S4 HMSC Paper: Health Indicators, Health Care Services And Social Care (100 Marks)

S4 HMSC Paper: Health Indicators, Health Care Services and Social Care (100 marks)

2026-11-11

S4 HMSC – health indicators and social care

S4 Health Management and Social Care paper: 6 questions, 24 parts, 100 marks, 75 minutes on BMI and energy balance, health care planning, poverty and health promotion.

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Paper details　S4 Health Management and Social Care practice paper with six questions on body mass index, stress scores, hospital bed planning, poverty rate, smoking cessation and nutrient energy ratios. Answers with working and marking notes included.

Section A: Body Mass Index and Energy Balance (15 marks)

A Secondary 4 student is 1.65 m tall and weighs 62.0 kg. The WHO Asian adult classification is used: below 18.5 is underweight, 18.5 to 22.9 is normal, 23.0 to 24.9 is overweight and 25.0 or above is obese. His doctor advises a body mass index of 22.0 over a 90-day period. Reducing 1 kg of body fat is estimated to require an energy deficit of 7700 kcal.

#	Question	Answer and marking notes
1	Calculate the student body mass index to one decimal place.	Answer: mass 62.0 / (1.65 x 1.65) = 62.0 / 2.7225 = 22.8, that is 22.8.
2	State his weight category and give the reason.	Answer: 22.8 falls between 18.5 and 22.9, so he is in the normal range. A doctor may still advise weight reduction because a normal index does not by itself prove that body fat or waist circumference is ideal.
3	Find the mass that gives an index of 22.0 (one decimal place) and the mass to be reduced.	Answer: target mass 22.0 x 2.7225 = 59.9 kg, so the reduction is 62.0 – 59.9 = 2.1 kg.
4	Assuming all the loss is fat, calculate the average daily energy deficit over 90 days, to the nearest whole kcal.	Answer: 2.1 x 7700 = 16170 kcal, then 16170 / 90 = 179.7, about 180 kcal per day. A very large daily deficit is hard to sustain, so diet and exercise should be combined.

Section B: Stress and Mental Health (15 marks)

A school uses a 40-mark stress scale with 12 Secondary 4 students; a higher score means more stress. The scores are 12, 18, 20, 25, 9, 31, 16, 22, 27, 14, 20 and 36. A score of 20 is the follow-up threshold.

#	Question	Answer and marking notes
5	Calculate the mean score to one decimal place.	Answer: the total is 250, so 250 / 12 = 20.8333, that is 20.8.
6	Find the median and show your working.	Answer: in order the scores are 9, 12, 14, 16, 18, 20, 20, 22, 25, 27, 31 and 36; with 12 values the median is the average of the sixth and seventh, (20 + 20) / 2 = 20.
7	Find the range.	Answer: the highest score is 36 and the lowest is 9, so the range is 36 – 9 = 27. A wide range shows the students differ a lot, so the mean alone is not enough.
8	Find the percentage of students at or above the threshold (one decimal place) and explain why the median describes this set better than the mean.	Answer: 7 / 12 x 100 = 58.3 per cent. The score of 36 is an extreme value that pulls the mean up, while the median of 20 is not affected by extreme values, so it represents the typical student better; the range then shows the spread.

Section C: Health Care Planning and Three Levels of Prevention (18 marks)

A hospital cluster has 8500 beds serving a population of 1 200 000. The population is forecast to rise to 1 320 000, and the government wants the number of beds per 1000 people to stay at the present level.

#	Question	Answer and marking notes
9	Calculate the present number of beds per 1000 people to two decimal places.	Answer: 8500 / 1 200 000 x 1000 = 7.08, that is 7.08 beds per 1000 people.
10	If the number of beds does not change and the population rises to 1 320 000, find the new beds per 1000 people (two decimal places).	Answer: 8500 / 1 320 000 x 1000 = 6.44, that is 6.44 beds per 1000, which is 0.64 fewer than now, so the service is under more pressure.
11	How many extra beds are needed to keep the present level per 1000 people, to the nearest whole bed?	Answer: the beds needed are 7.08 x 1 320 000 / 1000 = 9345.6, that is 9346 beds, so 9346 – 8500 = 846 extra beds are required.
12	Explain primary, secondary and tertiary prevention and give one example of each in a hospital cluster.	Answer: primary prevention lowers risk before disease starts, such as community vaccination and smoking cessation support. Secondary prevention finds disease early and treats it promptly, such as cervical screening and diabetic eye checks. Tertiary prevention limits disability and relapse after disease has occurred, such as rehabilitation after a stroke and long term follow-up for heart patients. All three must run together, because hospital beds alone cannot improve population health.

