Practice problems
One worked problem per calculator, grouped by category. Solutions are hidden until you open them, and every problem can be loaded straight into its calculator.
Measurement & Conversions
Liquid (volume) conversions
A discharge prescription directs "2 tablespoonfuls twice daily for 10 days." How many millilitres per dose, and what total volume should be dispensed?
Answer: 2 tbsp = 30 mL per dose. Two doses daily for 10 days = 20 doses × 30 mL = 600 mL to dispense.
Solid (weight) conversions
A patient weighs 176 lb. Convert to kilograms for weight-based dosing.
Answer: 176 lb ÷ 2.2 = 80 kg. This is the total body weight (TBW); check whether the drug should be dosed on TBW, IBW, or AdjBW before proceeding.
Height & length conversions
A patient is 5 ft 9 in tall. Express the height in centimetres for a Mosteller BSA calculation.
Answer: 5 ft 9 in = 69 in; 69 × 2.54 = 175.26 cm.
Temperature conversion (°F ↔ °C)
A patient reports a home temperature of 101.3 °F. Convert to Celsius and classify.
Answer: (101.3 − 32) ÷ 1.8 = 38.5 °C — febrile by the ≥38.0 °C definition.
Milliequivalents ↔ millimoles
A magnesium order is written for 20 mEq. The laboratory system requires the dose in millimoles. Magnesium is divalent.
Answer: 20 mEq ÷ 2 = 10 mmol. Reporting 20 mmol instead would represent a two-fold overdose.
Concentrations & Compounding
Percentage strength (w/v, v/v, w/w)
You need to compound 480 mL of a 2% w/v potassium permanganate soak. How many grams of potassium permanganate are required?
Answer: (2 × 480) ÷ 100 = 9.6 g.
Ratio strength
An older ampoule is labelled "epinephrine 1:1000". Express this as a percentage strength and as mg/mL.
Answer: 100 ÷ 1000 = 0.1% w/v, which is 1 mg/mL. The 1:10,000 presentation is 0.01% = 0.1 mg/mL.
Parts per million (ppm)
A municipal water supply is fluoridated to 0.7 ppm. Express this as a percentage strength and in mg/L.
Answer: 0.7 ÷ 10,000 = 0.00007%. In dilute aqueous solution this is 0.7 mg/L, which is the current US Public Health Service recommended fluoridation level.
Specific gravity
A formula calls for 60 mL of glycerin (specific gravity 1.25). What weight should you place on the balance?
Answer: 1.25 × 60 = 75 g. Measuring 60 g instead would under-deliver glycerin by 20%.
Dilution & concentration (Q₁C₁ = Q₂C₂)
You have 70% w/v stock solution and need 500 mL of a 20% w/v solution. How much stock is required, and how much diluent?
Answer: Q₁ = (500 × 20) ÷ 70 = 142.9 mL of stock, made up to 500 mL with 357.1 mL of diluent.
Alligation (alternate)
You stock 20% and 5% benzoyl peroxide gels. Prepare 240 g of an 8% gel.
Answer: 3 parts of 20% to 12 parts of 5%, a total of 15 parts. For 240 g: (3/15) × 240 = 48 g of 20% and (12/15) × 240 = 192 g of 5%.
Osmolarity (mOsmol/L)
Calculate the osmolarity of 0.9% sodium chloride injection, and the total milliosmoles in a 1 litre bag.
Answer: 0.9% = 9 g/L. (9 ÷ 58.5) × 2 × 1000 = 308 mOsmol/L; a 1 L bag delivers 308 mOsmol. This is why normal saline is described as isotonic with plasma.
Isotonicity — sodium chloride equivalent (E value)
Prepare 30 mL of a 1% w/v solution of a drug with molecular weight 337.4 that dissociates into two ions. How much sodium chloride is needed to make it isotonic?
Answer: E = (58.5 × 1.8) ÷ (337.4 × 1.8) = 0.173. The preparation contains 0.3 g of drug, contributing 0.052 g NaCl equivalent. Isotonic 30 mL needs 0.27 g NaCl, so add 0.27 − 0.052 = 0.218 g.
