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Alligation Practice Problems

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Alligation mixes two preparations of different strengths to get an intermediate strength. Alligation alternate starts from two known strengths and finds the quantity of each to mix; alligation medial starts from known quantities and finds the resulting strength. The alternate method uses a grid where you take diagonal differences to get the parts of each component.

Formulaparts of high = target − low  •  parts of low = high − target  •  volume of each = (its parts ÷ total parts) × final volume
Worked example

Make 250 mL of 25% dextrose from 70% and 10% stock. Parts of 70% = 25 − 10 = 15; parts of 10% = 70 − 25 = 45; total = 60. Volume of 70% = (15 ÷ 60) × 250 = 62.5 mL (and 187.5 mL of the 10%).

Don't confuse the two: medial = known quantities → find strength; alternate = known strengths → find quantities.

The alligation grid Two diagrams. The first is the general form: the higher strength sits top left, the lower strength bottom left, and the desired strength in the middle. Subtracting diagonally gives the parts of each. Parts of the higher strength equal desired minus lower; parts of the lower strength equal higher minus desired. Total parts is the sum. The second diagram works the same grid for 95 percent and 70 percent isopropyl alcohol mixed to 90 percent, giving 20 parts of 95 percent and 5 parts of 70 percent, 25 parts in total. THE GENERAL FORM Higher strength Lower strength Desired strength Desired − Lower = parts of the HIGHER Higher − Desired = parts of the LOWER Total parts = higher parts + lower parts WORKED: 95% AND 70% IPA TO MAKE 90% 95% IPA 70% IPA 90% desired 90 − 70 = 20 parts of the 95% IPA 95 − 90 = 5 parts of the 70% IPA Total = 25 parts, so 20/25 is 95% stock and 5/25 is 70% stock
Subtract along the diagonals. The number that lands on the right is the parts of the strength diagonally opposite it. That crossover is the step people get backwards.
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Alligation Medial
You mix 150 mL of 50% dextrose with 1000 mL of 10% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.
%

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10 Alligation practice problems with worked solutions

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Problem 1: Alligation Medial

You mix 250 mL of 70% dextrose with 750 mL of 5% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.

Answer 21.3 %

Solution
  1. Grams from each: 250 mL × 70% = 175 g; 750 mL × 5% = 37.5 g.
  2. Total drug = 212.5 g in total volume 1000 mL.
  3. Strength = 212.5 g ÷ 1000 mL × 100 = 21.25%.
  4. Rounded as the question asks: 21.3 %.

Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.

Problem 2: Alligation Alternate

You need 120 g of 2% 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 90.0 g

Solution
  1. Alligation grid: high (2.5%) top-left, low (0.5%) bottom-left, target (2%) center.
  2. Parts of 2.5% = 2 − 0.5 = 1.5 parts (diagonal difference).
  3. Parts of 0.5% = 2.5 − 2 = 0.5 parts.
  4. Total parts = 2.
  5. Amount of 2.5% = (1.5 parts ÷ 2 total parts) × 120 g = 90 g (and 30 g of 0.5%).
  6. Rounded as the question asks: 90.0 g.

Sanity check: 90×2.5% + 30×0.5% = 2.4 g = 120 g × 2% ✓

Problem 3: Alligation Medial

You mix 150 mL of 50% dextrose with 200 mL of 10% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.

Answer 27.1 %

Solution
  1. Grams from each: 150 mL × 50% = 75 g; 200 mL × 10% = 20 g.
  2. Total drug = 95 g in total volume 350 mL.
  3. Strength = 95 g ÷ 350 mL × 100 ≈ 27.143%.
  4. Rounded as the question asks: 27.1 %.

Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.

Problem 4: Alligation Alternate

You need 240 mL of 80% alcohol, prepared from 95% and 70% stock. How many mL of the 95% stock do you need? Round your answer to one decimal place.

Answer 96.0 mL

Solution
  1. Alligation grid: high (95%) top-left, low (70%) bottom-left, target (80%) center.
  2. Parts of 95% = 80 − 70 = 10 parts (diagonal difference).
  3. Parts of 70% = 95 − 80 = 15 parts.
  4. Total parts = 25.
  5. Amount of 95% = (10 parts ÷ 25 total parts) × 240 mL = 96 mL (and 144 mL of 70%).
  6. Rounded as the question asks: 96.0 mL.

