Conversions

A 72-year-old female (weight: 64 kg, SCr: 1.3 mg/dL) needs renal dosing guidance.
What is her estimated CrCl (mL/min)?

Round to nearest whole number.

Answer explanation

CrCl (Cockcroft-Gault)

Given:
Age = 72, Female, Weight = 64 kg, SCr = 1.3
Formula:
CrCl = [(140 - age) × weight] / (72 × SCr) × 0.85 (for female)
= [(140 - 72) × 64] / (72 × 1.3) × 0.85
= (68 × 64) / 93.6 × 0.85
= 4352 / 93.6 × 0.85 ≈ 46.5 × 0.85 ≈ 40 mL/min ✅
Answer: 40

A male patient is 5'10" and weighs 100 kg.
What is his Adjusted Body Weight (AdjBW) in kg?

Answer explanation

Adjusted Body Weight (AdjBW)

Given:
Height = 70 in, ABW = 100 kg
IBW (Male) = 50 + 2.3(70–60) = 73 kg
AdjBW = IBW + 0.4(ABW – IBW) = 73 + 0.4(27) = 83.8 kg ✅
Answer: 83.8

Lab values: Na = 141, Cl = 106, HCO₃ = 18
What is the anion gap?

Answer explanation

Formula:
AG = Na – Cl – HCO₃ = 141 – 106 – 18 = 17 ✅
Answer: 17

Albumin = 2.6 g/dL, Serum Ca = 7.8 mg/dL
What is the corrected calcium?

Answer explanation

Corrected Calcium

Formula:
Corrected Ca = Measured Ca + 0.8 × (4 – Albumin)
= 7.8 + 0.8 × (4 – 2.6)
= 7.8 + 0.8 × 1.4 = 8.92 mg/dL ✅
Answer: 8.9

A female patient is 65 inches tall and weighs 140 lb.
What is her BMI?
Round to 1 decimal place

Answer explanation

Convert height: 65 in = 1.65 m, Weight = 140 lb = 63.6 kg
BMI = kg / m² = 63.6 / (1.65²) ≈ 63.6 / 2.7225 ≈ 23.4 ✅
Answer: 23.4

A patient is 160 cm tall and weighs 70 kg.
What is the BSA?
Round to 2 decimal places

Answer explanation

Formula: √[(height cm × weight kg)/3600]
= √[(160 × 70)/3600] = √(11200 / 3600) = √3.11 = 1.76 m² ✅
Answer: 1.76

A patient weighs 76 kg.
What is the estimated daily fluid requirement in mL?

Answer explanation

For weight >20 kg: 1500 mL + 20 × (kg – 20)
= 1500 + 20 × (76 – 20) = 1500 + 20 × 56 = 1500 + 1120 = 2620 mL/day ✅
Answer: 2620

How many mL of 25% stock solution are needed to prepare 500 mL of a 10% solution?

Answer explanation

C1 = 25%, Q2 = 500 mL, C2 = 10%
Q1 = (Q2 × C2) / C1 = (500 × 10) / 25 = 200 mL ✅
Answer: 200

You need to make 60 mL of a 12% solution using 5% and 20% stock concentrations.
How many mL of the 20% solution will you use?

Answer explanation

Parts high = 12 – 5 = 7
Parts low = 20 – 12 = 8
Total parts = 15
Volume needed = 60 × (8/15) = 32 mL of 20% ✅
Answer: 32

You are preparing a TPN with the following:

• Dextrose 20% (200 mL)

• Amino acids 10% (250 mL)
  What is the total osmolarity (mOsm/L) of the TPN solution?
  Use:
  Dextrose = 5 × (%)(mL)
  Amino Acids = 10 × (%)(mL)
  Assume final volume = 1 L

Answer explanation

Dextrose: 5 × 20 × 200 = 20,000 mOsm
AA: 10 × 10 × 250 = 25,000 mOsm
Total = 45,000 mOsm
Final Volume = 1000 mL = 1 L
Osmolarity = 45,000 / 1000 = 450 mOsm/L ✅
Answer: 450