Worked Examples
One fully worked problem for every question type on the site — 42 of them, across 19 topics.
A topic is not one question. “Dispensing quantity” is five different questions depending on whether you are counting bottles, vials, IV bags, inhalers, or tablets crushed for a batch, and each has its own setup. Every distinct question type this site can generate appears below, once, solved start to finish. If a type here looks unfamiliar, that is the one to go practise.
Ratio Strength
Express a 1:5,000 ratio strength as a percentage (w/v). Round your answer to two decimal places.
Answer 0.02 %
Solution- 1:5,000 means 1 g of drug in 5,000 mL.
- Percent w/v = grams per 100 mL: (1 ÷ 5,000) × 100 = 0.02%.
- Rounded as the question asks: 0.02 %.
Anchors worth memorizing: 1:1,000 = 0.1% (epinephrine), 1:10,000 = 0.01%, 1:100,000 = 0.001%.
Percentage Strength
How many grams of active drug are in 300 mL of a 12.5% w/v solution? Round your answer to one decimal place.
Answer 37.5 g
Solution- 12.5% w/v = 12.5 g per 100 mL.
- Grams = 12.5 g/100 mL × 300 mL = 37.5 g.
- Rounded as the question asks: 37.5 g.
Practice percentage & ratio strength →
Dilution
How many mL of a 5% potassium permanganate stock solution are needed to prepare 490 mL of a 0.1% solution? Round your answer to two decimal places.
Answer 9.80 mL
Solution- Use C₁V₁ = C₂V₂.
- V₁ = (C₂ × V₂) ÷ C₁ = (0.1% × 490 mL) ÷ 5%.
- V₁ = 49 ÷ 5 = 9.8 mL.
- If the question asks for the final preparation, QS with diluent up to 490 mL total — the 9.8 mL of stock is only part of the final volume.
- Rounded as the question asks: 9.80 mL.
The drug amount is unchanged by dilution; only concentration and volume change. Don't forget to QS to the final volume when the question asks for it.
Practice dilution →
Alligation Alternate
You need 30 g of 1% hydrocortisone ointment, prepared from 2.5% and 0.5% stock. How many g of the 2.5% stock do you need? Round your answer to one decimal place.
Answer 7.5 g
Solution- Alligation grid: high (2.5%) top-left, low (0.5%) bottom-left, target (1%) center.
- Parts of 2.5% = 1 − 0.5 = 0.5 parts (diagonal difference).
- Parts of 0.5% = 2.5 − 1 = 1.5 parts.
- Total parts = 2.
- Amount of 2.5% = (0.5 ÷ 2) × 30 = 7.5 g (and 22.5 g of 0.5%).
- Rounded as the question asks: 7.5 g.
Sanity check: 7.5×2.5% + 22.5×0.5% = 0.3 g = 30 g × 1% ✓
Alligation Medial
You mix 300 mL of 50% dextrose with 250 mL of 20% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.
Answer 36.4 %
Solution- Grams from each: 300×50% = 150 g; 250×20% = 50 g.
- Total drug = 200 g in total volume 550 mL.
- Strength = 200 ÷ 550 × 100 ≈ 36.364%.
- Rounded as the question asks: 36.4 %.
Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.
Practice alligation →
Milliequivalents
How many milliequivalents are in 1 g of magnesium sulfate heptahydrate (MgSO₄·7H₂O)? (MW = 246.5, valence = 2) Round your answer to one decimal place.
Answer 8.1 mEq
Solution- mg: 1 g = 1000 mg.
- mmol = 1000 ÷ 246.5 ≈ 4.057 mmol.
- mEq = mmol × valence = 4.057 × 2 ≈ 8.11 mEq.
- Rounded as the question asks: 8.1 mEq.
Monovalent ions: mEq = mmol. Divalent (Ca²⁺, Mg²⁺): 2 mEq per mmol — forgetting valence is the classic error.
