,
,Chapter 1: Arithmetic Needed for Dosage
p2 p2 p2 p2 p2
MULTIPLE CHOICE p2
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
oz. What portion of the water remained?
p2 p2 p2 p2 p2 p2 p2
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A p 2
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
dividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
To reduce fractions, find the largest number that can be divided evenly into the numerator and the
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
denominator
p2
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
2/5.
p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Apply
p2 p2 p2
Difficulty: Moderate p2
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Teaching/Learning
p2 p2 p2
Objective: 1, 2 p2 p2
2. A patient/client was prescribed 240 W
p2 mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
100 mL. What portion of the Ensure remained?
p2 p2 p2 p2 p2 p2 p2 p2
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B p 2
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
140/240 can be reduced to its lowest terms of 7/12.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Apply
p2 p2 p2
Difficulty: Moderate p2
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Teaching/Learning
p2 p2 p2
Objective: 1, 2 p2 p2
1 | P a g e
p2 p2 p2 p2 p2
, 3. A patient/client consumed
p2 oz. of coffee, 2/3 oz. of ice cream, and oz. of beef broth.
p2 p 2 p 2 p2 p2 p2 p2 p2 p2 p2 p2 p 2 p 2 p2 p2 p2
What is the total number of ounces consumed that should be documented for the patient/client?
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
a. 3 3/4 p2
b. 4 5/12 p2
c. 4 2/3 p2
d. 4 4/9 p2
ANS: B p 2
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
multiplying the whole number by the denominator and then adding that total to the numerator. For
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
then multiply that result by the numerator of the fraction. The new fractions to be added are 27/12
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
are added together and the fraction is reduced to the lowest terms.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Analyze
p2 p2 p2
Difficulty: Difficult p2
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Communication and Documentation
p2 p2 p2 p2 p2
Objective: 1, 2
p2 p2 p2
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the nurse document as
p2 p2 p2
WWW.TBSM.WS
consumed?
p2
a. 360
b. 420
c. 510
d. 600
ANS: B p 2
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
milliliters consumed, multiply 180 7/3 (
p2 ). When a mixed number is present, change it to an
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
improper fraction by multiplying the whole number by the denominator and then adding that total to
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Analyze
p2 p2 p2
Difficulty: Difficult p2
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Communication and Documentation
p2 p2 p2 p2 p2
Objective: 1, 2
p2 p2 p2
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
were gained since admission?
p2 p2 p2 p2
a. 0.78
b. 0.88
2 | P a g e
p2 p2 p2 p2 p2
,Chapter 1: Arithmetic Needed for Dosage
p2 p2 p2 p2 p2
MULTIPLE CHOICE p2
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
oz. What portion of the water remained?
p2 p2 p2 p2 p2 p2 p2
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A p 2
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
dividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
To reduce fractions, find the largest number that can be divided evenly into the numerator and the
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
denominator
p2
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
2/5.
p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Apply
p2 p2 p2
Difficulty: Moderate p2
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Teaching/Learning
p2 p2 p2
Objective: 1, 2 p2 p2
2. A patient/client was prescribed 240 W
p2 mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
100 mL. What portion of the Ensure remained?
p2 p2 p2 p2 p2 p2 p2 p2
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B p 2
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
140/240 can be reduced to its lowest terms of 7/12.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Apply
p2 p2 p2
Difficulty: Moderate p2
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Teaching/Learning
p2 p2 p2
Objective: 1, 2 p2 p2
1 | P a g e
p2 p2 p2 p2 p2
, 3. A patient/client consumed
p2 oz. of coffee, 2/3 oz. of ice cream, and oz. of beef broth.
p2 p 2 p 2 p2 p2 p2 p2 p2 p2 p2 p2 p 2 p 2 p2 p2 p2
What is the total number of ounces consumed that should be documented for the patient/client?
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
a. 3 3/4 p2
b. 4 5/12 p2
c. 4 2/3 p2
d. 4 4/9 p2
ANS: B p 2
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
multiplying the whole number by the denominator and then adding that total to the numerator. For
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
then multiply that result by the numerator of the fraction. The new fractions to be added are 27/12
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
are added together and the fraction is reduced to the lowest terms.
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Analyze
p2 p2 p2
Difficulty: Difficult p2
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Communication and Documentation
p2 p2 p2 p2 p2
Objective: 1, 2
p2 p2 p2
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the nurse document as
p2 p2 p2
WWW.TBSM.WS
consumed?
p2
a. 360
b. 420
c. 510
d. 600
ANS: B p 2
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
milliliters consumed, multiply 180 7/3 (
p2 ). When a mixed number is present, change it to an
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
improper fraction by multiplying the whole number by the denominator and then adding that total to
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
Format: Multiple Choice p2 p2
Chapter: 1
p2 p2
Client Needs: Physiological Integrity: Basic Care and Comfort
p2 p2 p2 p2 p2 p2 p2
Cognitive Level: Analyze
p2 p2 p2
Difficulty: Difficult p2
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
p2 p2 p2 p2 p2 p2 p2 p2
Integrated Process: Communication and Documentation
p2 p2 p2 p2 p2
Objective: 1, 2
p2 p2 p2
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2 p2
were gained since admission?
p2 p2 p2 p2
a. 0.78
b. 0.88
2 | P a g e
p2 p2 p2 p2 p2
