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