HSRT Quantitative Reasoning and Numeracy 3 — Questions and Answers
Question 1: A pharmacy prepares a 0.9% NaCl (normal saline) solution. 0.9% means 0.9 g of NaCl per 100 mL. How many milligrams of NaCl are in a 500 mL bag?
- 4500 mg (Correct answer)
- 450 mg
- 900 mg
- 45 mg
Correct answer: 4500 mg
0.9 g/100 mL × 500 mL = 4.5 g = 4500 mg of NaCl.
Concentration calculation: 0.9 g/100 mL × 500 mL = 4.5 g. Converting to milligrams: 4.5 g × 1000 mg/g = 4500 mg. This is a fundamental pharmaceutical numeracy calculation. Normal saline (0.9% NaCl) is isotonic with plasma because its osmolarity (308 mOsm/L) approximates plasma osmolarity (~285–295 mOsm/L). Understanding the actual mass of electrolytes in IV fluids is essential for managing patients with electrolyte disorders, where cumulative sodium loads from multiple bags must be tracked.
Question 2: A hospital reports a mortality rate of 8.2 per 1000 patient admissions. If the hospital had 12,500 admissions last year, how many deaths occurred?
- 103 deaths (Correct answer)
- 82 deaths
- 820 deaths
- 1025 deaths
Correct answer: 103 deaths
Deaths = (8.2 / 1000) × 12,500 = 8.2 × 12.5 = 102.5 ≈ 103 deaths.
Rate application: Deaths = Rate × Population = (8.2 per 1000) × 12,500 = 8.2 × (12,500/1000) = 8.2 × 12.5 = 102.5 ≈ 103 deaths. This type of calculation translates epidemiological rates into absolute numbers — essential for resource planning, quality reporting, and communicating the real-world burden of mortality rates. A rate of 8.2/1000 sounds small in isolation, but 103 deaths contextualizes the hospital's annual mortality burden in actionable terms for quality improvement committees.
Question 3: A patient requires a potassium infusion. The order reads: 'Infuse 40 mEq KCl in 100 mL NS over 4 hours.' Potassium chloride is available as 2 mEq/mL. How many mL of KCl concentrate must be added to the NS bag?
- 20 mL (Correct answer)
- 40 mL
- 10 mL
- 80 mL
Correct answer: 20 mL
Volume of KCl = 40 mEq ÷ 2 mEq/mL = 20 mL.
Volume calculation: Volume = Amount (mEq) / Concentration (mEq/mL) = 40 mEq / 2 mEq/mL = 20 mL. This 20 mL of KCl concentrate is added to the NS bag, yielding a final volume of approximately 120 mL. Note: potassium infusion rates must not exceed 10–20 mEq/hour via peripheral IV to prevent cardiac arrhythmia. At 40 mEq over 4 hours = 10 mEq/hour, this is at the upper limit of peripheral administration and requires continuous monitoring. Concentration and infusion rate calculations are critical safety competencies in IV electrolyte replacement.
Question 4: A research paper reports that a new drug reduces LDL cholesterol by a mean of 38 mg/dL (SD = 12 mg/dL). A patient's baseline LDL is 165 mg/dL. After treatment, what is the expected LDL range for approximately 95% of patients like this one?
- 103–151 mg/dL (expected mean reduction 38, ± 2 SD = ± 24 mg/dL) (Correct answer)
- 127–165 mg/dL (38 mg/dL reduction only)
- 90–165 mg/dL (full range of responses)
- 115–141 mg/dL (mean ± 1 SD)
Correct answer: 103–151 mg/dL (expected mean reduction 38, ± 2 SD = ± 24 mg/dL)
Expected post-treatment LDL = 165 - 38 = 127 mg/dL mean. 95% range = mean ± 2SD = 127 ± 24 = 103–151 mg/dL.
Applying population statistics to individual patients: Mean post-treatment LDL = 165 - 38 = 127 mg/dL. 95% of individuals in a normal distribution fall within ±2 SD of the mean: 127 ± (2 × 12) = 127 ± 24 = 103–151 mg/dL. This range represents the expected variation in treatment response. Some patients will reduce by only 14 mg/dL (165-38-24=103 baseline adjustment; no — post-treatment range is 103–151). This quantitative reasoning helps clinicians set realistic expectations for individual patients, recognizing that the mean response will not be achieved by all and some patients may be non-responders or hyper-responders.
Question 5: A clinical nutrition calculation shows a patient needs 1.5 g protein/kg/day. The patient weighs 70 kg. Their current enteral formula provides 60 g protein per 1000 mL. If the patient is receiving 1200 mL/day, are protein needs being met?
- No — patient needs 105 g/day but receives only 72 g/day from the formula (Correct answer)
- Yes — 1200 mL provides 1.2 × 60 = 72 g, exceeding requirements
- Yes — the formula provides exactly the required protein
- Cannot be determined without the patient's serum albumin level
Correct answer: No — patient needs 105 g/day but receives only 72 g/day from the formula
Requirements = 1.5 g/kg × 70 kg = 105 g/day. Formula delivers: (60 g/1000 mL) × 1200 mL = 72 g/day. Deficit = 33 g/day.
Step 1: Calculate protein requirement = 1.5 g/kg × 70 kg = 105 g/day. Step 2: Calculate protein delivered = (60 g / 1000 mL) × 1200 mL = 72 g/day. Step 3: Compare: 72 g < 105 g → deficit of 33 g/day (31% shortfall). This shortfall is clinically significant, particularly for critically ill patients, surgical patients, or those with pressure injuries where protein requirements for wound healing and immune function are increased. The intervention would be to increase formula rate, switch to a higher-protein formula, or add protein supplements.
Question 6: A patient's INR on warfarin was 1.8 (target 2.0–3.0). The physician increases the weekly warfarin dose by 15%. The previous weekly dose was 35 mg. What is the new weekly dose?
- 40.25 mg (Correct answer)
- 40.5 mg
- 38 mg
- 42 mg
Correct answer: 40.25 mg
New dose = 35 mg × 1.15 = 40.25 mg per week.
Percentage increase calculation: New dose = Original dose × (1 + percent increase) = 35 mg × 1.15 = 40.25 mg/week. In clinical practice, this would typically be rounded to the nearest available tablet combination (e.g., 40 or 40.5 mg). Warfarin dosing adjustments must be precise due to the drug's narrow therapeutic index. A 15% increase is a conservative adjustment appropriate for an INR of 1.8 (sub-therapeutic but not dangerously low). Larger adjustments would risk overcorrection. Follow-up INR monitoring in 5–7 days is needed to assess the response to the adjustment.
HSRT Quantitative Reasoning and Numeracy 3
A pharmacy prepares a 0.9% NaCl (normal saline) solution. 0.9% means 0.9 g of NaCl per 100 mL.
How many milligrams of NaCl are in a 500 mL bag?