Millimoles
How many millimoles are in 13 g of sodium bicarbonate (NaHCO₃)? (MW = 84) Round your answer to one decimal place.
Answer 154.8 mmol
Solution- Convert to mg: 13 g = 13000 mg.
- mmol = mg ÷ MW = 13000 ÷ 84 ≈ 154.76 mmol.
- Rounded as the question asks: 154.8 mmol.
Milliosmoles
Calculate the osmolarity of a 10% dextrose monohydrate solution in mOsm/L. (MW = 198; does not dissociate (1 particle per molecule).) Round your answer to the nearest whole number.
Answer 505 mOsm/L
Solution- Step 1 — grams in 1,000 mL: 10% = 10 g/100 mL, so in 1,000 mL = 100 g.
- Step 2 — apply mOsm/L = (mass in g ÷ MW) × number of particles × 1,000.
- = (100 ÷ 198) × 1 × 1,000 ≈ 505.05 mOsm/L.
- Rounded as the question asks: 505 mOsm/L.
Landmarks: NS ≈ 308 mOsm/L, D5W ≈ 252 mOsm/L (dextrose monohydrate, MW 198) — both close enough to plasma (~285–295) to run peripherally. Particles: NaCl = 2, dextrose = 1 (nonelectrolyte).
Practice milliequivalent (meq), millimole & milliosmole →
IV Rate (mL/hr)
Dopamine 400 mg in 250 mL D5W is ordered at 5 mcg/kg/min for a 89 kg patient. What rate in mL/hr should the pump be set at? Round your answer to one decimal place.
Answer 16.7 mL/hr
Solution- Concentration: 400 mg × 1000 ÷ 250 mL = 1600 mcg/mL.
- Dose rate: 5 mcg/kg/min × 89 kg = 445 mcg/min.
- Per hour: 445 × 60 = 26700 mcg/hr.
- Pump rate: 26700 ÷ 1600 = 16.6875 mL/hr.
- Rounded as the question asks: 16.7 mL/hr.
The three-step chain (concentration → mcg/min → mL/hr) never changes; only the numbers do. Weight only enters when the order is written per kg — norepinephrine and nitroglycerin are commonly ordered as a flat mcg/min.
IV Drip Rate
An order calls for 1000 mL of NS over 3 hours with a set that delivers 20 drops/mL. What is the flow rate in drops per minute? Round your answer to the nearest whole number.
Answer 111 drops/min
Solution- Minutes: 3 × 60 = 180 min.
- Total drops: 1000 mL × 20 drops/mL = 20000 drops.
- 20000 ÷ 180 ≈ 111.11 → 111 drops/min.
- Rounded as the question asks: 111 drops/min.
Work in whole drops and whole minutes rather than a decimal mL/min — carrying a rounded mL/min into the next line is how a correct method produces a wrong number. Drops (gtt) are always whole; you can't give a partial drop.
Practice iv infusion & drip rate →
BSA Dosing
A patient is 170 cm tall and weighs 83 kg. The order is etoposide 100 mg/m². Using the Mosteller formula, what dose in mg should be prepared? Round your answer to the nearest whole number.
Answer 198 mg
Solution- Mosteller: BSA = √(ht[cm] × wt[kg] ÷ 3600).
- 170 × 83 = 14110; 14110 ÷ 3600 ≈ 3.91944.
- BSA = √3.91944 ≈ 1.97976 m².
- Dose = 100 mg/m² × 1.97976 m² ≈ 198 mg.
- Rounded as the question asks: 198 mg.
Keep BSA unrounded until the final multiplication — rounding BSA to 2 decimals early can shift the dose by several mg, which is why the steps above carry 5.
Practice body surface area (bsa) & chemo dosing →
TPN — Nonprotein kcal
A TPN contains 500 mL dextrose 50%, 250 mL amino acids 10%, and 250 mL of 20% lipid emulsion. What are the NONPROTEIN kilocalories? (3.4 kcal/g dextrose, 4 kcal/g protein, 20% lipid = 2 kcal/mL) Round your answer to the nearest whole number.
