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: very moist; activity: walks frequently; mobility: slightly limited; nutrition: adequate; friction and shear: no apparent 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 115 kg and is 164 cm tall. Calculate the BMI and click the WHO weight category.

115 kg164 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⁺ 148 mmol/L, glucose 4.2 mmol/L and urea 24 mmol/L. Calculate the serum osmolality Osm = 2·Na⁺ + glucose + urea to the nearest whole number.

Na⁺ 148 mmol/Lglucose 4.2 mmol/Lurea 24 mmol/L
Osm = mOsm/kg
4.Arterial blood gas interpretationHard
pH7.357.457.22PaCO₂mmHg354520HCO₃⁻mmol/L22268added normal-gap acidosispure high-gap acidosisadded metabolic alkalosis

A practice gas shows pH 7.22, PaCO₂ 20 mmHg (2.7 kPa) and HCO₃⁻ 8 mmol/L, with Na⁺ 143 and Cl⁻ 112 mmol/L. Use the anion gap and the delta ratio to click the best interpretation.

pH 7.22PaCO₂ 20 mmHg (2.7 kPa)HCO₃⁻ 8 mmol/LNa⁺ 143 mmol/LCl⁻ 112 mmol/L
  1. 1added normal-gap acidosis
  2. 2pure high-gap acidosis
  3. 3added metabolic alkalosis
5.Nursing fundamentalsMedium

A patient shouts at the nurse that nobody has told them when their operation will be. What is the best first response?

  1. AArgue back that the staff are very busy
  2. BWalk away until the patient calms down by themselves
  3. CStay calm, listen and acknowledge their frustration
  4. DTell the patient to stop shouting or leave the ward
6.Nursing fundamentalsHard

Which pair of details is best for confirming a patient's identity before giving a medicine?

  1. AThe name on the door and the patient's face
  2. BFull name and date of birth, checked against the wristband
  3. CBed number and the patient's first name, checked against the ward list
  4. DRoom number and diagnosis
7.Nursing math & unit conversionsMedium
0.50627 mL? mLround to 0.01 mL

Practice exercise (made-up values, not clinical guidance). A dose calculation for a small injection gives 0.50627 mL. The rule in this exercise is that volumes below 1 mL are rounded to the nearest hundredth of a mL. What value x do you record?

0.50627 mL
x = mL
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 four times a day 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 tspfour times a day14 days50 mL1 tsp = 5 mL
N =

3Solutions

1.Clinical assessment scores

Answer: B = 19

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: very moist → 2
activity: walks frequently → 4
mobility: slightly limited → 3
nutrition: adequate → 3
friction and shear: no apparent problem → 3

Add

Braden total = 4 + 2 + 4 + 3 + 3 + 3 = 19

Answer

B = 19
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 = 115/1.64² = 42.76

Category

obesity class III

Answer

obesity class III
3.Arterial blood gas interpretation

Answer: Osm = 324 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 = 2·148 + 4.2 + 24 = 324.2 ≈ 324 mOsm/kg

Answer

Osm = 324 mOsm/kg
4.Arterial blood gas interpretation

Answer: added normal-gap acidosis

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.

Anion gap

AG = 143 − (112 + 8) = 23

Delta ratio

(23 − 12)/(24 − 8) = 0.6875
added normal-gap acidosis

Answer

added normal-gap acidosis
5.Nursing fundamentals

Answer: C · Stay calm, listen and acknowledge their frustration

Why this is the answer

Listening calmly and acknowledging feelings often de-escalates anger. The nurse then finds out the answer or explains what will happen next. Staff safety is maintained throughout.

Answer

C · Stay calm, listen and acknowledge their frustration
6.Nursing fundamentals

Answer: B · Full name and date of birth, checked against the wristband

Why this is the answer

At least two personal identifiers, such as name, date of birth and hospital number, are checked with the patient and the wristband. Bed and room numbers change and are not identifiers.

Answer

B · Full name and date of birth, checked against the wristband
7.Nursing math & unit conversions

Answer: x = 0.51 mL

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 hundredths place

0.50627 → 0

Only the next digit decides

0.50627 → 6
6 is 5 or more, so round 0 up.
The digits after it (27) are ignored: round once, never step by step.

Round

x = 0.51 mL

Answer

x = 0.51 mL
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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