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
11965SBPDBPmmHgHR 133/min · SI = ?

Practice exercise (made-up values, not clinical guidance). A patient has a pulse of 133/min and a blood pressure of 119/65 mmHg. Find the shock index SI = HR/SBP. Round to two decimal places.

HR 133/minBP 119/65 mmHg
SI =
2.Clinical assessment scoresHard
underweighthealthy weightoverweightobesity class Iobesity class IIobesity class III

Practice exercise (made-up values, not clinical guidance). An adult weighs 112 kg and is 150 cm tall. Calculate the BMI and click the WHO weight category.

112 kg150 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

Practice values: Na⁺ 129 mmol/L, glucose 30.3 mmol/L and urea 7.8 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 129 mmol/Lglucose 30.3 mmol/Lurea 7.8 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 30% oxygen shows PaO₂ 58 mmHg (7.7 kPa) and PaCO₂ 40 mmHg (5.3 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 30%PaO₂ 58 mmHg (7.7 kPa)PaCO₂ 40 mmHg (5.3 kPa)
A–a gradient = mmHg
5.Nursing fundamentalsMedium

What does the Malnutrition Universal Screening Tool (MUST) assess?

  1. ASwallowing ability
  2. BThe level of pain
  3. CThe risk of malnutrition
  4. DThe risk of falling
6.Nursing fundamentalsHard

A patient with a valid DNACPR (DNR) decision, whose death was expected, stops breathing. What should the nurse do?

  1. AStart CPR but stop after five minutes
  2. BNot start CPR, note the time, inform the team and support the family
  3. CStart CPR and call the cardiac arrest team
  4. DLeave the patient and ask the family to wait outside until the doctor comes
7.Nursing math & unit conversionsMedium
67.819 gtt/min? gtt/minround to 1 gtt/min

Practice exercise (made-up values, not clinical guidance). A drip-rate calculation gives 67.819 gtt/min. The rule in this exercise is that drip rates are rounded to the nearest whole drop. What value x do you record?

67.819 gtt/min
x = gtt/min
8.Nursing math & unit conversionsHard
1 tsp · 4/day · 5 d? × 150 mL1 tsp = 5 mL

Practice exercise (made-up values, not clinical guidance). A prescription reads 1 tsp of an oral suspension every 6 hours for 5 days. The suspension comes in 150 mL bottles. Using 1 tsp = 5 mL, how many whole bottles N must be supplied so the course is complete?

1 tspevery 6 hours5 days150 mL1 tsp = 5 mL
N =

3Solutions

1.Clinical assessment scores

Answer: SI = 1.12

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 = 133/119 = 1.118 ≈ 1.12

Answer

SI = 1.12

1 or more: a warning sign of shock in this practice case.

2.Clinical assessment scores

Answer: obesity class III

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 = 112/1.5² = 49.78

Category

obesity class III

Answer

obesity class III
3.Arterial blood gas interpretation

Answer: Osm = 296 mOsm/kg

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.

Calculated osmolality

Osm = 2·129 + 30.3 + 7.8 = 296.1 ≈ 296 mOsm/kg

Answer

Osm = 296 mOsm/kg
4.Arterial blood gas interpretation

Answer: A–a gradient = 105.9 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 = 0.3·713 − 40/0.8 = 213.9 − 50 = 163.9 mmHg

Gradient

A–a = 163.9 − 58 = 105.9 mmHg

Answer

A–a gradient = 105.9 mmHg
5.Nursing fundamentals

Answer: C · The risk of malnutrition

Why this is the answer

MUST combines body mass index, unplanned weight loss and acute disease effect into a score that shows the risk of malnutrition and guides nutritional care.

Answer

C · The risk of malnutrition
6.Nursing fundamentals

Answer: B · Not start CPR, note the time, inform the team and support the family

Why this is the answer

CPR is not attempted when a valid DNACPR decision applies. The nurse notes the time, follows local policy for verifying death, informs the medical team and supports the family.

Answer

B · Not start CPR, note the time, inform the team and support the family
7.Nursing math & unit conversions

Answer: x = 68 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

67.819 → 7

Only the next digit decides

67.819 → 8
8 is 5 or more, so round 7 up.
The digits after it (19) are ignored: round once, never step by step.

Round

x = 68 gtt/min

Answer

x = 68 gtt/min
8.Nursing math & unit conversions

Answer: N = 1

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.

Total volume

5 mL × 4 × 5 = 100 mL

Bottles

100/150 = 0.6667 → N = 1

Answer

N = 1

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