,
,Chapter 1: Arithmetic Needed for Dosage
C C C C C
MULTIPLE CHOICE C
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
C C C C C C C C C C C C C C C C C C C
oz. What portion of the water remained?
C C C C C C C
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A C
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
C C C C C C C C C C C C C C C
C(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
C C C C C C C C C C C C C C C C C
Cdividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
C C C C C C C C C C C C C C C C
CTo reduce fractions, find the largest number that can be divided evenly into the numerator and the
C C C C C C C C C C C C C C C C
Cdenominator
(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.
C C C C C C C C C C C C C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Apply
C C C
Difficulty: Moderate C
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Teaching/Learning
C C C
Objective: 1, 2 C C
2. A patient/client was prescribed 240 W
C mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
C C C C C C C C C C
C100 mL. What portion of the Ensure remained?
C C C C C C C
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B C
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
C C C C C C C C C C C C C C
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
C C C C C C C C C C C C C C C C C
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
C C C C C C C C C C C C C C C
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
C C C C C C C C C C C C C C C C C C
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
C C C C C C C C C C C C C C C C C
140/240 can be reduced to its lowest terms of 7/12.
C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Apply
C C C
Difficulty: Moderate C
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Teaching/Learning
C C C
Objective: 1, 2 C C
, 3. A patient/client consumed
C oz. of coffee, 2/3 oz. of ice cream, and Coz. of beef broth. C C C C C C C C C C C C C C C
What is the total number of ounces consumed that should be documented for the patient/client?
C C C C C C C C C C C C C C C
a. 3 3/4 C
b. 4 5/12 C
c. 4 2/3 C
d. 4 4/9 C
ANS: B C
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
C C C C C C C C C C C C C C C
Cmultiplying the whole number by the denominator and then adding that total to the numerator. For the
C C C C C C C C C C C C C C C C
coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
C C C C C C C C C C C C C C C C C
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
C C C C C C C C C C C C C C C C C C C
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
C C C C C C C C C C C C C C
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
C C C C C C C C C C C C C C C C C
Cmultiply that result by the numerator of the fraction. The new fractions to be added are 27/12
C C C C C C C C C C C C C C C C
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
C C C C C C C C C C C C C C C
are added together and the fraction is reduced to the lowest terms.
C C C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Analyze
C C C
Difficulty: Difficult C
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Communication and Documentation
C C C C C
Objective: 1, 2
C C C
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
C C C C C C C C C C C C C C C C
the nurse document as
C C C
WWW.TBSM.WS
Cconsumed?
a. 360
b. 420
c. 510
d. 600
ANS: B C
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number
C C C C C C C C C C C C C C C C
of milliliters consumed, multiply 180 7/3 (
C C ). When a mixed number is present, change it to
C C C C C C C C C C C C C C C
an improper fraction by multiplying the whole number by the denominator and then adding that
C C C C C C C C C C C C C C C
total to the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 ×
C C C C C C C C C C C C C C C C C C C C C C C C C C C
7 = 420).
C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Analyze
C C C
Difficulty: Difficult C
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Communication and Documentation
C C C C C
Objective: 1, 2
C C C
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
C C C C C C C C C C C C C C
were gained since admission?
C C C C
a. 0.78
b. 0.88
,Chapter 1: Arithmetic Needed for Dosage
C C C C C
MULTIPLE CHOICE C
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
C C C C C C C C C C C C C C C C C C C
oz. What portion of the water remained?
C C C C C C C
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A C
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity
C C C C C C C C C C C C C C C
C(25 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by
C C C C C C C C C C C C C C C C C
Cdividing 10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25.
C C C C C C C C C C C C C C C C
CTo reduce fractions, find the largest number that can be divided evenly into the numerator and the
C C C C C C C C C C C C C C C C
Cdenominator
(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.
C C C C C C C C C C C C C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Apply
C C C
Difficulty: Moderate C
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Teaching/Learning
C C C
Objective: 1, 2 C C
2. A patient/client was prescribed 240 W
C mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only
C C C C C C C C C C
C100 mL. What portion of the Ensure remained?
C C C C C C C
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B C
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available
C C C C C C C C C C C C C C
quantity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a
C C C C C C C C C C C C C C C C C
fraction by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided
C C C C C C C C C C C C C C C
by 240 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the
C C C C C C C C C C C C C C C C C C
numerator and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction
C C C C C C C C C C C C C C C C C
140/240 can be reduced to its lowest terms of 7/12.
C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Apply
C C C
Difficulty: Moderate C
Page and Header: 2, Dividing Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Teaching/Learning
C C C
Objective: 1, 2 C C
, 3. A patient/client consumed
C oz. of coffee, 2/3 oz. of ice cream, and Coz. of beef broth. C C C C C C C C C C C C C C C
What is the total number of ounces consumed that should be documented for the patient/client?
C C C C C C C C C C C C C C C
a. 3 3/4 C
b. 4 5/12 C
c. 4 2/3 C
d. 4 4/9 C
ANS: B C
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by
C C C C C C C C C C C C C C C
Cmultiplying the whole number by the denominator and then adding that total to the numerator. For the
C C C C C C C C C C C C C C C C
coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
C C C C C C C C C C C C C C C C C
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators,
C C C C C C C C C C C C C C C C C C C
find the least common denominator (LCD). For 2, 3, and 4, the LCD =
C C C C C C C C C C C C C C
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
C C C C C C C C C C C C C C C C C
Cmultiply that result by the numerator of the fraction. The new fractions to be added are 27/12
C C C C C C C C C C C C C C C C
(coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators
C C C C C C C C C C C C C C C
are added together and the fraction is reduced to the lowest terms.
C C C C C C C C C C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Analyze
C C C
Difficulty: Difficult C
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Communication and Documentation
C C C C C
Objective: 1, 2
C C C
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
C C C C C C C C C C C C C C C C
the nurse document as
C C C
WWW.TBSM.WS
Cconsumed?
a. 360
b. 420
c. 510
d. 600
ANS: B C
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number
C C C C C C C C C C C C C C C C
of milliliters consumed, multiply 180 7/3 (
C C ). When a mixed number is present, change it to
C C C C C C C C C C C C C C C
an improper fraction by multiplying the whole number by the denominator and then adding that
C C C C C C C C C C C C C C C
total to the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 ×
C C C C C C C C C C C C C C C C C C C C C C C C C C C
7 = 420).
C C C
Format: Multiple Choice C C
Chapter: 1
C C
Client Needs: Physiological Integrity: Basic Care and Comfort
C C C C C C C
Cognitive Level: Analyze
C C C
Difficulty: Difficult C
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions
C C C C C C C C
Integrated Process: Communication and Documentation
C C C C C
Objective: 1, 2
C C C
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms
C C C C C C C C C C C C C C
were gained since admission?
C C C C
a. 0.78
b. 0.88
