SOLUTIONS FOR:

Errors, Omissions, and Calculations (EOAC) Exam: Calculations Practice Questions

APPE Ready (PHCY 3202)

1. What is the percentage strength of a 1:400 solution of an oil in alcohol?

Solution:
1mL/400mL = X mL /100 mL; X = 0.25 mL = 0.25% (v/v)

2. A viral vaccine is preserved with 1:10,000 solution of benzalkonium chloride (a solid). How do you
express this concentration as a percentage?
Solution:
1g/10,000mL = X g /100 mL; X = 0.01 g = 0.01% (w/v)

3. How many liters of a 1:1500 solution can be made by dissolving 4.80 g of Drug X in water?
Solution:
1g/1500mL = X g /100 mL; X = 0.067 g
4.8 g x 100 mL/0.067 g = 7164.2 mL = 7.2 L

4. How many g of lactose should be combined with 140 mg of a drug to make a 1:10 dilution?
Solution:
1 part = 140 mg of drug
9 parts = X mg of lactose
X = 1260 mg = 1.26 g of lactose

5. Calculate the number of mg of Drug X in 90 mL of a 1:500 w/v Drug X solution.

Solution:
90 mL x 1g/500mL x 1000 mg/g = 180 mg

6. How many mg of drug are there in 25 mL of 3.25% w/v solution? Round to the nearest whole
number.
Solution:
3.25 g/100mL x 25 mL = 0.8125 g = 813 mg

7. How many g of solute are there in 170 mL of 15.0% w/v solution?

Solution:
170 mL x 15g/100mL = 25.5 g

8. How many mL of a 10% w/v Drug X solution should be diluted with water to make 440 mL of a 0.25%
w/v Drug X solution?
Solution:
(10% x Q1) = (0.25% x 440 mL); Q1 = 11 mL

9. How many mL of water should be added to 200 mL of a topical solution containing 1:50 (v/v) methyl
salicylate to make a solution with 1:400 (v/v) methyl salicylate?
Solution:
1:50 = (1/50) x 100 = 2% and 1:400 = (1/400) x 100 = 0.25%
200 mL x 2% = (200 mL + X) x 0.25%; 350 = 0.25 (X); X = 1400mL of water
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10. Use alligation method to find how many mL of 95% alcohol should be diluted with water to make 65
mL of 40% alcohol.
Solution:
Water = 0%
40 parts = X mL ; X = 27.4 mL of 95% alcohol
95% 40 parts 95% 95 parts 65 mL
40% (Desired) 95 parts Total = 65 mL Total

0% 55 parts of water

40 parts/95 parts = x/65 mL total ; x=27.4 mL of 95% alcohol

11. Use alligation method to find how many mL of a 2% (w/v) lidocaine HCl solution and how many mL
of normal saline (NS) should be mixed to make 500mL of a diluted solution containing 4 mg of the
drug per mL.
Solution:
Convert final concentration to a percentage: 4 mg/mL = 400mg/100mL or 0.4g/100mL = 0.4%

0.4 parts = X mL ; X = 100 mL of 2% Lid. HCl

2% (Lid. HCl) 0.4 parts Lid. HCl 2 parts 500mL
0.4% (Desired) 2 parts Total = 500mL Total

0% (NS) 1.6 parts NS

500 mL Total – 100 mL = 400 mL of NS

0.4 parts/2 parts = x/500 mL total ; x=100 mL of Lid. HCl

500 mL total – 100 mL of Lid. HCl = 400 mL of NS

12. Use alligation method to find how many mL of sorbitol solution (specific gravity = 1.25) and how
many mL of water should mixed to prepare 200 mL of a sorbitol product with a specific gravity of
1.20.
Solution:
Specific gravity of water = 1.0

0.20 parts = X g ; X = 160 g Sorbitol solution

1.25 0.20 parts of 1.25 0.25 parts 200

1.20 (Desired) 0.25 parts Total = 200 mL Total

1.00 0.05 parts of water 200 Total – 160 g = 40 g (mL) of water

0.20 parts/0.25 parts = x/200 ; x=160 mL of sorbitol

200 total – 160 mL = 40 mL of water

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13. If 250 g of dextrose are dissolved in 300 mL of water, what is the percentage strength of dextrose in
the solution in w/w basis?
Solution:
  
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛  = 250 g/(250 g + 300 g) = 0.45 g x 100 = 45%
  

14. The formula for 1 liter of syrup contains 0.5 mL of orange oil. What is the percentage strength (v/v)
of the oil in syrup?
Solution:
= (0.5 mL/1,000 mL) x 100 = 0.05%

