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
H? / 25+D? / 15+Aid? / 30+IV? / 20+G? / 20+MS? / 15Σ?

Practice exercise (made-up values, not clinical guidance). A practice patient is assessed with the Morse fall scale: history of falling in the last 3 months: no; secondary diagnosis: yes; ambulatory aid: crutches, cane or walker; IV therapy or saline lock: no; gait: impaired; mental status: knows own limits. Find the Morse total MFS.

0–125
MFS =
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 16/min, SpO₂ 94% on room air, blood pressure 124/74 mmHg, pulse 38/min, alert, temperature 36.6 °C. Which clinical risk group does the patient fall into? Click it.

RR 16SpO₂ 94%airSBP 124HR 38AT 36.6 °C
  1. 1low
  2. 2low–medium
  3. 3medium
  4. 4high
3.Arterial blood gas interpretationMedium

A practice gas shows a metabolic acidosis with HCO₃⁻ 6 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.

HCO₃⁻ 6 mmol/L
PaCO₂ = mmHg
4.Arterial blood gas interpretationHard

At sea level, a practice gas taken on 28% oxygen shows PaO₂ 82 mmHg (10.9 kPa) and PaCO₂ 51 mmHg (6.8 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.

FiO₂ 28%PaO₂ 82 mmHg (10.9 kPa)PaCO₂ 51 mmHg (6.8 kPa)
A–a gradient = mmHg
5.Nursing fundamentalsMedium

Which care best protects the skin of a patient with frequent urinary incontinence?

  1. AGentle cleansing after each episode and a barrier product
  2. BLeaving the skin wet so it does not dry out
  3. CApplying talcum powder thickly after each episode
  4. DScrubbing the area with soap and hot water twice a day
6.Nursing fundamentalsHard

When is a "clean" (non-sterile) technique acceptable instead of an aseptic technique?

  1. AFor procedures that do not enter sterile body areas, such as giving an oral medicine
  2. BFor inserting a urinary catheter
  3. CFor inserting a central venous catheter
  4. DFor dressing a fresh surgical wound in theatre, provided that non-sterile gloves are worn throughout
7.Nursing math & unit conversionsMedium
4 ft 1 in1 ft = 12 in1 in = 2.54 cmh = ? cm

Practice exercise (made-up values, not clinical guidance). A patient is 4 ft 1 in tall. Using 1 ft = 12 in and 1 in = 2.54 cm, find the height h in centimetres, to the nearest whole centimetre.

4 ft 1 in1 ft = 12 in1 in = 2.54 cm
h = cm
8.Nursing math & unit conversionsHard
2 tsp · 4/day · 14 d? × 50 mL1 tsp = 5 mL

Practice exercise (made-up values, not clinical guidance). A prescription reads 2 tsp of an oral suspension every 6 hours for 14 days. The suspension comes in 50 mL bottles. Using 1 tsp = 5 mL, how many whole bottles N must be supplied so the course is complete?

2 tspevery 6 hours14 days50 mL1 tsp = 5 mL
N =

3Solutions

1.Clinical assessment scores

Answer: MFS = 50

Morse fall scale: six items. History of falling: no 0, yes 25. Secondary diagnosis: no 0, yes 15. Ambulatory aid: none, bed rest or nurse assistance 0; crutches, cane or walker 15; holding on to furniture 30. IV therapy or saline lock: no 0, yes 20. Gait: normal, bed rest or wheelchair 0; weak 10; impaired 20. Mental status: knows own limits 0; overestimates or forgets limits 15. The total runs from 0 to 125. Risk cut-offs differ between hospitals; a common set is 0–24 low, 25–44 moderate and 45 or more high.

Score each item

history of falling in the last 3 months: no → 0
secondary diagnosis: yes → 15
ambulatory aid: crutches, cane or walker → 15
IV therapy or saline lock: no → 0
gait: impaired → 20
mental status: knows own limits → 0

Add

Morse total = 0 + 15 + 15 + 0 + 20 + 0 = 50

Answer

MFS = 50
2.Clinical assessment scores

Answer: low–medium

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 16/min → 0
SpO₂ 94% → 1
on air → 0
SBP 124 mmHg → 0
HR 38/min → 3
alert (A) → 0
T 36.6 °C → 0

Total and red scores

NEWS2 = 0 + 1 + 0 + 0 + 3 + 0 + 0 = 4
At least one parameter scores 3.

Risk group

low–medium

Answer

low–medium

A single score of 3 needs an urgent review even when the total is low.

3.Arterial blood gas interpretation

Answer: PaCO₂ = 17 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.

Winter’s formula

PaCO₂ = 1.5·6 + 8 = 17 mmHg
range: 15–19 mmHg

Answer

PaCO₂ = 17 mmHg
4.Arterial blood gas interpretation

Answer: A–a gradient = 53.89 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 − 51/0.8 = 199.6 − 63.75 = 135.9 mmHg

Gradient

A–a = 135.9 − 82 = 53.89 mmHg

Answer

A–a gradient = 53.89 mmHg
5.Nursing fundamentals

Answer: A · Gentle cleansing after each episode and a barrier product

Why this is the answer

Moisture and urine break down the skin barrier. Prompt, gentle cleansing with a pH-balanced cleanser and a barrier cream or film protects against incontinence-associated dermatitis.

Answer

A · Gentle cleansing after each episode and a barrier product
6.Nursing fundamentals

Answer: A · For procedures that do not enter sterile body areas, such as giving an oral medicine

Why this is the answer

Aseptic technique is required when entering normally sterile sites or handling key parts of invasive devices. Routine non-invasive tasks only need standard precautions.

Answer

A · For procedures that do not enter sterile body areas, such as giving an oral medicine
7.Nursing math & unit conversions

Answer: h = 124 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

4·12 + 1 = 49 in

Inches to centimetres

h = 49·2.54 = 124.5 ≈ 124 cm

Answer

h = 124 cm
8.Nursing math & unit conversions

Answer: N = 12

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

10 mL × 4 × 14 = 560 mL

Bottles

560/50 = 11.2 → N = 12

Answer

N = 12

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