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Bioavailability (F) Practice Problems

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Absolute bioavailability compares how much drug reaches the systemic circulation from an oral dose versus an IV dose (which is 100% by definition). It is calculated from the dose-normalized areas under the concentration-time curve (AUC).

FormulaF = (AUC_oral ÷ Dose_oral) × (Dose_IV ÷ AUC_IV)
Worked example

IV 100 mg gives AUC 50; oral 200 mg gives AUC 60. First term: 60 ÷ 200 = 0.30; second term: 100 ÷ 50 = 2; F = 0.30 × 2 = 0.60 = 60%.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

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Bioavailability (F)
A 200 mg IV dose gives an AUC of 50 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.
%

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10 Bioavailability (F) practice problems with worked solutions

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Problem 1: Bioavailability (F)

A 50 mg IV dose gives an AUC of 60 mg·hr/L. A 400 mg oral dose gives an AUC of 240 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 50 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 240 mg·hr/L × Dose_iv 50 mg = 12000.
  4. Denominator: Dose_po 400 mg × AUC_iv 60 mg·hr/L = 24000.
  5. F = 12000 ÷ 24000 = 0.5, i.e. 50%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 50 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 2: Bioavailability (F)

A 200 mg IV dose gives an AUC of 40 mg·hr/L. A 100 mg oral dose gives an AUC of 18 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 90 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 18 mg·hr/L × Dose_iv 200 mg = 3600.
  4. Denominator: Dose_po 100 mg × AUC_iv 40 mg·hr/L = 4000.
  5. F = 3600 ÷ 4000 = 0.9, i.e. 90%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 90 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 3: Bioavailability (F)

A 100 mg IV dose gives an AUC of 80 mg·hr/L. A 100 mg oral dose gives an AUC of 72 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 90 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 72 mg·hr/L × Dose_iv 100 mg = 7200.
  4. Denominator: Dose_po 100 mg × AUC_iv 80 mg·hr/L = 8000.
  5. F = 7200 ÷ 8000 = 0.9, i.e. 90%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 90 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 4: Bioavailability (F)

A 50 mg IV dose gives an AUC of 30 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 22.5 mg·hr/L × Dose_iv 50 mg = 1125.
  4. Denominator: Dose_po 150 mg × AUC_iv 30 mg·hr/L = 4500.
  5. F = 1125 ÷ 4500 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 5: Bioavailability (F)

A 200 mg IV dose gives an AUC of 60 mg·hr/L. A 150 mg oral dose gives an AUC of 31.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 70 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 31.5 mg·hr/L × Dose_iv 200 mg = 6300.
  4. Denominator: Dose_po 150 mg × AUC_iv 60 mg·hr/L = 9000.
  5. F = 6300 ÷ 9000 = 0.7, i.e. 70%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 70 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 6: Bioavailability (F)

A 50 mg IV dose gives an AUC of 40 mg·hr/L. A 200 mg oral dose gives an AUC of 40 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 40 mg·hr/L × Dose_iv 50 mg = 2000.
  4. Denominator: Dose_po 200 mg × AUC_iv 40 mg·hr/L = 8000.
  5. F = 2000 ÷ 8000 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 7: Bioavailability (F)

A 100 mg IV dose gives an AUC of 50 mg·hr/L. A 150 mg oral dose gives an AUC of 22.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 30 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 22.5 mg·hr/L × Dose_iv 100 mg = 2250.
  4. Denominator: Dose_po 150 mg × AUC_iv 50 mg·hr/L = 7500.
  5. F = 2250 ÷ 7500 = 0.3, i.e. 30%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 30 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 8: Bioavailability (F)

A 100 mg IV dose gives an AUC of 25 mg·hr/L. A 250 mg oral dose gives an AUC of 15.625 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 15.625 mg·hr/L × Dose_iv 100 mg = 1562.5.
  4. Denominator: Dose_po 250 mg × AUC_iv 25 mg·hr/L = 6250.
  5. F = 1562.5 ÷ 6250 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 9: Bioavailability (F)

A 100 mg IV dose gives an AUC of 40 mg·hr/L. A 100 mg oral dose gives an AUC of 12 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 30 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 12 mg·hr/L × Dose_iv 100 mg = 1200.
  4. Denominator: Dose_po 100 mg × AUC_iv 40 mg·hr/L = 4000.
  5. F = 1200 ÷ 4000 = 0.3, i.e. 30%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 30 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

Problem 10: Bioavailability (F)

A 50 mg IV dose gives an AUC of 25 mg·hr/L. A 300 mg oral dose gives an AUC of 37.5 mg·hr/L. What is the absolute bioavailability (%)? Round your answer to the nearest whole number.

Answer 25 %

Solution
  1. F = (AUC_po ÷ Dose_po) × (Dose_iv ÷ AUC_iv).
  2. Rearranged as one fraction: F = (AUC_po × Dose_iv) ÷ (Dose_po × AUC_iv).
  3. Numerator: AUC_po 37.5 mg·hr/L × Dose_iv 50 mg = 1875.
  4. Denominator: Dose_po 300 mg × AUC_iv 25 mg·hr/L = 7500.
  5. F = 1875 ÷ 7500 = 0.25, i.e. 25%. the mg·hr/L and mg units cancel top and bottom, which is why F is a dimensionless fraction and never carries a unit.
  6. Rounded as the question asks: 25 %.

Always dose-normalize both AUCs. Comparing raw AUCs from different doses is the built-in trap.

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