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
14881SBPDBPmmHgHR 57/min · SI = ?

Practice exercise (made-up values, not clinical guidance). A patient has a pulse of 57/min and a blood pressure of 148/81 mmHg. Find the shock index SI = HR/SBP. Round to two decimal places.

HR 57/minBP 148/81 mmHg
SI =
2.Clinical assessment scoresHard
3210123RR≤89–1112–2021–24≥25SpO₂≤9192–9394–95≥96air / O₂O₂airSBP≤9091–100101–110111–219≥220HR≤4041–5051–9091–110111–130≥131ACVPUAC, V, P, UT °C≤35.035.1–36.036.1–38.038.1–39.0≥39.1lowlow–mediummediumhigh

Practice exercise (made-up values, not clinical guidance). Use the NEWS2 chart (SpO₂ scale 1) for these observations: respiration rate 12/min, SpO₂ 97% on room air, blood pressure 102/62 mmHg, pulse 82/min, alert, temperature 37.6 °C. Which clinical risk group does the patient fall into? Click it.

RR 12SpO₂ 97%airSBP 102HR 82AT 37.6 °C
  1. 1low
  2. 2low–medium
  3. 3medium
  4. 4high
3.Arterial blood gas interpretationMedium
pH7.357.457.42PaCO₂mmHg354529HCO₃⁻mmol/L222618uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.42, PaCO₂ 29 mmHg (3.9 kPa) and HCO₃⁻ 18 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.42PaCO₂ 29 mmHg (3.9 kPa)HCO₃⁻ 18 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 28% oxygen shows PaO₂ 69 mmHg (9.2 kPa) and PaCO₂ 29 mmHg (3.9 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 28%PaO₂ 69 mmHg (9.2 kPa)PaCO₂ 29 mmHg (3.9 kPa)
A–a gradient = mmHg
5.Nursing fundamentalsMedium

Teaching a patient how to inject insulin mainly involves which domain of learning?

  1. APsychomotor
  2. BSpiritual
  3. CCognitive
  4. DAffective
6.Nursing fundamentalsHard

A nurse realises that a dose of analgesia given two hours ago was not documented. What is correct?

  1. AWrite it in the space for two hours ago as if it was recorded then
  2. BWrite it now as a late entry, stating when it was given
  3. CAsk a colleague to sign that they gave it
  4. DLeave it out, as it is now too late to record it in the notes
7.Nursing math & unit conversionsMedium
5 ft 9 in1 ft = 12 in1 in = 2.54 cmh = ? cm

Practice exercise (made-up values, not clinical guidance). A patient is 5 ft 9 in tall. Using 1 ft = 12 in and 1 in = 2.54 cm, find the height h in centimetres, to the nearest whole centimetre.

5 ft 9 in1 ft = 12 in1 in = 2.54 cm
h = cm
8.Nursing math & unit conversionsHard
4 mg/h? mcg/min1 mg = 1000 mcg1 h = 60 min

Practice exercise (made-up values, not clinical guidance). An infusion delivers 4 mg/h. Express this dose rate D in mcg/min. Round to one decimal place.

4 mg/h1 mg = 1000 mcg1 h = 60 min
D = mcg/min

3Solutions

1.Clinical assessment scores

Answer: SI = 0.39

Blood pressure values: pulse pressure PP = SBP − DBP. Mean arterial pressure MAP ≈ (SBP + 2·DBP)/3 = DBP + PP/3, an estimate based on the heart spending about two thirds of each beat in diastole at a normal heart rate (less accurate when the heart rate is fast); a MAP of at least about 65 mmHg is a common target for organ perfusion. The shock index SI = heart rate/SBP is normally about 0.5–0.7; a value above about 1 is a warning sign of shock.

Shock index

SI = 57/148 = 0.3851 ≈ 0.39

Answer

SI = 0.39

Within or below the usual 0.5–0.7.

2.Clinical assessment scores

Answer: low

NEWS2 (National Early Warning Score 2, Royal College of Physicians 2017): score each of the seven parameters with the chart (respiration rate, SpO₂ on scale 1, air or supplemental oxygen, systolic blood pressure, pulse, consciousness on the ACVPU scale and temperature) and add the scores. Clinical risk: 0–4 low; a score of 3 in any single parameter low–medium; 5–6 medium; 7 or more high. A total of 5 or more needs an urgent response.

Score each parameter

RR 12/min → 0
SpO₂ 97% → 0
on air → 0
SBP 102 mmHg → 1
HR 82/min → 0
alert (A) → 0
T 37.6 °C → 0

Total and red scores

NEWS2 = 0 + 0 + 0 + 1 + 0 + 0 + 0 = 1
No parameter scores 3.

Risk group

low

Answer

low
3.Arterial blood gas interpretation

Answer: fully compensated

Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.

ROME, value by value

pH 7.42: within 7.35–7.45
PaCO₂ 29 mmHg: low (35–45)
HCO₃⁻ 18 mmol/L: low (22–26)

Compensation

fully compensated

Answer

fully compensated
4.Arterial blood gas interpretation

Answer: A–a gradient = 94.39 mmHg

Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.

Alveolar oxygen

PAO₂ = 0.28·713 − 29/0.8 = 199.6 − 36.25 = 163.4 mmHg

Gradient

A–a = 163.4 − 69 = 94.39 mmHg

Answer

A–a gradient = 94.39 mmHg
5.Nursing fundamentals

Answer: A · Psychomotor

Why this is the answer

The psychomotor domain covers physical skills, the cognitive domain covers knowledge and the affective domain covers attitudes and feelings. Skills are learned by demonstration and practice.

Answer

A · Psychomotor
6.Nursing fundamentals

Answer: B · Write it now as a late entry, stating when it was given

Why this is the answer

Late entries are clearly marked as such, showing both when care was given and when the note was written. This keeps the record accurate and prevents a second dose being given.

Answer

B · Write it now as a late entry, stating when it was given
7.Nursing math & unit conversions

Answer: h = 175 cm

Household measures and weight: nursing exams use rounded equivalents: 1 teaspoon (tsp) = 5 mL, 1 tablespoon (tbsp) = 15 mL = 3 tsp, 1 fluid ounce (fl oz) = 30 mL = 2 tbsp, and 1 cup = 8 fl oz = 240 mL. For weight, 1 kg = 2.2 lb and 1 lb = 16 oz; for length, 1 inch = 2.54 cm. Set the conversion up so that the unit you do not want cancels out.

Height in inches

5·12 + 9 = 69 in

Inches to centimetres

h = 69·2.54 = 175.3 ≈ 175 cm

Answer

h = 175 cm
8.Nursing math & unit conversions

Answer: D = 66.7 mcg/min

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.

Set up the chain

D = 4 mg/h × (1000 mcg/1 mg) × (1 h/60 min)

Calculate

D = 4000/60 = 66.67 ≈ 66.7 mcg/min

Answer

D = 66.7 mcg/min

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