15. How much boric acid (a solid) and how much camphor water (a liquid) is needed for the following
prescription?

Rx: Boric acid 2%

Camphor water 35%
Aqua pur qs ad 15.0 mL
Ft. sol
Solution:
Boric acid (w/v): 2g/100 mL x 15.0 mL = 0.3 g
Camphor water (v/v): 0.35 x 15.0 mL = 5.25 mL

16. An order for a patient weighing 121lb calls for 0.25 mg/kg of Drug X to be added to 200mL D5W.
Drug X is available in a concentration of 50 mg/mL. How many mL should be added to the D5W
solution? Round answer to the nearest tenth.
Solution:
121 lb x (1 kg/2.2 lb) x 0.25 mg/kg x mL/50mg = (121 x 0.25)/(2.2 x 50) = 0.275 mL = 0.3 mL

17. An IV admixture calls for addition of 20mEq of sodium bicarbonate. How many mL of 8.4%(w/v)
sodium bicarbonate solution should be added? (1 mEq = 84 mg)
Solution:
20mEq x 84 mg/mEq x 100mL/8400 mg = (20 x 84 x 100)/8400 = 20 mL

18. A patient receives a solution by intravenous infusion at a rate of 36 drops/min. How much solution is
infused in 3 hours if the infusion set delivers 30 drops/mL?
Solution:
36 drop/min x 1 mL/30 drops = (36/30) = 1.2 mL/min x 180 min = 216 mL

19. A solution contains 1.25 gm of drug per mL. At what rate should the solution be infused (drops/min)
if the drug is to be administered at a rate of 80mg/hr? (set delivers 30 drops/mL)
Solution:
30 drops/mL x mL/1.25 mg x 80 mg/hr x hr/60 min = 32 drops/min
SOLUTIONS FOR:
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20. A physician orders 50 µg/kg digoxin for a 10 lb newborn baby to be diluted five‐fold with D5W. How
many mL of 0.25 mg/mL digoxin injection should be used?
Solution:
mg  103 µg  0.25mg/mL = 250 µg/mL
50 µg/kg x 1 kg/2.2lb x 10lb x mL/250 µg/mL = (50 x 10)/(2.2 x 250) = 0.9 mL

21. Calculate the CrCl (mL/min) for a 57 year old black female (Ht. 64 in, Wt. 65.9 kg). Her most recent
serum creatinine is 1.4 mg/dL.
Solution:
(140 – 57) x 65.9 kg x (0.85 for a female) = 46.1 mL/min
72 x 1.4

22. Calculate the CrCl (mL/min) for a 42‐year‐old white male (Ht. 71 in, Wt. 82.7 kg). His most recent
serum creatinine is 0.6 mg/dL. Use Ideal body weight to calculate CrCl.
Solution:
IBW = 50 + 2.3 (11) = 75.3 kg

(140 – 42) x 75.3 kg = 170.8 mL/min

72 x 0.6

23. Calculate the CrCl (mL/min) for a 54‐year‐old Asian female (Ht. 61 in, Wt. 54.5 kg). Her most recent
serum creatinine is 1.1 mg/dL. Use Ideal body weight to calculate CrCl.
Solution:
IBW = 45.5 + 2.3 (1) = 47.8 kg

(140 – 54) x 47.8 kg x (0.85 for a female) = 44.1 mL/min

72 x 1.1

24. Calculate the adjusted body weight and body mass index (kg/m2) for a 27‐year‐old female with a
total body weight of 104.5 kg and height of 68 inches.
Solution:
IBW = 45.5 + 2.3 (8) = 63.9 kg

ABW = 63.9 kg + 0.4 (104.5 kg – 63.9 kg) = 80.1 kg

BMI = 104.5 kg/(68in x .0254)m2 = 35 kg/m2

25. Calculate the adjusted body weight and body mass index (kg/m2) for a 56‐year‐old male with a
total body weight of 95.5 kg and height of 71 inches.
Solution:
IBW = 50 + 2.3 (11) = 75.3 kg
ABW = 75.3 kg + 0.4 (95.5 kg – 75.3 kg) = 83.4 kg

BMI = 95.5 kg/(71in x .0254)m2 = 29.4 kg/m2

SOLUTIONS FOR:
Errors, Omissions, and Calculations (EOAC) Exam: Calculations Practice Questions
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26. Calculate the BSA (m2) for an adult patient weighing 142 lb and measuring 62 in.
Solution:
 /.    .
BSA (m2) = = 1.68 m2


27. Calculate BSA (m2) for an adult patient weighing 180 lb and measuring 68 in.
Solution:
 /.    .
BSA (m2) = = 1.98 m2


28. A patient receives a prescription for Humulin U‐100 10mL vial with the direction: 10 units TID 15
minutes before meals. The patient’s insurance will cover a 90‐day supply. How many vials are
needed for a 90‐day supply? How many extra days will this supply last the patient?
Solution:
U‐100 indicates 100units/mL ; therefore 1 vial = (100 units/mL x 10mL) = 1,000 units
The patient needs 30 units/day x 90 days = 2700 units; therefore, the patient needs 3 bottles for a
90‐day supply.