Answer 1,350 kcal
Solution- Dextrose: 500 mL × 50% = 250 g; × 3.4 = 850 kcal.
- Protein: 250 mL × 10% = 25 g; × 4 = 100 kcal — excluded: the question asks for nonprotein kcal.
- Lipid: 250 mL × 2 kcal/mL = 500 kcal.
- Nonprotein = 850 + 500 = 1350 kcal.
- Rounded as the question asks: 1,350 kcal.
Total vs NONPROTEIN kcal is decided by the question's wording — read the ask twice before adding.
TPN — Nitrogen
A TPN contains 250 mL of 8.5% amino acids. How many grams of nitrogen does it provide? (Protein is 16% nitrogen: N = protein ÷ 6.25) Round your answer to one decimal place.
Answer 3.4 g
Solution- Protein grams: 250 mL × 8.5% = 21.25 g.
- Nitrogen = 21.25 ÷ 6.25 = 3.4 g N.
- Rounded as the question asks: 3.4 g.
Lower-yield for the exam, kept for completeness. Used for nitrogen-balance and calorie:nitrogen ratio questions.
TPN — Total kcal
A TPN contains 1000 mL dextrose 15%, 250 mL amino acids 8.5%, and 500 mL of 20% lipid emulsion. What are the total kilocalories? (3.4 kcal/g dextrose, 4 kcal/g protein, 20% lipid = 2 kcal/mL) Round your answer to the nearest whole number.
Answer 1,595 kcal
Solution- Dextrose: 1000 mL × 15% = 150 g; × 3.4 = 510 kcal.
- Protein: 250 mL × 8.5% = 21.25 g; × 4 = 85 kcal.
- Lipid: 500 mL × 2 kcal/mL = 1000 kcal.
- Total = 510 + 85 + 1000 = 1595 kcal.
- Rounded as the question asks: 1,595 kcal.
Total vs NONPROTEIN kcal is decided by the question's wording — read the ask twice before adding.
TPN — Dextrose kcal
A TPN contains 250 mL of dextrose 20%. How many kilocalories does the dextrose provide? (IV dextrose = 3.4 kcal/g) Round your answer to the nearest whole number.
Answer 170 kcal
Solution- Grams of dextrose: 250 mL × 20% = 50 g.
- kcal = 50 g × 3.4 kcal/g = 170 kcal.
- Rounded as the question asks: 170 kcal.
IV (TPN) energy densities: dextrose 3.4 kcal/g, protein 4 kcal/g, lipid 10% = 1.1 kcal/mL, 20% = 2 kcal/mL. ORAL/enteral nutrition uses 4 kcal/g carbohydrate, 4 protein, 9 fat — don't mix the two systems.
Oral Nutrition kcal
An oral nutrition supplement contains 30 g carbohydrate, 8 g protein, and 7 g fat per serving. How many kilocalories per serving? (Oral: 4 kcal/g carbohydrate, 4 kcal/g protein, 9 kcal/g fat) Round your answer to the nearest whole number.
Answer 215 kcal
Solution- Carbohydrate: 30 g × 4 = 120 kcal.
- Protein: 8 g × 4 = 32 kcal.
- Fat: 7 g × 9 = 63 kcal.
- Total = 120 + 32 + 63 = 215 kcal.
- Rounded as the question asks: 215 kcal.
The oral/enteral system (4/4/9) differs from IV: oral carbohydrate is 4 kcal/g, but IV dextrose monohydrate is 3.4 kcal/g. Route determines the numbers.
Practice tpn & nutrition calculation →
Isotonicity (E-values)
How many grams of NaCl must be added to make 30 mL of a 0.125% phenylephrine HCl solution isotonic? (E-value of phenylephrine HCl = 0.32) Round your answer to three decimal places.
