Health Assessment: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Clinical assessment scores

Glasgow Coma Scale
GCS = E (1–4) + V (1–5) + M (1–6) = 3–15
APGAR
5 signs × 0–2 = 0–10; 7–10 reassuring, 4–6 moderately abnormal, 0–3 low
Braden scale
5 × (1–4) + 1 × (1–3) = 6–23; ≤ 18 at risk
Morse fall scale
falls 25, diagnosis 15, aid 15/30, IV 20, gait 10/20, mental status 15
NEWS2 risk
0–4 low; single 3 low–medium; 5–6 medium; ≥ 7 high
Mean arterial pressure
MAP = (SBP + 2·DBP)/3 = DBP + PP/3
Pulse pressure and shock index
PP = SBP − DBP, SI = HR/SBP
Body mass index
BMI = kg/m² = 703·lb/in²

Method

  1. Find the score sheet that the question uses and its range (GCS 3–15, APGAR 0–10, Braden 6–23, Morse 0–125).
  2. Score every item separately, matching the description to its level.
  3. Add the item scores and check that the total lies inside the range.
  4. Apply the cut-offs: for NEWS2 also look for any single parameter scoring 3.
  5. For derived values use MAP = (SBP + 2·DBP)/3, PP = SBP − DBP, SI = HR/SBP and BMI = kg/m².

Arterial blood gas interpretation

Normal ranges
pH 7.35–7.45, PaCO₂ 35–45 mmHg (4.7–6.0 kPa), HCO₃⁻ 22–26 mmol/L
ROME
Respiratory: PaCO₂ opposite to pH; Metabolic: HCO₃⁻ equal to pH
Winter’s formula
PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2
Metabolic alkalosis
PaCO₂ ≈ 0.7·HCO₃⁻ + 21
Anion gap
AG = Na⁺ − (Cl⁻ + HCO₃⁻) ≈ 8–12
Albumin-corrected gap
AG + 2.5·(4.0 − alb g/dL) = AG + 0.25·(40 − alb g/L)
Delta ratio
(AG − 12)/(24 − HCO₃⁻)
P/F ratio
PaO₂/FiO₂
Alveolar gas equation
PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8
Serum osmolality
2·Na⁺ + glucose + urea (2·Na⁺ + glucose/18 + BUN/2.8)

Method

  1. Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
  2. Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
  3. Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
  4. In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
  5. Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.

Nursing math & unit conversions

Metric steps
1 kg = 1000 g, 1 g = 1000 mg, 1 mg = 1000 mcg, 1 L = 1000 mL
Weight and length
1 kg = 2.2 lb, 1 lb = 16 oz, 1 in = 2.54 cm, 1 ft = 12 in
Household measures
1 tsp = 5 mL, 1 tbsp = 15 mL, 1 fl oz = 30 mL, 1 cup = 8 fl oz = 240 mL
Temperature
°C = (°F − 32)·5/9, °F = °C·9/5 + 32
24-hour clock
PM hours 1–11: add 12; 12 AM = 00:xx; 12 PM = 12:xx
Dose rate
mcg/min × 60/1000 = mg/h
Laboratory factors
glucose 1 mmol/L = 18 mg/dL; creatinine 1 mg/dL = 88.4 µmol/L; Hb 1 g/dL = 10 g/L

Method

  1. Write the given quantity with its unit and the unit you want.
  2. Pick the equivalent: 1000 between kg, g, mg and mcg or L and mL; 1 kg = 2.2 lb; 1 tsp = 5 mL, 1 tbsp = 15 mL, 1 fl oz = 30 mL, 1 cup = 240 mL.
  3. Write it as a fraction so that the unwanted unit cancels; for a rate, use one fraction per unit that changes.
  4. Check the direction: a smaller unit gives a bigger number.
  5. Round only the final answer, to the precision the question names.

2Practice problems

1.Clinical assessment scoresMedium
S? / 4+M? / 4+A? / 4+Mob? / 4+N? / 4+F? / 3Σ?

Practice exercise (made-up values, not clinical guidance). A practice patient is assessed with the Braden scale: sensory perception: slightly limited; moisture: rarely moist; activity: walks occasionally; mobility: very limited; nutrition: very poor; friction and shear: potential problem. Find the Braden total B.

6–23
B =
2.Clinical assessment scoresHard
underweighthealthy weightoverweightobesity class Iobesity class IIobesity class III

Practice exercise (made-up values, not clinical guidance). An adult weighs 77 kg and is 158 cm tall. Calculate the BMI and click the WHO weight category.

77 kg158 cm
  1. 1underweight
  2. 2healthy weight
  3. 3overweight
  4. 4obesity class I
  5. 5obesity class II
  6. 6obesity class III
3.Arterial blood gas interpretationMedium

A practice gas shows PaCO₂ 32 mmHg (4.3 kPa). Use the rule that in acute respiratory alkalosis HCO₃⁻ falls by 2 mmol/L for every 10 mmHg that PaCO₂ is below 40, starting from HCO₃⁻ 24 mmol/L, to find the expected HCO₃⁻.

PaCO₂ 32 mmHg (4.3 kPa)baseline: 40 mmHg, 24 mmol/L
HCO₃⁻ = mmol/L
4.Arterial blood gas interpretationHard

A practice gas shows a metabolic alkalosis with HCO₃⁻ 35 mmol/L. Use expected PaCO₂ = 0.7·HCO₃⁻ + 21 to find the expected PaCO₂.