The day supply for 3 bottles is 3,000 units/30 units = 100 days, so the patient will have 10 extra days.

29. A patient receives a prescription for Lantus Solostar U‐100 pens with directions 35 units SQ QHS.
Calculate the day supply. How many boxes will the patient need for 90 days?

Solution: 1 pen = 3ml, 1 box = 5 pens, 3x5mL =15mL x100 units = 1,500 units / 35 = 42 days

2 boxes = 3,000 units/ 35 = 85 days

30. A patient receives a prescription for Prednisolone ophthalmic drops with the directions: 2 gtt OU BID
for 7 days. How many mL of Prednisolone ophthalmic drops will the patient need for the entire
course of the prescription?

Solution:
The patient needs 8 drops/day x 7 days = 56 drops; 1 mL/20 drops x 56 drops = 2.8 mL

31. A patient with a calcium level of 8.1 mg/dL and an albumin level of 2.2 is admitted to ED with
dehydration. Find the corrected calcium level.

Solution:
Corrected Calcium = 8.1 + [ 0.8 x (4 — 2.2) = 9.54 mg/dL
SOLUTIONS FOR:
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32. Each gallon of a dilute alcohol solution contains 500 ml of absolute ethanol. What is the v/v% of
ethanol in the solution?

Solution:
1 𝑔𝑎𝑙𝑙𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 500 𝑚𝑙 𝐸𝑡𝑂𝐻
    100%  13.2%
3785 𝑚𝑙 1 𝑔𝑎𝑙𝑙𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

33. How much hydrocortisone powder is needed to prepare 2 pounds of a 2% hydrocortisone ointment?

Solution:
454 𝑔𝑟𝑎𝑚𝑠
2 𝑙𝑏 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡     908 𝑔𝑟𝑎𝑚𝑠 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑝𝑟𝑒𝑝𝑎𝑟𝑒𝑑
1𝑙𝑏
2 𝑔 ℎ𝑦𝑑𝑟𝑜𝑐𝑜𝑟𝑡𝑖𝑠𝑜𝑛𝑒 𝑥 𝑔 ℎ𝑦𝑑𝑟𝑜𝑐𝑜𝑟𝑡𝑖𝑠𝑜𝑛𝑒
  
100 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡 908 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡
X = 18.2 g hydrocortisone

34. A sodium fluoride solution is prepared by dissolving 50 mg of sodium fluoride in enough water to
make 100 L of solution. Express the concentration of the solution as ratio strength.

Solution:
50 𝑚𝑔 𝑠𝑜𝑑𝑖𝑢𝑚 𝑓𝑙𝑢𝑜𝑟𝑖𝑑𝑒 1𝑔
    0.05 𝑔 𝑠𝑜𝑑𝑖𝑢𝑚 𝑓𝑙𝑢𝑜𝑟𝑖𝑑𝑒
1000 𝑚𝑔
100 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1000 𝑚𝑙
    100,000 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1𝐿

0.05 𝑔 𝑠𝑜𝑑𝑖𝑢𝑚 𝑓𝑙𝑢𝑜𝑟𝑖𝑑𝑒

 100%  0.00005% 𝑤/𝑤
100,000 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0.00005 𝑔 1 𝑝𝑎𝑟𝑡

100 𝑚𝑙 𝑥 𝑝𝑎𝑟𝑡𝑠
x = 2,000,000; thus, a ratio strength of 1:2,000,000

35. What is the ratio strength of a 0.25% solution?

Solution:
0.25 𝑔 1 𝑝𝑎𝑟𝑡

100 𝑚𝑙 𝑥 𝑝𝑎𝑟𝑡𝑠
X = 400; thus, a ratio strength of 1:400
SOLUTIONS FOR:
Errors, Omissions, and Calculations (EOAC) Exam: Calculations Practice Questions
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36. The typical dose of lithium carbonate (Li2CO3; MW = 74) in the treatment of bipolar disorder is 600
mg three times a day. How many milliequivalents of lithium will a patient on taking this dose receive
each day?