Sanity check: 96×95% + 144×70% = 192 g = 240 mL × 80% ✓

Problem 5: Alligation Medial

You mix 250 mL of 50% dextrose with 750 mL of 10% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.

Answer 20.0 %

Solution
  1. Grams from each: 250 mL × 50% = 125 g; 750 mL × 10% = 75 g.
  2. Total drug = 200 g in total volume 1000 mL.
  3. Strength = 200 g ÷ 1000 mL × 100 = 20%.
  4. Rounded as the question asks: 20.0 %.

Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.

Problem 6: Alligation Alternate

You need 1000 mL of 85% alcohol, prepared from 95% and 70% stock. How many mL of the 95% stock do you need? Round your answer to one decimal place.

Answer 600.0 mL

Solution
  1. Alligation grid: high (95%) top-left, low (70%) bottom-left, target (85%) center.
  2. Parts of 95% = 85 − 70 = 15 parts (diagonal difference).
  3. Parts of 70% = 95 − 85 = 10 parts.
  4. Total parts = 25.
  5. Amount of 95% = (15 parts ÷ 25 total parts) × 1000 mL = 600 mL (and 400 mL of 70%).
  6. Rounded as the question asks: 600.0 mL.

Sanity check: 600×95% + 400×70% = 850 g = 1000 mL × 85% ✓

Problem 7: Alligation Medial

You mix 300 mL of 50% dextrose with 250 mL of 5% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.

Answer 29.5 %

Solution
  1. Grams from each: 300 mL × 50% = 150 g; 250 mL × 5% = 12.5 g.
  2. Total drug = 162.5 g in total volume 550 mL.
  3. Strength = 162.5 g ÷ 550 mL × 100 ≈ 29.545%.
  4. Rounded as the question asks: 29.5 %.

Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.

Problem 8: Alligation Alternate

You need 250 mL of 20% dextrose, prepared from 70% and 10% stock. How many mL of the 70% stock do you need? Round your answer to one decimal place.

Answer 41.7 mL

Solution
  1. Alligation grid: high (70%) top-left, low (10%) bottom-left, target (20%) center.
  2. Parts of 70% = 20 − 10 = 10 parts (diagonal difference).
  3. Parts of 10% = 70 − 20 = 50 parts.
  4. Total parts = 60.
  5. Amount of 70% = (10 parts ÷ 60 total parts) × 250 mL ≈ 41.667 mL (and 208.333 mL of 10%).
  6. Rounded as the question asks: 41.7 mL.

Sanity check: 41.667×70% + 208.333×10% = 50 g = 250 mL × 20% ✓

Problem 9: Alligation Medial

You mix 100 mL of 50% dextrose with 750 mL of 10% dextrose. What is the percentage strength of the final mixture? Round your answer to one decimal place.

Answer 14.7 %

Solution
  1. Grams from each: 100 mL × 50% = 50 g; 750 mL × 10% = 75 g.
  2. Total drug = 125 g in total volume 850 mL.
  3. Strength = 125 g ÷ 850 mL × 100 ≈ 14.706%.
  4. Rounded as the question asks: 14.7 %.

Alligation MEDIAL = known quantities → find resulting strength. Alligation ALTERNATE = known strengths → find quantities.

Problem 10: Alligation Alternate

You need 240 mL of 85% alcohol, prepared from 95% and 50% stock. How many mL of the 95% stock do you need? Round your answer to one decimal place.

Answer 186.7 mL

Solution
  1. Alligation grid: high (95%) top-left, low (50%) bottom-left, target (85%) center.
  2. Parts of 95% = 85 − 50 = 35 parts (diagonal difference).
  3. Parts of 50% = 95 − 85 = 10 parts.
  4. Total parts = 45.
  5. Amount of 95% = (35 parts ÷ 45 total parts) × 240 mL ≈ 186.667 mL (and 53.333 mL of 50%).
  6. Rounded as the question asks: 186.7 mL.

Sanity check: 186.667×95% + 53.333×50% = 204 g = 240 mL × 85% ✓

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