Moles & millimoles
How many millimoles of sodium chloride are in 500 mg?
Answer: 500 ÷ 58.5 = 8.55 mmol. Since sodium chloride is monovalent, this is also 8.55 mEq of sodium.
Milliequivalents from a salt weight
A potassium chloride tablet contains 750 mg. How many milliequivalents of potassium does it supply?
Answer: (750 × 1) ÷ 74.5 = 10 mEq. This is why 750 mg KCl tablets are labelled as 10 mEq.
Minimum weighable quantity & aliquot planner
A Class III torsion balance has a sensitivity requirement of 6 mg. A formula requires 8 mg of drug. Can it be weighed directly?
Answer: MWQ = 6 ÷ 0.05 = 120 mg. 8 mg is far below that, so an aliquot is mandatory. Weigh 120 mg of drug, dilute to 1800 mg total with 1680 mg of lactose, then weigh 120 mg of the mixture — which contains 8 mg of drug.
Dosing & Body Metrics
Body mass index (BMI)
A patient weighs 95 kg and is 170 cm tall. Calculate BMI and classify.
Answer: 1.70 m² = 2.89; 95 ÷ 2.89 = 32.9 kg/m² — Obesity class I. Since BMI ≥ 30, aminoglycoside dosing would use adjusted body weight.
Ideal body weight (Devine)
A 178 cm male weighs 110 kg. Calculate his ideal and adjusted body weights.
Answer: 178 cm = 70.1 in, which is 10.1 in over 5 feet. IBW = 50 + (2.3 × 10.1) = 73.2 kg. AdjBW = 73.2 + 0.4 × (110 − 73.2) = 87.9 kg.
Adjusted body weight (AdjBW₀.₄)
A 165 cm female weighs 120 kg (BMI 44). Calculate the adjusted body weight for gentamicin dosing.
Answer: 165 cm = 64.96 in, 4.96 in over 5 ft. IBW = 45.5 + (2.3 × 4.96) = 56.9 kg. AdjBW = 56.9 + 0.4 × (120 − 56.9) = 82.1 kg.
Body surface area (Mosteller & Du Bois)
A patient is 170 cm and 70 kg. Calculate the body surface area for a chemotherapy dose.
Answer: (170 × 70) ÷ 3600 = 3.3056; √3.3056 = 1.82 m². Du Bois gives 1.81 m² — a 0.5% difference.
Which weight to use for dosing
A 175 cm, 130 kg male (BMI 42.4) is starting gentamicin. Which weight should the dose be based on?
Answer: IBW = 50 + 2.3 × 8.9 = 70.5 kg. BMI 42.4 means obese, so aminoglycosides use AdjBW = 70.5 + 0.4 × (130 − 70.5) = 94.3 kg.
Renal & Fluids
Creatinine clearance (Cockcroft–Gault)
A 72-year-old female is 160 cm and weighs 65 kg, with a serum creatinine of 1.4 mg/dL. Estimate her creatinine clearance for renal dose adjustment.
Answer: [(140 − 72) ÷ (72 × 1.4)] × 65 × 0.85 = (68 ÷ 100.8) × 65 × 0.85 = 37.3 mL/min — moderate impairment, so most renally cleared drugs will need adjustment. The height is supplied so the result panel can also show what the estimate would be using ideal and adjusted body weight.
Maintenance fluid requirements
Calculate maintenance fluid requirements for a 70 kg adult.
Answer: 1500 + 20 × (70 − 20) = 1500 + 1000 = 2500 mL/day, or about 104 mL/hr. The 30–40 mL/kg range gives 2100–2800 mL/day, which brackets it.
Infusion rate & drop factor
Infuse 1 litre over 8 hours using tubing with a drop factor of 20 gtt/mL. What rate in mL/hr, and how many drops per minute?
Answer: 1000 ÷ 8 = 125 mL/hr. Drops: 20 × 125 ÷ 60 = 41.7, so set about 42 drops per minute — roughly one drop every 1.4 seconds.
BUN : creatinine ratio
A patient admitted with several days of vomiting has a BUN of 40 mg/dL and a creatinine of 1.4 mg/dL.
