,
,Chapter 1: Arithmetic Needed for Dosage
p7 p7 p7 p7 p7
MULTIPLE CHOICE p7
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
oz. What portion of the water remained?
p7 p7 p7 p7 p7 p7 p7
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A p 7
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
dividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
To reduce fractions, find the largest number that can be divided evenly into the numerator and the
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
denominator
p7
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
2/5.
p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Apply
p7 p7 p7
Difficulty: Moderate p7
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Teaching/Learning
p7 p7 p7
Objective: 1, 2 p7 p7
2. A patient/client was prescribed 240 W
p7 mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
100 mL. What portion of the Ensure remained?
p7 p7 p7 p7 p7 p7 p7 p7
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B p 7
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
140/240 can be reduced to its lowest terms of 7/12.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Apply
p7 p7 p7
Difficulty: Moderate p7
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Teaching/Learning
p7 p7 p7
Objective: 1, 2 p7 p7
1 | P a g e
p7 p7 p7 p7 p7
, 3. A patient/client consumed
p7 oz. of coffee, 2/3 oz. of ice cream, and oz. of beef broth.
p7 p 7 p 7 p7 p7 p7 p7 p7 p7 p7 p7 p 7 p 7 p7 p7 p7
What is the total number of ounces consumed that should be documented for the patient/client?
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
a. 3 3/4 p7
b. 4 5/12 p7
c. 4 2/3 p7
d. 4 4/9 p7
ANS: B p 7
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
multiplying the whole number by the denominator and then adding that total to the numerator. For
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
then multiply that result by the numerator of the fraction. The new fractions to be added are 27/12
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
are added together and the fraction is reduced to the lowest terms.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Analyze
p7 p7 p7
Difficulty: Difficult p7
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Communication and Documentation
p7 p7 p7 p7 p7
Objective: 1, 2
p7 p7 p7
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the nurse document as
p7 p7 p7
WWW.TBSM.WS
consumed?
p7
a. 360
b. 420
c. 510
d. 600
ANS: B p 7
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
milliliters consumed, multiply 180 7/3 (
p7 ). When a mixed number is present, change it to an
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
improper fraction by multiplying the whole number by the denominator and then adding that total to
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Analyze
p7 p7 p7
Difficulty: Difficult p7
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Communication and Documentation
p7 p7 p7 p7 p7
Objective: 1, 2
p7 p7 p7
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
were gained since admission?
p7 p7 p7 p7
a. 0.78
b. 0.88
2 | P a g e
p7 p7 p7 p7 p7
,Chapter 1: Arithmetic Needed for Dosage
p7 p7 p7 p7 p7
MULTIPLE CHOICE p7
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
oz. What portion of the water remained?
p7 p7 p7 p7 p7 p7 p7
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A p 7
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
dividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
To reduce fractions, find the largest number that can be divided evenly into the numerator and the
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
denominator
p7
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
2/5.
p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Apply
p7 p7 p7
Difficulty: Moderate p7
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Teaching/Learning
p7 p7 p7
Objective: 1, 2 p7 p7
2. A patient/client was prescribed 240 W
p7 mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
100 mL. What portion of the Ensure remained?
p7 p7 p7 p7 p7 p7 p7 p7
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B p 7
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
140/240 can be reduced to its lowest terms of 7/12.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Apply
p7 p7 p7
Difficulty: Moderate p7
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Teaching/Learning
p7 p7 p7
Objective: 1, 2 p7 p7
1 | P a g e
p7 p7 p7 p7 p7
, 3. A patient/client consumed
p7 oz. of coffee, 2/3 oz. of ice cream, and oz. of beef broth.
p7 p 7 p 7 p7 p7 p7 p7 p7 p7 p7 p7 p 7 p 7 p7 p7 p7
What is the total number of ounces consumed that should be documented for the patient/client?
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
a. 3 3/4 p7
b. 4 5/12 p7
c. 4 2/3 p7
d. 4 4/9 p7
ANS: B p 7
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
multiplying the whole number by the denominator and then adding that total to the numerator. For
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
then multiply that result by the numerator of the fraction. The new fractions to be added are 27/12
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
are added together and the fraction is reduced to the lowest terms.
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Analyze
p7 p7 p7
Difficulty: Difficult p7
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Communication and Documentation
p7 p7 p7 p7 p7
Objective: 1, 2
p7 p7 p7
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the nurse document as
p7 p7 p7
WWW.TBSM.WS
consumed?
p7
a. 360
b. 420
c. 510
d. 600
ANS: B p 7
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
milliliters consumed, multiply 180 7/3 (
p7 ). When a mixed number is present, change it to an
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
improper fraction by multiplying the whole number by the denominator and then adding that total to
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
Format: Multiple Choice p7 p7
Chapter: 1
p7 p7
Client Needs: Physiological Integrity: Basic Care and Comfort
p7 p7 p7 p7 p7 p7 p7
Cognitive Level: Analyze
p7 p7 p7
Difficulty: Difficult p7
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p7 p7 p7 p7 p7 p7 p7 p7
Integrated Process: Communication and Documentation
p7 p7 p7 p7 p7
Objective: 1, 2
p7 p7 p7
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7 p7
were gained since admission?
p7 p7 p7 p7
a. 0.78
b. 0.88
2 | P a g e
p7 p7 p7 p7 p7