Answer 0.258 g
Solution- Step 1 — drug in the formulation: 30 mL × 0.125% = 0.0375 g.
- Step 2 — NaCl equivalent of the drug: 0.0375 g × E(0.32) = 0.012 g.
- Step 3 — NaCl to make 30 mL isotonic on its own: 30 × 0.009 = 0.27 g.
- Step 4 — NaCl to add = 0.27 − 0.012 = 0.258 g.
- Rounded as the question asks: 0.258 g.
E-value = grams of NaCl with the same osmotic effect as 1 g of drug. The exam gives you the E-value — the skill is the 4-step setup. The steps above carry 5 decimals on purpose: truncating an intermediate here is enough to move the final gram figure.
Practice isotonicity & e-value →
Creatinine Clearance
Calculate the creatinine clearance (Cockcroft-Gault) for a 71-year-old female, 5'6" tall, weighing 92 kg, with a serum creatinine of 2.2 mg/dL. Select the appropriate body weight: use actual body weight if it is below IBW, adjusted body weight if it exceeds 125% of IBW, and IBW otherwise. Round your answer to the nearest whole number.
Answer 27 mL/min
Solution- IBW (female): 45.5 + 2.3 × (inches over 60) = 45.5 + 2.3 × 6 = 59.3 kg.
- Choose the weight: Actual BW (92 kg) > 125% of IBW (74.13 kg) → patient is obese → AdjBW = IBW + 0.4 × (actual − IBW) = 59.3 + 0.4 × (92 − 59.3) = 72.38 kg.
- Cockcroft-Gault: CrCl = [(140 − age) × wt] ÷ (72 × SCr) × 0.85 (female).
- Numerator: (140 − 71) × 72.38 = 4994.22; denominator: 72 × 2.2 = 158.4.
- 4994.22 ÷ 158.4 ≈ 31.5292; × 0.85 (female) ≈ 26.8 mL/min (using AdjBW).
- Rounded as the question asks: 27 mL/min.
The weight-selection algorithm: always calculate IBW first. Underweight (actual < IBW) → use actual body weight. Obese (actual > 125% of IBW) → use AdjBW = IBW + 0.4(actual − IBW). Otherwise → use IBW.
Practice creatinine clearance (cockcroft-gault) →
PK — Volume of Distribution
A 500 mg IV bolus produces an initial plasma concentration (C₀) of 25 mg/L. What is the volume of distribution? Round your answer to one decimal place.
Answer 20.0 L
Solution- Vd = Dose ÷ C₀.
- Vd = 500 mg ÷ 25 mg/L = 20 L.
- Rounded as the question asks: 20.0 L.
Vd is theoretical — a Vd far above total body water (~42 L) means extensive tissue binding (e.g., digoxin ~500 L).
PK — Clearance
A drug has an elimination rate constant of 0.2 hr⁻¹ and a Vd of 80 L. What is its total body clearance? Round your answer to one decimal place.
Answer 16.0 L/hr
Solution- CL = k × Vd.
- CL = 0.2 × 80 = 16 L/hr.
- Rounded as the question asks: 16.0 L/hr.
Also rearranges to k = CL/Vd and t½ = 0.693·Vd/CL — same triangle, three disguises.
PK — Elimination Rate Constant
Two plasma levels are measured for a drug: 25 mg/L at hour 0 and 5 mg/L at hour 4. What is the elimination rate constant (k)? Round your answer to four decimal places.
Answer 0.4024 hr⁻¹
Solution- k = ln(C₁ ÷ C₂) ÷ (t₂ − t₁).
- C₁ ÷ C₂ = 25 ÷ 5 = 5; ln(5) ≈ 1.60944.
- k = 1.60944 ÷ 4 ≈ 0.40236 hr⁻¹.
- Rounded as the question asks: 0.4024 hr⁻¹.
First-order elimination: plot ln(concentration) vs time and k is the negative slope. Half-life then follows from t½ = 0.693 ÷ k. Carry k to at least 4 decimals — it is small, so a truncated k throws every downstream figure off.