Answer: 40 ÷ 1.4 = 28.6:1. Above the 20:1 threshold, consistent with a prerenal pattern from volume depletion.
Nutrition Support
Enteral / dietary calories
A day's tube feeding delivers 200 g carbohydrate, 80 g protein, and 60 g fat to a 70 kg patient. Calculate the total calories and the protein per kilogram.
Answer: (200 × 4) + (80 × 4) + (60 × 9) = 800 + 320 + 540 = 1660 kcal, which is 23.7 kcal/kg. Protein is 1.14 g/kg/day.
Parenteral nutrition calorie worksheet
A 2 litre bag is compounded to a final concentration of 15% dextrose and 4% amino acids, with 250 mL of 20% lipid emulsion. Calculate the daily calories for a 70 kg patient.
Answer: dextrose 300 g × 3.4 = 1020 kcal; amino acids 80 g × 4 = 320 kcal; lipid 250 × 2 = 500 kcal. Total 1840 kcal, which is 26.3 kcal/kg. Protein 1.14 g/kg/day. Non-protein calories 1520, nitrogen 12.8 g, so NPC:N is 119:1.
Nitrogen from protein
A patient receives 80 g of protein and 1520 non-protein calories per day. Their 24-hour urinary urea nitrogen is 12 g. Assess protein provision.
Answer: nitrogen intake = 80 ÷ 6.25 = 12.8 g. NPC:N = 1520 ÷ 12.8 = 119:1, within the usual range. Balance = 12.8 − (12 + 4) = −3.2 g/day, which is negative and suggests protein provision is insufficient for the current level of catabolism.
Total energy expenditure (TEE)
A 55-year-old male, 70 kg and 175 cm, is on bed rest following skeletal trauma. Estimate his total energy expenditure.
Answer: Mifflin–St Jeor BEE = (10 × 70) + (6.25 × 175) − (5 × 55) + 5 = 700 + 1094 − 275 + 5 = 1524 kcal. TEE = 1524 × 1.2 × 1.3 = 2377 kcal/day, or about 34 kcal/kg.
Pharmacokinetics
Pharmacokinetic workspace (linked solver)
A drug has a volume of distribution of 50 L and a clearance of 5 L/h. Derive the elimination rate constant and half-life.
Answer: ke = Cl ÷ Vd = 5 ÷ 50 = 0.1 h⁻¹. t½ = 0.693 ÷ 0.1 = 6.93 hours. Steady state is reached after about 5 half-lives, roughly 35 hours.
Absolute bioavailability (F)
A 500 mg oral dose gives an AUC of 32 mg·h/L; a 500 mg intravenous dose gives 40 mg·h/L. What is the absolute bioavailability?
Answer: 100 × (32/40) × (500/500) = 80%. One fifth of the oral dose does not reach the systemic circulation, lost to incomplete absorption and first-pass metabolism.
Predicting drug concentrations
A vancomycin concentration is 20 mg/L. The patient's elimination rate constant is 0.1 h⁻¹. What will the concentration be 8 hours later?
Answer: C₂ = 20 × e^(−0.1 × 8) = 20 × 0.449 = 8.99 mg/L. That is 1.15 half-lives, so a little more than half has been eliminated.
Loading dose
A drug has a volume of distribution of 0.7 L/kg. A 70 kg patient needs a plasma concentration of 20 mg/L. Calculate the intravenous loading dose.
Answer: Vd = 0.7 × 70 = 49 L. LD = (20 × 49) ÷ 1 = 980 mg.
Acid-Base & Laboratory
Arterial blood gas interpreter
A patient with a COPD exacerbation has pH 7.28, pCO₂ 55 mmHg, HCO₃ 25 mEq/L. Interpret the blood gas.
Answer: pH 7.28 is acidosis. pCO₂ 55 is above 45, so respiratory acidosis. HCO₃ 25 is normal. The respiratory value matches the pH direction, so this is an uncompensated respiratory acidosis.
Anion gap
Sodium 140, chloride 100, bicarbonate 24 mEq/L. Calculate the anion gap.
Answer: 140 − 100 − 24 = 16 mEq/L, which is above the usual range and indicates a high anion gap acidosis warranting a search for an added acid.
