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20 Drops-per-Minute Practice Questions

Practise the arithmetic behind fictional manual drop-rate questions. The exact value is shown before any exercise-specific whole-number rounding.

Educational boundary

Use fictional or tutor-provided values only. These pages explain arithmetic and do not select, recommend, verify or administer treatment. Do not enter patient information or use DosePractice in a clinical workflow.

Formula

gtt/min = (mL × gtt/mL) ÷ minutes

How to use these questions

Work each question on paper or in a separate note before opening the answer. Keep the units beside every number and write the formula before substituting values.

The questions are deliberately fictional and test arithmetic only. Real medicines, patient factors, limits, local policy and equipment checks are outside DosePractice’s purpose.

A repeatable checking method

20 practice questions

Try each fictional arithmetic question before opening its worked answer. Each item has a stable question ID and reporting link.

  1. seo-dr-001Content 0.9.2.4-prelaunchReport this question

    A fictional 500 mL volume uses a 20 gtt/mL set over 240 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    500 × 20 ÷ 240 = 41.67 gtt/min; a whole-drop exercise would round according to its stated rule

    Mathematical result for this practice example: 41.67 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  2. seo-dr-002Content 0.9.2.4-prelaunchReport this question

    A fictional 1,000 mL volume uses a 15 gtt/mL set over 480 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    1,000 × 15 ÷ 480 = 31.25 gtt/min

    Mathematical result for this practice example: 31.25 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  3. seo-dr-003Content 0.9.2.4-prelaunchReport this question

    A fictional 250 mL volume uses a 20 gtt/mL set over 120 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    250 × 20 ÷ 120 = 41.67 gtt/min

    Mathematical result for this practice example: 41.67 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  4. seo-dr-004Content 0.9.2.4-prelaunchReport this question

    A fictional 600 mL volume uses a 20 gtt/mL set over 300 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    600 × 20 ÷ 300 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  5. seo-dr-005Content 0.9.2.4-prelaunchReport this question

    A fictional 100 mL volume uses a 20 gtt/mL set over 60 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    100 × 20 ÷ 60 = 33.33 gtt/min

    Mathematical result for this practice example: 33.33 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  6. seo-dr-006Content 0.9.2.4-prelaunchReport this question

    A fictional 1,000 mL volume uses a 20 gtt/mL set over 600 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    1,000 × 20 ÷ 600 = 33.33 gtt/min

    Mathematical result for this practice example: 33.33 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  7. seo-dr-007Content 0.9.2.4-prelaunchReport this question

    A fictional 500 mL volume uses a 15 gtt/mL set over 300 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    500 × 15 ÷ 300 = 25 gtt/min

    Mathematical result for this practice example: 25 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  8. seo-dr-008Content 0.9.2.4-prelaunchReport this question

    A fictional 250 mL volume uses a 15 gtt/mL set over 150 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    250 × 15 ÷ 150 = 25 gtt/min

    Mathematical result for this practice example: 25 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  9. seo-dr-009Content 0.9.2.4-prelaunchReport this question

    A fictional 120 mL volume uses a 20 gtt/mL set over 60 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    120 × 20 ÷ 60 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  10. seo-dr-010Content 0.9.2.4-prelaunchReport this question

    A fictional 300 mL volume uses a 20 gtt/mL set over 180 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    300 × 20 ÷ 180 = 33.33 gtt/min

    Mathematical result for this practice example: 33.33 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  11. seo-dr-011Content 0.9.2.4-prelaunchReport this question

    A fictional 1,000 mL volume uses a 10 gtt/mL set over 480 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    1,000 × 10 ÷ 480 = 20.83 gtt/min

    Mathematical result for this practice example: 20.83 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  12. seo-dr-012Content 0.9.2.4-prelaunchReport this question

    A fictional 750 mL volume uses a 20 gtt/mL set over 360 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    750 × 20 ÷ 360 = 41.67 gtt/min

    Mathematical result for this practice example: 41.67 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  13. seo-dr-013Content 0.9.2.4-prelaunchReport this question

    A fictional 500 mL volume uses a 10 gtt/mL set over 240 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    500 × 10 ÷ 240 = 20.83 gtt/min

    Mathematical result for this practice example: 20.83 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  14. seo-dr-014Content 0.9.2.4-prelaunchReport this question

    A fictional 240 mL volume uses a 15 gtt/mL set over 120 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    240 × 15 ÷ 120 = 30 gtt/min

    Mathematical result for this practice example: 30 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  15. seo-dr-015Content 0.9.2.4-prelaunchReport this question

    A fictional 100 mL volume uses a 15 gtt/mL set over 90 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    100 × 15 ÷ 90 = 16.67 gtt/min

    Mathematical result for this practice example: 16.67 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  16. seo-dr-016Content 0.9.2.4-prelaunchReport this question

    A fictional 200 mL volume uses a 20 gtt/mL set over 100 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    200 × 20 ÷ 100 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  17. seo-dr-017Content 0.9.2.4-prelaunchReport this question

    A fictional 400 mL volume uses a 20 gtt/mL set over 200 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    400 × 20 ÷ 200 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  18. seo-dr-018Content 0.9.2.4-prelaunchReport this question

    A fictional 900 mL volume uses a 15 gtt/mL set over 540 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    900 × 15 ÷ 540 = 25 gtt/min

    Mathematical result for this practice example: 25 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  19. seo-dr-019Content 0.9.2.4-prelaunchReport this question

    A fictional 150 mL volume uses a 20 gtt/mL set over 75 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    150 × 20 ÷ 75 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

  20. seo-dr-020Content 0.9.2.4-prelaunchReport this question

    A fictional 60 mL volume uses a 20 gtt/mL set over 30 minutes. What is the exact drop rate?

    Show worked answer

    Formula or method: gtt/min = (mL × gtt/mL) ÷ minutes

    Substitution and working:

    60 × 20 ÷ 30 = 40 gtt/min

    Mathematical result for this practice example: 40 gtt/min

    Check the operation order, unit conversion and final unit before moving on.

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