PK — Loading Dose
Target plasma concentration is 18 mg/L and the drug's Vd is 30 L. What IV loading dose is required? (Assume S = 1, F = 1.) Round your answer to the nearest whole number.
Answer 540 mg
Solution- LD = (Cp_target × Vd) ÷ (S × F).
- LD = 18 × 30 ÷ 1 = 540 mg.
- Rounded as the question asks: 540 mg.
Loading dose depends on Vd; maintenance dose depends on clearance. Exam questions love to make you mix these up.
PK — Half-life
A drug has an elimination rate constant of 0.087 hr⁻¹. What is its half-life? Round your answer to one decimal place.
Answer 8.0 hr
Solution- t½ = 0.693 ÷ k.
- t½ = 0.693 ÷ 0.087 ≈ 7.966 hr.
- Rounded as the question asks: 8.0 hr.
~94% of a drug is eliminated after 4 half-lives; steady state is reached after ~4–5 half-lives of dosing.
Practice pharmacokinetics →
Bioavailability (F)
A 100 mg IV dose gives an AUC of 50 mg·hr/L. A 200 mg oral dose gives an AUC of 90 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.
Answer 90 %
Solution- F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
- Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
- Numerator: 90 × 100 = 9000.
- Denominator: 200 × 50 = 10000.
- F = 9000 ÷ 10000 = 0.9, i.e. 90%.
- Rounded as the question asks: 90 %.
Always dose-normalize both AUCs — comparing raw AUCs from different doses is the built-in trap.
Practice bioavailability (f) →
Insulin Day Supply
A patient uses Novolin N (10 mL vial, 100 units/mL) at 40 units/day. Once in use, this product is stable for 42 days, after which any remaining insulin must be discarded. What is the day supply of one vial? Round down to the nearest whole number.
Answer 25 days
Solution- Units in vial: 10 mL × 100 units/mL = 1,000 units.
- Raw days: 1,000 ÷ 40 = 25 days.
- Novolin N is stable 42 days once in use. Since 25 ≤ 42, no waste.
- Day supply = 25 days.
- Rounded as the question asks: 25 days.
Insulin day supply = total units ÷ daily dose, then capped at the product's in-use stability. You can never dispense a partial day, so this always goes DOWN.
Insulin Switch
A patient takes Toujeo (glargine U-300) 40 units daily. Convert to Basaglar. What is the new daily dose? Round your answer to the nearest whole number.
Answer 32 units
Solution- Toujeo (U-300) → Basaglar (U-100 glargine) uses a 20% reduction.
- 40 × 0.8 = 32 units of Basaglar daily.
- Rounded as the question asks: 32 units.
Switches are 1:1 EXCEPT two cases at −20%: NPH (BID) → basal glargine, and Toujeo (U-300) → Lantus/Basaglar.
Practice insulin day supply & switching →
Equianalgesic Conversion
A patient is receiving 1.5 mg/day of hydromorphone (IV) and is being switched to codeine (PO). Using the equianalgesic table below, convert the dose and then reduce it by 50% for incomplete cross-tolerance. What is the starting daily dose of codeine (PO) in mg/day?
| Opioid | PO (mg) | IV (mg) |
|---|
| morphine | 30 | 10 |
| hydromorphone | 7.5 | 1.5 |
| oxycodone | 20 | — |
| hydrocodone | 30 | — |
| codeine | 200 | — |
| meperidine | 300 | 75 |
| fentanyl | — | 0.1 (= 100 mcg) |
Round your answer to one decimal place.
Answer 100.0 mg/day
Solution- Pull the two relevant rows — drug AND route: hydromorphone (IV) = 1.5 mg ≈ codeine (PO) = 200 mg.
- Raw equivalent: 1.5 × 200 ÷ 1.5 = 200 mg/day.