Solution:

     

     48.6 𝑚𝐸𝑞/day
   

37. A solution is prepared by adding 50 g of calcium chloride (CaCl2; MW = 111) to enough water to
make 500 ml of solution. Express the concentration of the solution as mOsm / L.

Solution:

𝑚𝑂𝑠𝑚𝑜𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑔⁄𝐿

  # 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠  1000
𝐿 𝑀𝑊 𝑔⁄𝑚𝑜𝑙𝑒
𝑚𝑂𝑠𝑚𝑜𝑙 50 𝑔⁄0.5 𝐿
  3  1000  2703 𝑚𝑂𝑠𝑚/𝐿
𝐿 111 𝑔/𝑚𝑜𝑙𝑒

38. How much amoxicillin should a 40 lb male child receive in each dose when the usual dose is 30
mg/kg q 8 hrs?

Solution:

(40 lb) x (1 kg/ 2.2 lb) = 18.18 kg

Dose = (30 mg/kg) x (18.18 kg) = 545.45 mg

39. How much phenytoin is needed for a 75 kg female when the usual loading dose is 18 mg/kg?

Solution:

Dose = (18 mg/kg) x (75 kg) = 1350 mg

40. How many 300 mg lithium carbonate tablets (Li2CO3 ; MW = 73.89) are needed to provide a 16 mEq
lithium dose?

Solution:
16 𝑚𝐸𝑞 74 𝑚𝑔 𝑡𝑎𝑏𝑙𝑒𝑡
     2 𝑡𝑎𝑏𝑙𝑒𝑡𝑠/𝑑𝑜𝑠𝑒
𝑑𝑜𝑠𝑒 2 𝑚𝐸𝑞 300 𝑚𝑔
41. How much of a 1:1000 atropine solution is needed to give a 1 mg dose?

Solution:
1 part in 1000 parts = 1 g/1000 mL
1 𝑚𝑔 1𝑔 1000 𝑚𝑙
     1 𝑚𝑙/𝑑𝑜𝑠𝑒
𝑑𝑜𝑠𝑒 1000 𝑚𝑔 1 𝑔
SOLUTIONS FOR:
Errors, Omissions, and Calculations (EOAC) Exam: Calculations Practice Questions
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42. If 350ml of a 25% (v/v) solution is diluted to 2000ml, what is the new percentage strength?

Solution:

25% v/v = 25 ml/100 ml

350 ml solution 25 ml x 100% = 4.375% v/v
100 ml 2000 ml (new total volume)

43. How many grams of 10% (w/w) ammonia solution can be made from 2000 g of 28% (w/w) strong
ammonia solution?

Solution:

Knowing that 10% w/w = 10 g/100 g and 28% w/w = 28 g/100 g,

2000 g strong solution 28 g 100 g of 10% solution = 5600 g of 10% w/w
100 g strong solution 10 g ammonia solution

44. How many milliliters of a 60% (w/v) stock solution need to be used to prepare 3 pints of a 5% (w/v)
solution?

Solution:

3 pints 473 ml = 1419 ml of new 5% solution

1 pint
Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(0.6)(x) = (0.05)(1419)
x = 118.3 ml

Dilution Problems

Remember that dilution problems can be solved by a variety of approaches, including:

1. Using inverse proportions

𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑐𝑜𝑛𝑐.

𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑐𝑜𝑛𝑐.
2. Algebraically using equality of products of concentration and quantity
Old Conc. (C1) x Old Vol. (V1) = New Conc. (C2) x New Vol. (V2)
3. Using a two‐step process
a. Solve for amount of drug contained in the undiluted product OR in the undiluted
product (i.e., multiply quantity x concentration where you know both for the same
product), then
b. Divide the amount of drug by the new (diluted) volume to determine the new
concentration OR divide by the desired concentration to determine the new volume to
which the product should be diluted
4. Using alligation methods (see practice problems 10‐12)
SOLUTIONS FOR:
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45. How much zinc oxide should be added to 1 pound of 3% zinc oxide ointment to make a 10% zinc
oxide paste?

Solution:

Grams of 3% zinc oxide = 454 (remember, 1 lb = 454 g)

Grams of pure zinc oxide powder (100%) needed = x
Final product concentration = 10%
Final product weight (in g) = (454+x)  Remember the new weight of the compound includes the zinc oxide being added!

