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,Chapter 1: Arithmetic Needed for Dosage
MULTIPLE CHOICE
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15 oz. What J J J J J J J J J J J J J J J J J J J J J
portion of the water remained? J J J J
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A J
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25 oz):
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10 oz remain. To determine the portion of the water that remains, create a fraction by dividing 10 oz
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(remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce fractions, find
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the largest number that can be divided evenly into the numerator and the denominator
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(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of 2/5.
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Format:MultipleChoice J J J
Chapter: 1 J
Client Needs: Physiological Integrity: Basic Care and Comfort
J J J J J J J J
Cognitive Level: Apply J J
Difficulty: Moderate J
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
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Integrated Process: Teaching/Learning J J
Objective: 1, 2 J J
2. A patient/client was prescribed 240 W
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suSreMb.yWmSouth as a supplement but consumed only 100 mL. J J J J J J J J J J J J J
What portion of the Ensure remained?
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a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B J
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available quantity
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(240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fraction by
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dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 240 = 7/12. To
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reduce fractions, find the largest number that can be divided evenly into the numerator and the denominator
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(20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can be reduced to its lowest terms
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of 7/12.
J
Format:MultipleChoice J J J
Chapter: 1 J
Client Needs: Physiological Integrity: Basic Care and Comfort
J J J J J J J J
Cognitive Level: Apply J J
Difficulty: Moderate J
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
J J J J J J J J J
Integrated Process: Teaching/Learning J J
Objective: 1, 2 J J
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, 3. A patient/client consumed
J oz. of coffee, 2/3 oz. of ice cream, and oz. of beef broth. What is the J J J J J J J J J J J J J J J J J J J J
total number of ounces consumed that should be documented for the patient/client?
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a. 3 3/4 J
b. 4 5/12 J
c. 4 2/3 J
d. 4 4/9 J
ANS: B J
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
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multiplying the whole number by the denominator and then adding that total to the numerator. For the coffee,
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4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
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= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, find the least
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common denominator (LCD). For 2, 3, and 4, the LCD = J J J J J J J J J J
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
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multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coffee), 8/12
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(ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are added together and
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the fraction is reduced to the lowest terms.
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Format:MultipleChoice J J J
Chapter: 1 J
Client Needs: Physiological Integrity: Basic Care and Comfort
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Cognitive Level: Analyze J J
Difficulty: Difficult J
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
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Integrated Process: Communication and Documentation Objective: J J J J J J
1, 2 J
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
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the nurse document as consumed?
J J J J
WWW.TBSM.WS
a. 360
b. 420
c. 510
d. 600
ANS: B J
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of milliliters
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consumed, multiply 180 7/3 ( ). When a mixed number is present, change it to an improper fraction by
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multiplying the whole number by the denominator and then adding that total to J J J J J J J J J J J J
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
J J J J J J J J J J J J J J J J J J J J J J J J J J J
Format:MultipleChoice J J J
Chapter: 1 J
Client Needs: Physiological Integrity: Basic Care and Comfort
J J J J J J J J
Cognitive Level: Analyze J J
Difficulty: Difficult J
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
J J J J J J J J J
Integrated Process: Communication and Documentation Objective: J J J J J J
1, 2 J
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms were
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gained since admission? J J
a. 0.78
b. 0.88
2|Page
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