one example to explain how intrinsic and extrinsic motivation affect continued participation in sport.	Answer: for example, a student who enjoys the game trains weekly for the fun of it and keeps going without prizes, whereas one who trains only for medals or to please parents tends to drop out once the reward stops. Marking: 2 marks for the intrinsic example, 2 for the extrinsic one, 1 for the comparison.

Section E: Sports Injury and First Aid (17 marks)

Source A: Among 500 athletes at a school there were 45 injuries in one season: 18 at the ankle, 12 at the knee and 15 at other sites.
Source B: For acute soft-tissue injury, ice is applied every two hours for 15 minutes during the first 48 hours.
Source C: Acute care follows the RICE principle: rest, ice, compression and elevation.

#	Question	Answer and marking notes
17	(a) From Source A, calculate the injury rate as a percentage of athletes, and the share of injuries at the ankle.	Answer: the injury rate is 45/500 x 100 = 9.0 per cent, and the ankle accounts for 18/45 x 100 = 40.0 per cent. Marking: 2 marks for the rate, 1 for the ankle share.
18	(b) From Source B, calculate the number of icing sessions in the first 48 hours and the total icing time in minutes.	Answer: 48/2 = 24 sessions, and 24 x 15 = 360 minutes in total, that is six hours. Marking: 2 marks for the count, 2 for the total time, 1 for units.
19	(c) From Source A, calculate the percentage of all injuries that occurred at sites other than the knee, that is ankle plus other sites (one decimal place).	Answer: ankle plus other sites is 18 + 15 = 33 injuries, which is 33/45 x 100 = 73.3 per cent. Marking: 2 marks for the count, 3 for the percentage.
20	(d) Using Source C, explain the purpose of each part of the RICE principle.	Answer: rest stops the injured part from being loaded again, ice constricts blood vessels and reduces swelling and pain, compression with a bandage limits swelling, and elevation aids venous return and lessens oedema. Marking: 1 mark per measure plus 1 for the overall explanation.

Section F: Sport in Society and Fair Play (17 marks)

Source A: A sports meet has 1,200 entrants, of whom 480 are youths and 720 adults; 540 are female.
Source B: The event takes in 180,000 dollars and spends 150,000 dollars; the surplus goes to community sport and is split equally among six schemes.
Source C: Of 120 doping samples, 3 test positive.

#	Question	Answer and marking notes
21	(a) From Source A, calculate the youth and female shares of all entrants.	Answer: youths are 480/1200 = 40.0 per cent and females 540/1200 = 45.0 per cent. Marking: 1 mark each, plus 1 for noting adults take 60 per cent.
22	(b) From Source B, calculate the event surplus and the amount each community scheme receives.	Answer: the surplus is 180000 – 150000 = 30000 dollars, so each of the six schemes receives 30000/6 = 5000 dollars. Marking: 2 marks for the surplus, 2 for the share, 1 for units.
23	(c) From Source C, calculate the positive rate and estimate the number of positives per 1,000 samples.	Answer: the positive rate is 3/120 x 100 = 2.5 per cent, about 3/120 x 1000 = 25 per 1,000 samples. Marking: 2 marks for the rate, 2 for the extrapolation, 1 for the conclusion.
24	(d) Explain the link between anti-doping policy and fair play, and give one social benefit of community sport.	Answer: anti-doping policy uses testing and sanctions so that athletes cannot gain an unfair advantage from drugs; results then reflect training and talent, which protects fair play and athlete health. Community sport builds social contact and cohesion and encourages regular activity for better health. Marking: 3 marks for the fair-play argument, 2 for the social benefit.

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