Now, can solve either algebraically or using alligation method (algebraic method shown below):
454  0.03  𝑥  1  454  𝑥 0.1
13.62  𝑥   45.4  0.1𝑥
0.9𝑥  31.78
𝑥  35.3 𝑔

46. How much plain ointment base should be added to 50 grams of a 5% coal tar ointment to make a 2%
coal tar ointment.

Solution:

Grams of 5% coal tar = 50

Grams of ointment base (0%) needed = x
Final product concentration = 2%
Final product weight (in g) = (50+x)
Remember the new weight of the compound includes the base being added!

Now, can solve using various methods (algebraic method shown below):
50  0.05  𝑥  0  50  𝑥 0.02
2.5  0  1  0.02𝑥
1.5  0.02𝑥
𝑥  75 𝑔

47. A parenteral nutrition formula calls for amino acids at a final concentration of 3.5%. How much of an
8.5% amino acid stock solution is needed to prepare 2L of the final formulation?

Solution:

Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(0.085)(x)=(0.035)(2000)
0.085x=70
x=823.5 ml
SOLUTIONS FOR:
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48. If 200 ml of a 15% betadine solution are diluted to 500 ml, what will be the percentage
concentration of the resulting solution?
Solution:

Volume (ml) of 15% betadine = 200

Volume (ml) of water (0%) added = 300 (final volume of 500, started with 200, so 300 ml water were added)
Final product concentration = x
Final product volume (ml) = 500

Using algebraic method to solve:

200  0.15  300  0  500𝑥 
30  0  500𝑥
𝑥  0.06, 𝑠𝑜 6%

49. Use the following order to answer the questions (a‐c) below.

IV med order:
D5W 1000ml
Containing
Magnesium sulfate, USP 20 mEq
Infuse over 8 hours.

D5W = 5% dextrose in water, MW = 180

Magnesium Sulfate, USP = MgSO4 • 7 H2O, MW = 246.47
a) How much of a 50% solution of magnesium sulfate, USP is needed to prepare the bag? Round
your answer to the nearest hundredth.
Solution:
Remember that magnesium is a divalent cation, so MW is per 2 mEq.
  .     
      4.93 𝑚𝑙 of 50% magnesium solution
      

b) What is the flow rate in ml/min? Round your answer to the nearest tenth.
Solution:
1000 ml infused over 8 hours,
1000 𝑚𝑙 1 ℎ𝑟
    2.1 𝑚𝑙/𝑚𝑖𝑛
8 ℎ𝑟𝑠 60 𝑚𝑖𝑛
c) What is the flow rate in drops/min if the infusion set delivers 20 drops/ml? Round your
answer to the nearest whole number.
Solution:
1000 𝑚𝑙 1 ℎ𝑟 20 𝑑𝑟𝑜𝑝𝑠
     42 𝑑𝑟𝑜𝑝𝑠/𝑚𝑖𝑛
8 ℎ𝑟𝑠 60 𝑚𝑖𝑛 𝑚𝑙
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50. How many 1‐gram Carafate tablets should be used in preparing the prescription below?

Rx
Carafate 400mg/5ml
Cherry Syrup 60 ml
Sorbitol Solution 60 ml
Flavor q.s.
Purified Water ad 165 ml
Solution:
165 𝑚𝑙 400 𝑚𝑔
    13200 𝑚𝑔 𝐶𝑎𝑟𝑎𝑓𝑎𝑡𝑒 𝑛𝑒𝑒𝑑𝑒𝑑
5 𝑚𝑙
1 g = 1000 mg, so 13.2 g Carafate needed
Thus, use 14 tablets (and scale to proper final volume for desired 400 mg/5ml concentration)

51. The only atropine sulfate tablets available are 0.4 mg tablets. How many tablets are needed to
compound this prescription?

Rx
Atropine Sulfate 300 μg
Phenobarbital 30 mg
Dextroamphetamine Sulfate 3 mg
Make 20 such capsules
Sig: One capsule as directed.
Solution:
20 𝑐𝑎𝑝𝑠𝑢𝑙𝑒𝑠 300 𝑚𝑐𝑔 𝑚𝑔
     6 𝑚𝑔 𝑎𝑡𝑟𝑜𝑝𝑖𝑛𝑒 𝑛𝑒𝑒𝑑𝑒𝑑
𝑐𝑎𝑝𝑠𝑢𝑙𝑒 1000 𝑚𝑐𝑔
6 mg * (1 tablet/0.4 mg) = 15 tablets of 0.4 mg atropine needed

52. You needed to prepare a 1:5000 solution of benzalkonium chloride. The only commercially available
product is a 17% w/v solution. How much of the 17% solution is needed to prepare 8.5L of 1:5000
solution?