Corrected calcium for albumin
A hospitalised patient has a total calcium of 7.8 mg/dL and an albumin of 2.5 g/dL. Is the patient truly hypocalcaemic?
Answer: 7.8 + [(4.0 − 2.5) × 0.8] = 7.8 + 1.2 = 9.0 mg/dL, which is within the normal range. The apparent hypocalcaemia is entirely explained by the low albumin.
Corrected phenytoin for albumin (Sheiner–Tozer)
A patient with an albumin of 2.0 g/dL has a total phenytoin of 8 mcg/mL, which looks subtherapeutic. Should the dose be increased?
Answer: 8 ÷ [(0.2 × 2.0) + 0.1] = 8 ÷ 0.5 = 16 mcg/mL corrected — well within the therapeutic range. Increasing the dose on the uncorrected value would risk toxicity.
Absolute neutrophil count (ANC)
A patient receiving chemotherapy has a WBC of 3.5 × 10³/mm³ with 40% segs and 5% bands. Calculate the ANC.
Answer: 3500 × (45 ÷ 100) = 1575 cells/mm³ — above the 1500 threshold, so not neutropenic.
Mean arterial pressure (MAP)
A septic patient has a blood pressure of 110/60 mmHg. Calculate the MAP and compare it to the resuscitation target.
Answer: [(2 × 60) + 110] ÷ 3 = 230 ÷ 3 = 76.7 mmHg — above the 65 mmHg target.
Physical Pharmacy
Henderson–Hasselbalch equation
An acetate buffer contains 0.1 M sodium acetate and 0.05 M acetic acid. Acetic acid has a pKa of 4.76. What is the pH?
Answer: pH = 4.76 + log(0.1 ÷ 0.05) = 4.76 + log(2) = 4.76 + 0.30 = 5.06.
Percentage ionisation
A weak acid has a pKa of 3.5. What percentage is ionised in plasma at pH 7.4?
Answer: 100 ÷ (1 + 10^(3.5 − 7.4)) = 100 ÷ (1 + 10^−3.9) = 100 ÷ 1.000126 = 99.99% ionised. Almost none is available to cross membranes at plasma pH.
Biostatistics & Outcomes
2 × 2 table — risk, RR, RRR, ARR, NNT, OR
In a five-year trial, 15 of 100 treated patients and 25 of 100 control patients had a myocardial infarction. Calculate the effect measures.
Answer: risk 15% vs 25%. ARR = 10%. RRR = 10/25 = 40%. RR = 0.15/0.25 = 0.60. NNT = 1/0.10 = 10 — treat 10 patients for five years to prevent one MI. OR = (15 × 75) ÷ (85 × 25) = 0.53.
Hazard ratio
A survival analysis reports a hazard rate of 0.08 events per person-year in the treatment arm and 0.12 in the control arm.
Answer: HR = 0.08 ÷ 0.12 = 0.667, a 33% relative reduction in the instantaneous rate of the event.
Incremental cost-effectiveness ratio (ICER)
A new therapy costs $35,000 and yields 4.0 QALYs; the standard of care costs $10,000 and yields 3.2 QALYs. Calculate the ICER against a $100,000 per QALY threshold.
Answer: ΔC = $25,000, ΔE = 0.8 QALYs. ICER = 25,000 ÷ 0.8 = $31,250 per QALY — well below a $100,000 threshold, so cost-effective at that threshold.
Descriptive statistics
Calculate the descriptive statistics for the data set 12, 15, 15, 18, 22, 24, 31.
Answer: n = 7, sum = 137, mean = 19.57, median = 18 (the fourth of seven ordered values), mode = 15 (appears twice), range = 19, sample SD = 6.55.
Therapeutic Conversions
Corticosteroid equivalency
A patient on prednisone 40 mg daily is being switched to intravenous methylprednisolone. What is the equivalent dose?
Answer: 40 mg × (4 ÷ 5) = 32 mg of methylprednisolone. In practice this is often given as 30 mg or 40 mg depending on available vial sizes and the clinical situation.
Statin dose equivalency
A patient on simvastatin 40 mg is switching to atorvastatin because of a drug interaction. What is the approximately equivalent dose?
