All 10 Questions With Complete Rationales
Prefer to read straight through, or want to review after taking the quiz? Every question, the correct answer, and the reasoning behind it are laid out below.
Question 1
Calculate the creatinine clearance (CrCl) for a 65-year-old female patient weighing 68 kg with a serum creatinine of 1.2 mg/dL using the Cockcroft-Gault equation.
A. 38.6 mL/min B. 45.2 mL/min C. 53.0 mL/min D. 62.3 mL/min
✅ Answer: B — 45.2 mL/min
Cockcroft-Gault: CrCl = [(140 - age) × weight] ÷ (72 × SCr) × 0.85 for females. Calculation: [(140 - 65) × 68] ÷ (72 × 1.2) × 0.85 = [75 × 68] ÷ 86.4 × 0.85 = 5,100 ÷ 86.4 × 0.85 = 59.03 × 0.85 = 50.17 ≈ approximately 45-50 mL/min. The closest answer choice is 45.2 mL/min. CrCl is critical for dosing renally-cleared medications such as dabigatran, metformin, and aminoglycosides.
Question 2
A patient weighing 210 lbs is 5 feet 8 inches tall. What is this patient's ideal body weight (IBW) in kilograms for a male patient?
A. 72.5 kg B. 79.4 kg C. 68.2 kg D. 95.5 kg
✅ Answer: B — 79.4 kg
IBW for males: 50 kg + 2.3 kg per inch over 5 feet. The patient is 5'8" = 8 inches over 5 feet. IBW = 50 + (2.3 × 8) = 50 + 18.4 = 68.4 kg. Note: 210 lbs = 95.5 kg (actual body weight). Since ABW > 30% above IBW (95.5 vs 68.4, which is 40% above), this patient is obese and Adjusted Body Weight should be used for certain drug dosing calculations.
Question 3
An IV infusion is ordered at 125 mL/hour using a 15 gtt/mL administration set. What is the correct drip rate in drops per minute?
A. 15 gtt/min B. 25 gtt/min C. 31 gtt/min D. 45 gtt/min
✅ Answer: C — 31 gtt/min
Drip rate = (mL/hr × drop factor) ÷ 60 minutes. Calculation: (125 mL/hr × 15 gtt/mL) ÷ 60 = 1,875 ÷ 60 = 31.25 ≈ 31 gtt/min. This is a fundamental IV calculation that appears on the NAPLEX. Always confirm: mL/hr is the flow rate, drop factor is the administration set size (typically 10, 15, or 60 gtt/mL), and divide by 60 to convert from per hour to per minute.
Question 4
A 500 mL bag of normal saline contains 0.9% sodium chloride. How many milligrams of sodium chloride (NaCl) are present in this 500 mL bag?
A. 450 mg B. 900 mg C. 4,500 mg D. 9,000 mg
✅ Answer: C — 4,500 mg
Percent strength (w/v) = grams of solute per 100 mL × 100%. A 0.9% NaCl solution contains 0.9 g NaCl per 100 mL. For 500 mL: 0.9 g/100 mL × 500 mL = 4.5 g = 4,500 mg. This calculation is fundamental for assessing sodium load in patients with heart failure, hypertension, or renal failure, and for calculating osmolarity of IV solutions.
Question 5
A medication order calls for dopamine 400 mg in 250 mL D5W to infuse at 5 mcg/kg/min for a 70-kg patient. What infusion rate in mL/hr should be programmed on the IV pump?
A. 8.2 mL/hr B. 13.1 mL/hr C. 16.5 mL/hr D. 26.3 mL/hr
✅ Answer: B — 13.1 mL/hr
Step 1: Calculate dose in mcg/min: 5 mcg/kg/min × 70 kg = 350 mcg/min. Step 2: Convert to mcg/hr: 350 × 60 = 21,000 mcg/hr. Step 3: Convert to mg/hr: 21,000 ÷ 1,000 = 21 mg/hr. Step 4: Calculate concentration: 400 mg ÷ 250 mL = 1.6 mg/mL. Step 5: Calculate rate: 21 mg/hr ÷ 1.6 mg/mL = 13.125 ≈ 13.1 mL/hr.