Solution:

First, need to convert ratio strength to % strength using a proportion:

1 𝑝𝑎𝑟𝑡 𝑥𝑔

5000 𝑝𝑎𝑟𝑡𝑠 100 𝑚𝑙
x=0.02 g, thus 0.02% w/v
Now use: Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(0.17)(x) = (0.0002)(8500)
X=10 ml of 17% w/v solution needed
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53. A prescription calls for 10.0 g of citric acid syrup, a liquid whose specific gravity is 1.17. What volume
(in mL) of citric acid syrup should be used? Round your answer to the nearest hundredth.

Solution:

Specific gravity = g/ml

10 g citric acid 1 ml = 8.55 ml
1.17 g

54. Phosphoric acid contains 86.5% (w/w) of H3PO4. How many grams of H3PO4 are contained in 55.0
ml of phosphoric acid? (Specific gravity of phosphoric acid = 1.71) Round your answer to the nearest
tenth.

Solution:

Specific gravity = g/ml

55 ml 1.71 g 86.5 g =81.4 g of H3PO4
ml 100 g

55. A patient is receiving an IV drip of the following:

Sodium heparin 25,000 units

Sodium chloride injection (0.45%) 500 ml

a. How many ml/hr must be administered to achieve a rate of 1200 units of sodium heparin
per hour?

Solution:
1200 𝑢𝑛𝑖𝑡𝑠 500 𝑚𝑙
    24 𝑚𝑙/ℎ𝑟
ℎ𝑟 25000 𝑢𝑛𝑖𝑡𝑠

b. If the set delivers 15 drops/ml, how many drops/min should be administered?

Solution:

1200 𝑢𝑛𝑖𝑡𝑠 500 𝑚𝑙 ℎ𝑟 15 𝑑𝑟𝑜𝑝𝑠

      6 𝑑𝑟𝑜𝑝𝑠/𝑚𝑖𝑛
ℎ𝑟 25000 𝑢𝑛𝑖𝑡𝑠 60 𝑚𝑖𝑛 𝑚𝑙
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56. If a patient takes 0.5 mg of IV hydromorphone every 4 hours, what would be the equivalent total
daily dose of hydromorphone PO? (Hydromorphone oral to parenteral ratio is 7.5:1.5.)

Solution:
Set up a proportion of IV:PO daily dose (IV dose is given 6 times/day, since q4h schedule)

0.5 𝑚𝑔  6 1.5

𝑥 7.5
𝑥  15 𝑚𝑔 𝑃𝑂 daily

57. An order is received to start milrinone at 0.75 mcg/kg/min. Milrinone is available as 20 mg/100ml in
D5W. What infusion rate (in mL/hr) should be given to a patient weighing 115 kg? Round your
answer to the nearest tenth.

Solution:
0.75 𝑚𝑐𝑔/𝑘𝑔 60 𝑚𝑖𝑛 1 𝑚𝑔 100 𝑚𝑙
115 𝑘𝑔       25.9 𝑚𝑙/ℎ𝑟
𝑚𝑖𝑛 ℎ𝑟 1000 𝑚𝑐𝑔 20 𝑚𝑔

58. The bioavailability of ciprofloxacin is 70%. If a patient is prescribed 400 mg IV q8h, which would be
the best oral regimen for this patient, given that oral ciprofloxacin comes as 250 mg, 500 mg, and
750 mg tablets?

Solution:
Difference between IV and PO bioavailability = 100% ‐ 70% = 30 % (we will need 30% more of PO
drug to receive the same daily dose of ciprofloxacin)
400 mg q8h  1200 mg/day IV
30% of 1200 mg/day is: 1200mg x 0.30 = 360 mg
Add the 30% (360 mg) to current daily dose (1200 mg) to determine equivalent PO amount needed:
360mg + 1200mg = 1560 mg/day, or 750 mg BID (answer D)

59. A parenteral nutrition (PN) solution is to contain 3% amino acids. How many milliliters of a 10%
amino acid stock solution should be used to prepare 4 L of the parenteral nutrition solution?

Solution:
Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(0.1)(x)=(0.03)(4000)
0.1x=120
X=1200 ml
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60. A PN solution is to contain 35 mEq of sodium ions per liter. How many milliliters of a stock solution
containing 2.5 mEq of sodium (as NaCl) per mL should be used to prepare 1 L?

Solution:
Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(2.5 mEq/mL)(x mL)=(35 mEq/L)(1 L)
X=14 mL

61. A PN formula includes 22% dextrose. How many milliliters of a 70% dextrose solution should be used
for each liter of the PN mixture?