HCO₃⁻ 35 mmol/L
PaCO₂ = mmHg
5.Nursing fundamentalsMedium

A patient with capacity wants to leave hospital against medical advice. What should the nurse do?

  1. AExplain the risks, inform the doctor and document the discussion
  2. BHide the patient's clothes so they cannot leave
  3. CLet them go without telling anyone
  4. DLock the ward door to keep the patient in until the doctor arrives
6.Nursing fundamentalsHard

Which patient should the nurse see first?

  1. AA patient who is waiting for their discharge letter and medicines
  2. BA postoperative patient with SpO₂ 89% on air and a respiratory rate of 28
  3. CA patient who is asking for help to walk to the toilet now
  4. DA patient with a pain score of 3/10 whose analgesia is due in an hour
7.Nursing math & unit conversionsMedium
139.684 gtt/min? gtt/minround to 1 gtt/min

Practice exercise (made-up values, not clinical guidance). A drip-rate calculation gives 139.684 gtt/min. The rule in this exercise is that drip rates are rounded to the nearest whole drop. What value x do you record?

139.684 gtt/min
x = gtt/min
8.Nursing math & unit conversionsHard
187 mg/dL? mmol/L1 mmol/L = 18 mg/dL

Practice exercise (made-up values, not clinical guidance). A blood glucose result is 187 mg/dL. Using 1 mmol/L = 18 mg/dL for glucose, find c in mmol/L. Round to one decimal place.

187 mg/dL1 mmol/L = 18 mg/dL
c = mmol/L

3Solutions

1.Clinical assessment scores

Answer: B = 15

Braden scale: six subscales for pressure-injury risk. Sensory perception, moisture, activity, mobility and nutrition score 1 to 4, from the worst level (1) to no problem (4); friction and shear scores 1 to 3. The total runs from 6 to 23, and a lower total means a higher risk. Common bands: 19–23 not at risk, 15–18 mild risk, 13–14 moderate risk, 10–12 high risk and 9 or below very high risk.

Score each subscale

sensory perception: slightly limited → 3
moisture: rarely moist → 4
activity: walks occasionally → 3
mobility: very limited → 2
nutrition: very poor → 1
friction and shear: potential problem → 2

Add

Braden total = 3 + 4 + 3 + 2 + 1 + 2 = 15

Answer

B = 15
2.Clinical assessment scores

Answer: obesity class I

Body mass index: BMI = weight (kg)/height (m)², with the height converted to metres first. With pounds and inches, BMI = 703 × weight (lb)/height (in)². WHO adult categories: below 18.5 underweight, 18.5–24.9 healthy weight, 25.0–29.9 overweight, 30.0–34.9 obesity class I, 35.0–39.9 obesity class II and 40 or more obesity class III.

BMI

BMI = 77/1.58² = 30.84

Category

obesity class I

Answer

obesity class I
3.Arterial blood gas interpretation

Answer: HCO₃⁻ = 22.4 mmol/L

Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.

Change in PaCO₂

|32 − 40| = 8 mmHg → 0.8 × 10 mmHg

Expected HCO₃⁻

HCO₃⁻ = 24 − 2·0.8 = 22.4 mmol/L

Answer

HCO₃⁻ = 22.4 mmol/L
4.Arterial blood gas interpretation

Answer: PaCO₂ = 45.5 mmHg

Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.

Expected PaCO₂

0.7·35 + 21 = 45.5 mmHg

Answer

PaCO₂ = 45.5 mmHg

The rise has a limit: PaCO₂ rarely goes above about 55 mmHg from compensation alone.

5.Nursing fundamentals

Answer: A · Explain the risks, inform the doctor and document the discussion

Why this is the answer

An adult with capacity can discharge themselves. The nurse explains the risks, involves the doctor, offers follow-up and documents the decision; detaining them would be unlawful.

Answer

A · Explain the risks, inform the doctor and document the discussion
6.Nursing fundamentals

Answer: B · A postoperative patient with SpO₂ 89% on air and a respiratory rate of 28

Why this is the answer

Low saturation with a high respiratory rate shows a breathing problem that could worsen quickly. The other needs are real but not urgent.

Answer

B · A postoperative patient with SpO₂ 89% on air and a respiratory rate of 28
7.Nursing math & unit conversions

Answer: x = 140 gtt/min

Rounding: look at the digit just after the place you round to: 5 or more rounds up, 4 or less leaves the digit as it is. Round only the final answer, never the steps. Typical exam rules: volumes of 1 mL or more to the nearest tenth, volumes below 1 mL to the nearest hundredth, drip rates to whole drops and weights to one decimal place; always follow the rule the question states.

Digit in the ones place

139.684 → 9

Only the next digit decides

139.684 → 6
6 is 5 or more, so round 9 up.
The digits after it (84) are ignored: round once, never step by step.

Round

x = 140 gtt/min

Answer

x = 140 gtt/min
8.Nursing math & unit conversions

Answer: c = 10.4 mmol/L

Dimensional analysis: write the given quantity, then multiply by conversion factors written as fractions (for example 60 min/1 h or 1 mg/1000 mcg) so that each unwanted unit appears once on top and once underneath and cancels. When only the wanted unit is left, calculate. Use the equivalents the question gives and round only the final answer.

Convert

c = 187/18 = 10.39 ≈ 10.4 mmol/L

Answer

c = 10.4 mmol/L

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