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: no impairment; moisture: constantly moist; activity: chairfast; mobility: slightly limited; nutrition: very poor; friction and shear: problem. Find the Braden total B.

6–23
B =
2.Clinical assessment scoresHard
not at riskmild riskmoderate riskhigh riskvery high risk

Practice exercise (made-up values, not clinical guidance). A practice patient is assessed with the Braden scale: sensory perception: completely limited; moisture: constantly moist; activity: bedfast; mobility: very limited; nutrition: probably inadequate; friction and shear: problem. Use the bands 19–23 not at risk, 15–18 mild, 13–14 moderate, 10–12 high and 9 or below very high. Click the risk level.

  1. 1not at risk
  2. 2mild risk
  3. 3moderate risk
  4. 4high risk
  5. 5very high risk
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
pH7.357.457.28PaCO₂mmHg354524HCO₃⁻mmol/L222611appropriate compensationadded respiratory acidosisadded respiratory alkalosis

A practice gas shows a metabolic acidosis: pH 7.28, PaCO₂ 24 mmHg (3.2 kPa), HCO₃⁻ 11 mmol/L. Compare the measured PaCO₂ with Winter’s formula. Click your interpretation.

pH 7.28PaCO₂ 24 mmHg (3.2 kPa)HCO₃⁻ 11 mmol/L
  1. 1appropriate compensation
  2. 2added respiratory acidosis
  3. 3added respiratory alkalosis
5.Nursing fundamentalsMedium

A dying patient is no longer drinking and breathes through the mouth. What mouth care is most appropriate?

  1. AClean the mouth only once a day with a hard toothbrush
  2. BGive large drinks of water every hour
  3. CAvoid mouth care so the patient is not disturbed
  4. DKeep the mouth and lips moist with frequent gentle care
6.Nursing fundamentalsHard

Why must a needlestick injury be reported straight away rather than at the end of the shift?

  1. AThe source patient will be discharged soon
  2. BPost-exposure prophylaxis works best when it is started as soon as possible
  3. CThe injury must be documented within ten minutes for the report to be valid
  4. DReporting late means the nurse has to pay for the tests
7.Nursing math & unit conversionsMedium
94969810010210410635363738394041°F°C? °F36.8 °C

Practice exercise (made-up values, not clinical guidance). A patient’s temperature is 36.8 °C. Convert it to degrees Fahrenheit T with °F = °C × 9/5 + 32. Round to one decimal place.

36.8 °C
T = °F
8.Nursing math & unit conversionsHard
50 mL/h? L/day1 day = 24 h1 L = 1000 mL

Practice exercise (made-up values, not clinical guidance). A maintenance infusion runs at 50 mL/h without a break. How many litres V does it deliver in 24 hours?

50 mL/h24 h
V = L

3Solutions

1.Clinical assessment scores

Answer: B = 12

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: no impairment → 4
moisture: constantly moist → 1
activity: chairfast → 2
mobility: slightly limited → 3
nutrition: very poor → 1
friction and shear: problem → 1

Add

Braden total = 4 + 1 + 2 + 3 + 1 + 1 = 12

Answer

B = 12
2.Clinical assessment scores

Answer: very high risk

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: completely limited → 1
moisture: constantly moist → 1
activity: bedfast → 1
mobility: very limited → 2
nutrition: probably inadequate → 2
friction and shear: problem → 1

Add

Braden total = 1 + 1 + 1 + 2 + 2 + 1 = 8

Risk level

8 → very high risk

Answer

very high risk
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: appropriate compensation

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

1.5·11 + 8 = 24.5 → 22.5–26.5 mmHg

Compare

measured: 24 mmHg → appropriate compensation

Answer

appropriate compensation
5.Nursing fundamentals

Answer: D · Keep the mouth and lips moist with frequent gentle care

Why this is the answer

Frequent gentle mouth care with moist swabs, water sprays and lip balm keeps the patient comfortable. Forcing fluids risks choking and aspiration.

Answer

D · Keep the mouth and lips moist with frequent gentle care
6.Nursing fundamentals

Answer: B · Post-exposure prophylaxis works best when it is started as soon as possible

Why this is the answer

After a high-risk exposure, for example to HIV or hepatitis B, preventive treatment is most effective when started early. Prompt reporting also allows the source patient to be assessed.

Answer

B · Post-exposure prophylaxis works best when it is started as soon as possible
7.Nursing math & unit conversions

Answer: T = 98.2 °F

Temperature: °C = (°F − 32) × 5/9 and °F = °C × 9/5 + 32. Subtract 32 before you multiply when going from °F, and add 32 after you multiply when going from °C. A quick check: 37 °C is 98.6 °F, and every 1 °C is 1.8 °F.

Multiply by 9/5

36.8·9/5 = 66.24

Add 32

T = 66.24 + 32 = 98.24 ≈ 98.2 °F

Answer

T = 98.2 °F
8.Nursing math & unit conversions

Answer: V = 1.2 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.

Volume in a day

50 mL/h × 24 h = 1200 mL

In litres

V = 1200/1000 = 1.2 L

Answer

V = 1.2 L

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