Solution:
Old Concentration (C1) x Old Volume (V1) = New Concentration (C2) x New Volume (V2)
(22 g/100 mL)(1000 mL)=(70 g/100 mL)(x mL)
X=314 mL

62. If 500 mL of a 10% lipid emulsion were used to prepare 1 L of a total nutrient admixture solution,
how many kcal were contributed by the emulsion?

Solution:
500 mL lipids x 1.1 kcal/mL lipids = 550 kcal

63. How many kcal would be supplied by 2 L of a 25% dextrose solution?

Solution:
25 𝑔 3.4 𝑘𝑐𝑎𝑙
2000 𝑚𝐿    1700 𝑘𝑐𝑎𝑙
100 𝑚𝑙 𝑔

64. A 72‐year‐old male is receiving the following PN. Calculate his total daily calorie intake.

FreAmine 8.5% 500 mL

D25W 500 mL
Rate: 110 ml/hr
Intralipid 10%, 200 mL/day, IV infusion, 20 ml/hr

Solution:
.   
Amino acids: 500 𝑚𝐿    170 𝑘𝑐𝑎𝑙
  
  . 
Dextrose: 500 𝑚𝐿    425 𝑘𝑐𝑎𝑙
  
. 
Fats (lipid emulsion): 200 𝑚𝐿   220 𝑘𝑐𝑎𝑙

Total daily caloric intake: 170 + 425 + 220 = 815 kcal
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65. Drug X has a half‐life of 6 hours. If the initial plasma concentration of Drug X, given as a single dose,
is 600 mg/L, what will its plasma concentration (in mg/L) be after 18 hours? Round your answer to
the nearest whole number.
Solution:
Half‐life is the time it takes for the concentration of a drug to decrease by 50%.
18 hours = 3 half‐lives, so concentration of drug will decrease by 50% three times:

Initial concentration After 1 half‐life (6 h) After 2 half‐lives (12 h) After 3 half‐lives (18 h)
600 mg/L 300 mg/L 150 mg/L 75 mg/L

66. A patient (wt = 70 kg) is receiving Drug Y for pain. According to the literature, Drug Y has a clearance
of 200 mL/min, a volume of distribution of 1.5 L/kg, a half‐life of 6 hours, and follows a one‐
compartment model. He is currently receiving an infusion of 50 mg/hr of Drug Y.
a. Approximately how long will it take for the drug to reach steady state in this patient?

Solution:
Steady state is achieved after 5 half‐lives
5 x 6 hr = 30 hr

b. What loading dose would you recommend to reach a steady state concentration of 6 mg/L?

Solution:
𝐿
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛  𝑉𝑑 6 𝑚𝑔/𝐿  1.5 𝑘𝑔  70𝑘𝑔
𝐿𝐷    630 𝑚𝑔
𝐹 1
67. Kelly is started on 500 mg q12h of Drug A given as a 30‐minute infusion. At steady state, a blood
sample is drawn 1 hour after the infusion was stopped and had a concentration of 14.2 mg/L. A
second blood concentration, drawn right before the next infusion, had a concentration of 1.6 mg/L.
a. What is the elimination constant (ke) of Drug A?

Solution:
Need to first determine time between two concentrations drawn:
Time 0 = start of infusion
0.5 hr = end of infusion
1.5 hr = blood sample of 14.2 mg/L
12 hr = blood sample of 1.6 mg/L
12 hr – 1.5 hr = 10.5 hr (t)

Now calculate eliminate rate constant:

ln𝐶 ⁄𝐶  ln14.2 𝑚𝑔/𝐿⁄1.6 𝑚𝑔/𝐿
𝑘𝑒    0.2079 ℎ𝑟 
𝑡 10.5 ℎ𝑟

b. What is the half‐life (t1/2) of Drug A?

. .
Solution: 𝑡/    3.3 ℎ𝑟
 .
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68. A patient (wt = 62.5 kg) is taking Drug M which has 100% renal elimination. The patient has normal
renal function. Use the PK data below to answer the following questions (a‐c).
IV bolus dose: 500 mg
Peak level: 8 mcg/mL
Trough level (drawn 12 hours after peak): 0.8 mcg/mL
Therapeutic MIC level: 1 mcg/mL

a. What is the volume of distribution (in L/kg) of Drug M in this patient?