Section D: Poverty and Social Care (18 marks)

The monthly household income of a four person household in a district has a median of 24 000 dollars, and the government uses 50 per cent of the median as the relative poverty line. The district has 8000 households and 1440 of them have income below the line. A community support scheme helps 240 of these households raise their income above the line.

#	Question	Answer and marking notes
13	Calculate the monthly relative poverty line for a four person household in the district.	Answer: 24 000 x 50 per cent = 12000 dollars. The line moves with the overall median income, which is the idea of relative poverty.
14	Calculate the poverty rate of the district to one decimal place.	Answer: 1440 / 8000 x 100 = 18.0 per cent.
15	After the scheme, find the new poverty rate (one decimal place) and state how many percentage points it fell.	Answer: (1440 – 240) / 8000 x 100 = 15.0 per cent, a fall of 18.0 – 15.0 = 3.0 percentage points.
16	Compare absolute and relative poverty and explain why cash assistance alone cannot solve poverty.	Answer: absolute poverty is defined by whether a household can afford basic necessities such as food, shelter and health care, while relative poverty is defined by the gap from the median income, so a household may be counted as poor even when basic needs are met. Cash assistance quickly reduces income shortfall but does not deal with the causes, such as job opportunities, education, housing, health and poverty across generations. It must be combined with employment support, learning support for children, community care and health services, with the poverty line and data reviewed regularly.

Section E: Health Promotion and School Intervention (18 marks)

A school has 1250 students, of whom 75 smoke. The school enrols 40 smoking students in a six month cessation programme and 25 of them quit. The school aims to bring the smoking rate down to 4.0 per cent.

#	Question	Answer and marking notes
17	Calculate the school smoking rate before the programme to one decimal place.	Answer: 75 / 1250 x 100 = 6.0 per cent.
18	Calculate the success rate of the programme to one decimal place.	Answer: 25 / 40 x 100 = 62.5 per cent. The 35 smoking students who did not join are still untreated and are the next target group.
19	After the programme, find the new school smoking rate (one decimal place) and decide whether the 4.0 per cent target is met.	Answer: 75 – 25 = 50 students still smoke, so 50 / 1250 x 100 = 4.0 per cent, which meets the 4.0 per cent target.
20	From a health promotion perspective, propose two measures the school should keep running and explain how each changes behaviour.	Answer: first, extend the smoke free zone to the area around the school with student leaders on patrol, which cuts environmental cues and peer pressure and makes smoking inconvenient. Second, give the 35 students who did not join personal counselling plus referral for nicotine replacement therapy, and use peer sharing of success stories to raise self efficacy. The measures should be paired with regular measurement of the smoking rate, parent involvement and referral to community cessation services. A one off activity lowers the rate, but without continued measures relapse will push it back up.

Section F: Dietary Energy and Nutrient Proportions (16 marks)

A Secondary 4 student takes 260 g of carbohydrate, 75 g of protein and 60 g of fat in one day. Energy values are 4 kcal per gram of carbohydrate, 4 kcal per gram of protein and 9 kcal per gram of fat. The usual guidance is 50 to 60 per cent of energy from carbohydrate, 10 to 15 per cent from protein and not more than 30 per cent from fat.

#	Question	Answer and marking notes
21	Calculate the total energy of the diet in kcal.	Answer: carbohydrate 260 x 4 = 1040 kcal, protein 75 x 4 = 300 kcal and fat 60 x 9 = 540 kcal, giving 1040 + 300 + 540 = 1880 kcal in total.
22	Calculate the percentage of energy from carbohydrate to two decimal places.	Answer: 1040 / 1880 x 100 = 55.32 per cent, which is inside the suggested 50 to 60 per cent range.
23	Calculate the percentage of energy from protein and from fat, each to two decimal places.	Answer: protein 300 / 1880 x 100 = 15.96 per cent and fat 540 / 1880 x 100 = 28.72 per cent; the three add up to 55.32 + 15.96 + 28.72 = 100.00 per cent, close to 100 per cent.
24	Compared with the guidance, identify one aspect that should be improved and suggest a specific food swap.	Answer: fat at 28.72 per cent meets the limit of not more than 30 per cent, while protein at 15.96 per cent is slightly above the 10 to 15 per cent range. Part of the meat can be replaced by beans or fish and fried food reduced, keeping carbohydrate inside the range instead of replacing it with high fat food.

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