Answer: 40 mg × (10 ÷ 20) = 20 mg of atorvastatin. Note that atorvastatin 20 mg is moderate-intensity therapy; if the patient requires high-intensity therapy, 40–80 mg would be needed regardless of the equivalency.
Loop diuretic equivalency & route conversion
A patient on intravenous furosemide 40 mg is being switched to intravenous bumetanide. What is the equivalent dose?
Answer: 40 mg × (1 ÷ 40) = 1 mg of bumetanide. The classic equivalency to memorise is furosemide 40 mg = torsemide 20 mg = bumetanide 1 mg.
Insulin product conversion
A patient takes NPH insulin 24 units in the morning and 16 units in the evening, a total of 40 units daily. Convert to once-daily glargine.
Answer: 40 × 0.8 = 32 units of glargine once daily. The reduction reflects glargine's flatter, longer profile and reduces the risk of nocturnal hypoglycaemia during the switch.
IV ↔ PO route conversion
A patient is receiving metoprolol 5 mg intravenously and is being switched to oral therapy. What is the equivalent oral dose?
Answer: 5 mg × 2.5 = 12.5 mg orally. In practice this would be rounded to an available strength, typically 12.5 mg using half of a 25 mg tablet.
Aminophylline ↔ theophylline
A patient is receiving aminophylline 500 mg. What is the equivalent theophylline dose?
Answer: 500 mg × 0.8 = 400 mg of theophylline.
Calcium salt ↔ elemental calcium
A calcium carbonate tablet is labelled 1250 mg. How much elemental calcium does it provide?
Answer: 1250 mg × 0.40 = 500 mg of elemental calcium. This is why 1250 mg carbonate tablets are commonly described as "500 mg calcium".
Potassium chloride 10% solution ↔ mEq
A patient is prescribed 30 mL of potassium chloride 10% oral solution. How many milliequivalents is that?
Answer: 30 mL × (20 mEq ÷ 15 mL) = 40 mEq, equivalent to two 20 mEq tablets.
Lithium citrate syrup ↔ carbonate ↔ mEq
A patient takes lithium carbonate 900 mg daily but cannot swallow tablets. What volume of lithium citrate solution provides the equivalent dose?
Answer: 900 mg ÷ 300 mg = 3 units; 3 × 5 mL = 15 mL of citrate solution, supplying 24 mEq of lithium.
Diabetes & Lipids
Insulin initiation & adjustment worksheet
A newly diagnosed 70 kg adult with type 1 diabetes is starting rapid-acting basal-bolus insulin. Their glucose is currently 250 mg/dL with a target of 120 mg/dL.
Answer: TDD = 70 × 0.5 = 35 units. Basal 17.5 units, mealtime 17.5 units total, so roughly 6 units per meal. ICR = 500 ÷ 35 = 1 unit per 14 g carbohydrate. Correction factor = 1800 ÷ 35 = 51 mg/dL per unit. Correction dose = (250 − 120) ÷ 51 = 2.5 units.
LDL cholesterol (Friedewald equation)
A lipid panel shows total cholesterol 200 mg/dL, HDL 50 mg/dL, triglycerides 150 mg/dL. Calculate the LDL.
Answer: VLDL = 150 ÷ 5 = 30. LDL = 200 − 50 − 30 = 120 mg/dL. Non-HDL cholesterol is 200 − 50 = 150 mg/dL.
Public Health & Misc
Pack-year smoking history
A 62-year-old patient has smoked one pack (20 cigarettes) a day for 30 years and still smokes. Calculate the pack-year history and check screening eligibility.
Answer: 20 ÷ 20 = 1 pack/day; 1 × 30 = 30 pack-years. Age 62 is within 50–80, exposure exceeds 20 pack-years, and they currently smoke — all three USPSTF criteria are met, so annual low-dose CT screening is recommended.
Sun protection — time to burn (SPF)
A patient burns after about 15 minutes of unprotected midday sun. How long could SPF 30 theoretically extend that?
Answer: 30 × 15 = 450 minutes, or 7.5 hours in theory. In practice, under-application and rubbing off mean protection is far shorter — reapply every two hours.