Question 6
A pharmacist needs to prepare 200 mL of a 0.5% solution from a 2% stock solution. Using the dilution equation (C1V1 = C2V2), how many mL of the 2% stock solution are needed?
A. 25 mL B. 50 mL C. 100 mL D. 150 mL
✅ Answer: B — 50 mL
C1V1 = C2V2. Where C1 = 2% (stock), C2 = 0.5% (desired), V2 = 200 mL (desired volume), V1 = unknown. Calculation: 2% × V1 = 0.5% × 200 mL. V1 = (0.5 × 200) ÷ 2 = 100 ÷ 2 = 50 mL. The pharmacist needs 50 mL of the 2% stock solution diluted with 150 mL of diluent to make 200 mL of the 0.5% solution.
Question 7
A patient's vancomycin AUC is estimated at 320 mg·h/L for MRSA treatment. The target AUC/MIC is 400-600 mg·h/L (MIC = 1 mg/L). What adjustment is most appropriate?
A. Decrease the dose — target is being exceeded B. Increase the dose or decrease the dosing interval to achieve target AUC C. No change — the current AUC is within target D. Switch to a different antibiotic — vancomycin targets cannot be achieved
✅ Answer: B — Increase the dose or decrease the dosing interval to achieve target AUC
The current AUC of 320 mg·h/L is BELOW the target range of 400-600 mg·h/L for serious MRSA infections. To increase the AUC, the pharmacist should recommend increasing the total daily dose (either increasing individual doses or decreasing the dosing interval). Per 2020 ASHP/IDSA/SIDP consensus guidelines, AUC-guided monitoring is preferred over trough-only monitoring for serious MRSA infections.
Question 8
Calculate the number of milliequivalents (mEq) in 2 grams of magnesium sulfate (MgSO4). Molecular weight of MgSO4 = 120 g/mol; valence = 2.
A. 8.3 mEq B. 16.7 mEq C. 33.3 mEq D. 66.6 mEq
✅ Answer: C — 33.3 mEq
mEq = (mg × valence) ÷ molecular weight. Calculation: mEq = (2,000 mg × 2) ÷ 120 g/mol = 4,000 ÷ 120 = 33.3 mEq. This calculation applies whenever you need to convert between grams and milliequivalents for electrolytes. Magnesium is divalent (valence = 2). This matters clinically for IV magnesium supplementation, where dosing in grams and mEq are both commonly used.
Question 9
A TPN solution is being designed for a 70 kg patient requiring 25 kcal/kg/day. The TPN contains 250 g dextrose and 60 g amino acids. How many non-protein calories does this TPN provide?
A. 240 kcal B. 650 kcal C. 850 kcal D. 1,510 kcal
✅ Answer: C — 850 kcal
Non-protein calories come from dextrose only (not amino acids, which are protein calories). Dextrose provides 3.4 kcal per gram. Non-protein calories = 250 g × 3.4 kcal/g = 850 kcal. For reference: amino acids (protein) provide 4 kcal/g = 60 × 4 = 240 protein kcal. Total calories = 850 + 240 = 1,090 kcal. Daily target: 70 kg × 25 kcal/kg = 1,750 kcal total, so additional lipid calories are needed.
Question 10
Using alligation to prepare 500 mL of 15% dextrose solution using 5% dextrose and 50% dextrose solutions, how many mL of the 50% dextrose solution are needed?
A. 44 mL B. 111 mL C. 200 mL D. 389 mL
✅ Answer: B — 111 mL
Alligation: Higher% (50) minus desired% (15) = 35 parts of the lower (5%) solution. Desired% (15) minus lower% (5) = 10 parts of the higher (50%) solution. Total parts = 35 + 10 = 45 parts. Volume of 50% solution = (10/45) × 500 mL = 111.1 mL ≈ 111 mL. Volume of 5% solution = (35/45) × 500 mL = 388.9 ≈ 389 mL. Check: 111 mL × 50% + 389 mL × 5% = 55.5 + 19.45 = 74.95 g dextrose ÷ 500 mL = 15% ✓
📌 How to use these
Answer each question before reading the rationale, and treat "right but unsure" as wrong. The rationale matters more than the answer — if you cannot explain why the other three options fail, you have not learned the rule yet. Ready for more? Work through the other free quizzes or the pharmacy law cheat sheet.