- Apply the 50% cross-tolerance reduction: 200 × 0.50 = 100 mg/day = 100 mg/day.
- Rounded as the question asks: 100.0 mg/day.
Two-step questions are the norm: convert on the table first, THEN reduce 25–50%. The reduction exists because tolerance to one opioid does not transfer completely to another — skipping it is how a conversion overdoses a patient. Reduce toward the larger end (50%) when the patient is elderly, frail, or the new opioid is methadone.
Fentanyl Patch
A patient's opioid regimen totals 205 mg/day of oral morphine equivalents. Using the conversion table below, which fentanyl patch strength (mcg/hr) is appropriate? This table comes from the product labeling and already builds in the conservative conversion factor, so no further cross-tolerance reduction is applied.
| Oral morphine (mg/day) | Fentanyl patch (mcg/hr) |
|---|
| 60–134 | 25 |
| 135–224 | 50 |
| 225–314 | 75 |
| 315–404 | 100 |
Round your answer to the nearest whole number.
Answer 50 mcg/hr
Solution- Find the range containing 205 mg/day → 135–224 mg/day.
- That row corresponds to the 50 mcg/hr patch.
- Rounded as the question asks: 50 mcg/hr.
Patch selection is a table LOOKUP, not a proportion — never interpolate between rows. The Duragesic table is deliberately conservative, so you do NOT subtract a further 25–50% on top of it. Patches are changed every 72 hours and are for opioid-tolerant patients only — both favorite exam distractors.
MME
A patient takes codeine (PO) 60 mg, 2 times daily. Using the equianalgesic table below, what is the total daily dose in morphine milligram equivalents (MME/day)?
| Opioid | PO (mg) | IV (mg) |
|---|
| morphine | 30 | 10 |
| hydromorphone | 7.5 | 1.5 |
| oxycodone | 20 | — |
| hydrocodone | 30 | — |
| codeine | 200 | — |
| meperidine | 300 | 75 |
| fentanyl | — | 0.1 (= 100 mcg) |
Round your answer to the nearest whole number.
Answer 18 MME/day
Solution- Find the right cell — drug AND route: 200 mg codeine (PO) ≈ 30 mg morphine (PO).
- Daily dose: 60 mg × 2 = 120 mg/day.
- Proportion: 200 mg / 30 mg = 120 mg / x → x = 120 × 30 ÷ 200 = 18 MME/day.
- Rounded as the question asks: 18 MME/day.
MME is just an equianalgesic conversion with 30 mg oral morphine as the anchor. 90 MME/day is the common review threshold in opioid safety guidance. MME factors are defined for ORAL opioids; a parenteral dose must first be converted to its oral equivalent, which is why the CDC table has no IV column.
Practice opioid conversion & mme →
Days Supply
An ipratropium/albuterol Respimat inhaler contains 60 actuations (puffs). The sig is 1 puff four times daily (QID). What is the days supply? Round down to the nearest whole number.
Answer 15 days
Solution- Puffs per day: 1 × 4 = 4 puffs/day.
- Days supply: 60 ÷ 4 = 15 → round down = 15 days.
- Rounded as the question asks: 15 days.
Inhaler days supply uses total actuations (not mL). Rescue/PRN inhalers are billed on maximum daily puffs.
Practice days supply calculation →
Weight-Based Dosing
A pediatric patient weighs 70.4 lb. The order is gentamicin 7.5 mg/kg/day divided three times daily (q8h). How many mg per dose? (Use 1 kg = 2.2 lb.) Round your answer to one decimal place.
Answer 80.0 mg/dose
Solution- Convert: 70.4 ÷ 2.2 = 32 kg.
- Daily dose: 7.5 × 32 = 240 mg/day.
- Per dose: 240 ÷ 3 = 80 mg/dose.
- Rounded as the question asks: 80.0 mg/dose.
Watch for mg/kg/DAY vs mg/kg/DOSE — the #1 trap in this family. Conversion factors also vary: 2.2 lb/kg is the exam convention, 2.2046 is the exact figure, so the question states which to use.
