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
mildmoderatesevere

Practice exercise (made-up values, not clinical guidance). A practice patient after a head injury does not open the eyes at all, makes no sounds and shows as the best motor response: no movement at all. Work out the GCS and classify the head injury by it. Click your answer.

  1. 1mild
  2. 2moderate
  3. 3severe
2.Clinical assessment scoresHard
underweighthealthy weightoverweightobesity class Iobesity class IIobesity class III

Practice exercise (made-up values, not clinical guidance). An adult weighs 102 kg and is 172 cm tall. Calculate the BMI and click the WHO weight category.

102 kg172 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
pH7.357.457.28PaCO₂mmHg354564HCO₃⁻mmol/L222629uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.28, PaCO₂ 64 mmHg (8.5 kPa) and HCO₃⁻ 29 mmol/L. The primary disorder is respiratory acidosis. How far is it compensated? Click your answer.

pH 7.28PaCO₂ 64 mmHg (8.5 kPa)HCO₃⁻ 29 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 135, Cl⁻ 112, HCO₃⁻ 11 mmol/L and albumin 30 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 135 mmol/LCl⁻ 112 mmol/LHCO₃⁻ 11 mmol/Lalbumin 30 g/L
AGc = mmol/L
5.Nursing fundamentalsMedium

Which adverse effect is a particular risk with NSAIDs such as ibuprofen?

  1. ABreathing slowing down
  2. BStomach irritation and bleeding
  3. CLow blood sugar
  4. DSevere constipation that lasts for weeks
6.Nursing fundamentalsHard

A postoperative wound drain that had drained little suddenly fills with bright red blood. What should the nurse do?

  1. ACheck the patient's vital signs and report it urgently
  2. BClamp the drain and tell no one
  3. CEmpty the drain and record it at the next routine round
  4. DRemove the drain because it is not working
7.Nursing math & unit conversionsMedium
127.902 gtt/min? gtt/minround to 1 gtt/min

Practice exercise (made-up values, not clinical guidance). A drip-rate calculation gives 127.902 gtt/min. The rule in this exercise is that drip rates are rounded to the nearest whole drop. What value x do you record?

127.902 gtt/min
x = gtt/min
8.Nursing math & unit conversionsHard
½ tbsp · 4/day · 7 d? mL1 tbsp = 15 mL

Practice exercise (made-up values, not clinical guidance). A prescription reads ½ tbsp of an oral suspension every 6 hours for 7 days. Using 1 tbsp = 15 mL, how many mL V are needed for the whole course?

½ tbspevery 6 hours7 days1 tbsp = 15 mL
V = mL

3Solutions

1.Clinical assessment scores

Answer: severe

Glasgow Coma Scale: add the best response in three areas. Eye opening E: 4 spontaneous, 3 to sound or speech, 2 to pressure, 1 none. Verbal V: 5 oriented, 4 confused, 3 inappropriate words, 2 incomprehensible sounds, 1 none. Motor M: 6 obeys commands, 5 localises, 4 normal flexion (withdraws), 3 abnormal flexion, 2 extension, 1 none. The total runs from 3 to 15. After a head injury, 13–15 is usually called mild, 9–12 moderate and 3–8 severe.

Score each area

E: does not open the eyes at all → 1
V: makes no sounds → 1
M: no movement at all → 1

Add

GCS = 1 + 1 + 1 = 3

Severity

GCS 3 → severe

Answer

severe
2.Clinical assessment scores

Answer: obesity class I

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 = 102/1.72² = 34.48

Category

obesity class I

Answer

obesity class I
3.Arterial blood gas interpretation

Answer: partially 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 7.28: acidaemia (below 7.35)
PaCO₂ 64 mmHg: high (35–45)
HCO₃⁻ 29 mmol/L: high (22–26)

Compensation

partially compensated

Answer

partially compensated
4.Arterial blood gas interpretation

Answer: AGc = 14.5 mmol/L

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 = 135 − (112 + 11) = 12

Correction for albumin

12 + 0.25·(40 − 30) = 14.5 mmol/L

Answer

AGc = 14.5 mmol/L

Above about 12 after correction: a high anion gap.

5.Nursing fundamentals

Answer: B · Stomach irritation and bleeding

Why this is the answer

NSAIDs can irritate the stomach lining and cause ulcers and bleeding. They are taken with food, avoided in some patients and may also affect the kidneys.

Answer

B · Stomach irritation and bleeding
6.Nursing fundamentals

Answer: A · Check the patient's vital signs and report it urgently

Why this is the answer

A sudden increase in fresh blood may mean active bleeding. The nurse assesses for signs of shock, such as a rising pulse and falling blood pressure, and escalates immediately.

Answer

A · Check the patient's vital signs and report it urgently
7.Nursing math & unit conversions

Answer: x = 128 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

127.902 → 7

Only the next digit decides

127.902 → 9
9 is 5 or more, so round 7 up.
The digits after it (02) are ignored: round once, never step by step.

Round

x = 128 gtt/min

Answer

x = 128 gtt/min
8.Nursing math & unit conversions

Answer: V = 210 mL

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 per dose

½ tbsp = 0.5·15 = 7.5 mL

Whole course

V = 7.5 mL × 4/day × 7 = 210 mL

Answer

V = 210 mL

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