Solution:
𝑑𝑜𝑠𝑒 𝑜𝑟 𝑎𝑚𝑡 𝑜𝑓 𝑑𝑟𝑢𝑔 𝑖𝑛 𝑏𝑜𝑑𝑦 𝑚𝑔 500 𝑚𝑔
𝑉𝑑 𝐿    62.5 𝐿
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑟𝑢𝑔 𝑖𝑛 𝑝𝑙𝑎𝑠𝑚𝑎 𝑚𝑔⁄𝐿 8 𝑚𝑔/𝐿
62.5 𝐿  62.5 𝑘𝑔  1 𝐿/𝑘𝑔

b. What is the elimination rate constant (ke) of Drug M?

Solution:

ln𝐶 ⁄𝐶  ln8 𝑚𝑔/𝐿⁄0.8 𝑚𝑔/𝐿

𝑘𝑒    0.1919 ℎ𝑟 
𝑡 12 ℎ𝑟

c. What is the clearance of Drug M?

Solution:
𝐶𝐿 𝐿⁄ℎ𝑟  𝑘𝑒 ℎ𝑟    𝑉𝑑 𝐿  0.1919  62.5  12 𝐿/ℎ𝑟

69. 120 mg of Drug X is administered to the body as IV bolus and a blood sample is taken immediately
after drug administration. This sample had a concentration of 1.5 mg/L. What is the apparent
volume of distribution of Drug X in Liters?

Solution:
𝑑𝑜𝑠𝑒 𝑜𝑟 𝑎𝑚𝑡 𝑜𝑓 𝑑𝑟𝑢𝑔 𝑖𝑛 𝑏𝑜𝑑𝑦 𝑚𝑔 120 𝑚𝑔
𝑉𝑑 𝐿    80 𝐿
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑟𝑢𝑔 𝑖𝑛 𝑝𝑙𝑎𝑠𝑚𝑎 𝑚𝑔⁄𝐿 1.5 𝑚𝑔/𝐿

70. In a study where rivaroxaban was compared to enoxaparin to find total venous thromboembolism
(VTE) following hip replacement surgery, there were 17 total VTE out of 1513 patients in the
rivaroxaban group and 57 total VTE out of 1473 patients in the enoxaparin group.
a. What is the relative risk reduction of using rivaroxaban over enoxaparin?

Solution:

Relative risk = (Event rate in rivaroxaban group)/(Event rate in enoxaparin group)

= (17/1513)/(57/1473) = 0.2903
Relative risk reduction: 1 – (relative risk) = 1 – 0.2903 = 0.7097 = 0.71
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b. How many patients would you need to treat with rivaroxaban rather than enoxaparin to
prevent 1 VTE event?
Solution:
 
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑅𝑖𝑠𝑘 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝐴𝑅𝑅   𝐴𝑅 𝑐𝑜𝑛𝑡𝑟𝑜𝑙  𝐴𝑅 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡      0.02746
 
 
𝑁𝑢𝑚𝑏𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 𝑡𝑜 𝑇𝑟𝑒𝑎𝑡 𝑁𝑁𝑇    36.42, so 36 patients
 .

71. A study investigating mortality after dronedarone therapy for severe heart failure found that 24
patients died in the dronedarone group (n=310) while 9 patients died in the placebo group (n=317).
a. What is the absolute risk of cardiovascular death with dronedarone therapy for severe heart
failure?

Solution:

Absolute risk = (Cardiovascular death with Dronedarone)/(Total in Dronedarone group)

= 24/310
= 0.08

b. What is the relative risk of cardiovascular death with dronedarone therapy for severe heart
failure as compared to placebo?

Solution:

Relative risk = (Event rate in Dronedarone group)/(Event rate in Placebo group)

= (24/310)/(9/317)
= 2.7

c. What is the relative risk increase using Dronedarone compared to placebo in severe heart
failure patients?

Solution:

Relative risk reduction (or increase, if negative value results) = 1 – (relative risk)
= 1 – 2.72688
= ‐ 1.72688
Relative risk increase of 1.7 (or 170%)

72. A study looking at osteoporosis in women found 80 out of 100 did not use calcium plus vitamin D
developed osteoporosis by age 75 years, compared to 15 out of 100 did use calcium plus vitamin D
developed osteoporosis. What is the odds ratio?

Solution:

Odds ratio = (exposed cases / unexposed cases) / (exposed non‐cases / unexposed non‐cases)
Odds ratio = (15/80) / (85/20) = 0.04
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73. You are interested in observing the effects a particular beta blocker has on heart rate (HR). From six
different patients, you collect the following HR values (in beats per minute): 96, 65, 69, 65, 59, 64.
What is the mode HR?

Solution:

65 (the mode is the value that occurs most frequently)