Practice weight-based (mg/kg) dosing →
Reconstitution
A vial contains 1 g of powdered drug. Adding 9.7 mL of diluent yields a final concentration of 100 mg/mL. What is the powder (displacement) volume? Round your answer to two decimal places.
Answer 0.30 mL
Solution- Final volume needed: 1 g = 1000 mg; 1000 mg ÷ 100 mg/mL = 10 mL.
- Powder volume = final volume − diluent added = 10 − 9.7 = 0.3 mL.
- Rounded as the question asks: 0.30 mL.
The dry powder occupies space (displacement). Powder volume = final reconstituted volume − volume of diluent added.
Practice reconstitution & powder volume →
Specific Gravity
A solution has a specific gravity of 1.15. What volume in mL contains 225 g? Round your answer to one decimal place.
Answer 195.7 mL
Solution- volume (mL) = weight (g) ÷ specific gravity.
- volume = 225 ÷ 1.15 ≈ 195.652 mL.
- Rounded as the question asks: 195.7 mL.
Rearranged from weight = volume × specific gravity. A liquid denser than water (SG > 1) occupies less volume per gram.
Practice specific gravity →
Anion Gap
A patient has Na⁺ 132, Cl⁻ 102, and HCO₃⁻ 18 (all mEq/L). What is the anion gap? Round your answer to the nearest whole number.
Answer 12 mEq/L
Solution- Anion gap = Na⁺ − (Cl⁻ + HCO₃⁻).
- = 132 − (102 + 18) = 132 − 120 = 12 mEq/L.
- Rounded as the question asks: 12 mEq/L.
Normal anion gap ≈ 8–12 mEq/L (potassium omitted). Elevated gap points to MUDPILES causes (lactic acidosis, DKA, salicylates, etc.).
Corrected Calcium
A patient has a measured serum calcium of 8.5 mg/dL and an albumin of 2.8 g/dL. What is the corrected calcium? Round your answer to one decimal place.
Answer 9.5 mg/dL
Solution- Corrected Ca = measured Ca + 0.8 × (4.0 − albumin).
- = 8.5 + 0.8 × (4.0 − 2.8) = 8.5 + 0.8 × 1.2 = 9.46 mg/dL.
- Rounded as the question asks: 9.5 mg/dL.
Only correct when albumin is low. Each 1 g/dL of albumin below 4.0 masks ~0.8 mg/dL of calcium.
ANC
A patient has a WBC of 12,000 cells/mm³ with 20% segs and 2% bands. What is the absolute neutrophil count (ANC)? Round your answer to the nearest whole number.
Answer 2,640 cells/mm³
Solution- ANC = WBC × (% segs + % bands) ÷ 100.
- = 12,000 × (20 + 2) ÷ 100 = 12,000 × 0.22 = 2640 cells/mm³.
- Rounded as the question asks: 2,640 cells/mm³.
Neutrophils only — segs plus bands. Neutropenia: ANC < 1500 mild, < 1000 moderate, < 500 severe (high infection risk).
Corrected Phenytoin
A patient's measured total phenytoin level is 12 mcg/mL with an albumin of 3 g/dL and normal renal function. What is the corrected phenytoin level? (Sheiner-Tozer) Round your answer to one decimal place.
Answer 17.1 mcg/mL
Solution- Corrected = measured ÷ [(0.2 × albumin) + 0.1].
- = 12 ÷ [(0.2 × 3) + 0.1] = 12 ÷ 0.7 ≈ 17.1429 mcg/mL.
- Rounded as the question asks: 17.1 mcg/mL.
Sheiner-Tozer adjusts total phenytoin for hypoalbuminemia. For CrCl < 10 mL/min, the 0.2 factor becomes 0.1.
Practice clinical lab calculation →
IV Bags Needed
A maintenance 0.9% sodium chloride infusion is running at 100 mL/hr. Pharmacy supplies it in 1000 mL bags. How many bags must be sent to cover the next 72 hours without the line running dry? Round up to the next whole number.
Answer 8 bags
Solution- Bags = (rate × hours) ÷ volume per bag.
- Total volume: 100 mL/hr × 72 hr = 7200 mL.
- Bags: 7200 mL ÷ 1000 mL = 7.2 bags.
- Rounded as the question asks: 8 bags.
A partly used bag cannot cover the gap at the end of the window, so the last bag has to be a whole one. This counts bags needed to cover the time — the final bag will still have volume in it when the window ends, and IV admixtures also carry their own hang-time limits that can force a bag change before it empties.
Vials per Dose
A patient weighing 80 kg is prescribed liposomal amphotericin B 5 mg/kg IV for one dose. The drug is stocked as 50 mg single-dose vials. How many vials must be opened to prepare this dose? Round up to the next whole number.
Answer 8 vials
Solution- Vials = (weight × dose per kg) ÷ vial size.
- Dose: 80 kg × 5 mg/kg = 400 mg.
- Vials: 400 mg ÷ 50 mg = 8 vials.
- Rounded as the question asks: 8 vials.
A single-dose vial that is entered is a single-dose vial that is spent, so any fraction costs a whole extra vial and the remainder is discarded. This is also the number to quote for cost and for waste — not the calculated dose.
Bottles to Dispense
A prescription is written for prednisolone 15 mg/5 mL oral solution, 7.5 mL twice daily (BID) for 10 days. The product is supplied in 120 mL bottles. How many bottles must be dispensed to complete the course? Round up to the next whole number.
Answer 2 bottles
Solution- Bottles = (mL per dose × doses per day × days) ÷ bottle size.
- Per day: 7.5 mL × 2 = 15 mL/day.
- Whole course: 15 mL × 10 days = 150 mL.
- Bottles: 150 mL ÷ 120 mL = 1.25 bottles.
- Rounded as the question asks: 2 bottles.
Suspensions are dispensed in whole manufacturer bottles, so a course that needs 2.25 bottles is dispensed as 3. Sending home 2 guarantees the patient runs out before the antibiotic course is finished — the single most common version of this error.
Inhalers to Dispense
A prescription for budesonide/formoterol inhaler reads: 2 puffs twice daily. Each inhaler is labelled 120 actuations. How many inhalers must be dispensed for a 60-day supply? Round up to the next whole number.
Answer 2 inhalers
Solution- Inhalers = (puffs per day × days) ÷ actuations per inhaler.
- Per day: 2 puffs × 2 doses = 4 puffs/day.
- Whole supply: 4 puffs/day × 60 days = 240 puffs.
- Inhalers: 240 puffs ÷ 120 actuations = 2 inhalers.
- Rounded as the question asks: 2 inhalers.
Inhalers are dispensed whole, so a 90-day supply needing 2.7 inhalers is billed as 3. Watch the actuation count on the label rather than assuming: rescue HFAs are commonly 200 actuations while most maintenance HFAs are 120, and the same sig gives a different quantity for each.
Tablets to Compound
A compounding order calls for 45 capsules, each containing 2.5 mg of spironolactone. The powder will be prepared by crushing commercial 25 mg spironolactone tablets. How many tablets are needed to compound this prescription? Round up to the next whole number.
Answer 5 tablets
Solution- Tablets = (number of capsules × mg per capsule) ÷ mg per tablet.
- Total drug needed: 45 × 2.5 mg = 112.5 mg.
- Tablets: 112.5 mg ÷ 25 mg = 4.5 tablets.
- Rounded as the question asks: 5 tablets.
Whole tablets go into the mortar, so the count is a ceiling and the surplus powder is waste. Note this is the minimum — a real compounder adds an overage on top for loss during trituration and filling, which is why the batch is often prepared from one or two more tablets than the calculation alone demands.
Practice dispensing quantity →