,
,Chapter 1: Arithmetic Needed for Dosage
mf mf mf mf mf
MULTIPLE CHOICE mf
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
oz. What portion of the water remained?
mf mf mf mf mf mf
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A m f
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (2
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
5 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce f
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ractions, find the largest number that can be divided evenly into the numerator and the denominator
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of 2/
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
5.
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Apply mf mf
Difficulty: Moderate mf
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integ
mf mf mf mf mf mf mf mf mf
rated Process: Teaching/Learning
mf mf
Objective: 1, 2 mf mf
2. A patient/client was prescribed 240 W
mf mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
mf mf mf mf mf mf mf mf mf mf mf
0 mL. What portion of the Ensure remained?
mf mf mf mf mf mf mf
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B m f
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qu
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
antity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fract
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ion by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 24
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
0 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can b
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
e reduced to its lowest terms of 7/12.
mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Apply mf mf
Difficulty: Moderate mf
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integ
mf mf mf mf mf mf mf mf mf
rated Process: Teaching/Learning
mf mf
Objective: 1, 2 mf mf
1|Pag e
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, 3. A patient/client consumed
mf oz. of coffee, 2/3 oz. of ice cream, and mf m f m f mf mf mf mf mf mf mf mf m f
oz. of beef broth. What is the total number of ounces consumed that should be documented for the
m f mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
patient/client?
a. 3 3/4 mf
b. 4 5/12 mf
c. 4 2/3 mf
d. 4 4/9 mf
ANS: B m f
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mu
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ltiplying the whole number by the denominator and then adding that total to the numerator. For the co
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, fin
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
d the least common denominator (LCD). For 2, 3, and 4, the LCD =
mf mf mf mf mf mf mf mf mf mf mf mf mf
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and the
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
n multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coff
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are ad
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ded together and the fraction is reduced to the lowest terms.
mf mf mf mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Analyze mf mf
Difficulty: Difficult mf
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions In
mf mf mf mf mf mf mf mf mf
tegrated Process: Communication and Documentation Objective:
mf mf mf mf mf m
1, 2
f mf
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
the nurse document as consumed
mf mf mf mf
WWW.TBSM.WS
?
a. 360
b. 420
c. 510
d. 600
ANS: B m f
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
milliliters consumed, multiply 180 7/3 ( mf mf mf mf mf mf
). When a mixed number is present, change it to an improper fraction by multiplying the whole numb
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
er by the denominator and then adding that total to
mf mf mf mf mf mf mf mf mf
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Analyze mf mf
Difficulty: Difficult mf
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions In
mf mf mf mf mf mf mf mf mf
tegrated Process: Communication and Documentation Objective:
mf mf mf mf mf m
1, 2
f mf
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms wer
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
e gained since admission?
mf mf mf
a. 0.78
b. 0.88
2|Pag e
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,Chapter 1: Arithmetic Needed for Dosage
mf mf mf mf mf
MULTIPLE CHOICE mf
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
oz. What portion of the water remained?
mf mf mf mf mf mf
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A m f
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (2
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
5 oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
10 oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce f
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ractions, find the largest number that can be divided evenly into the numerator and the denominator
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
(5). Ten divided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be reduced to its lowest terms of 2/
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
5.
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Apply mf mf
Difficulty: Moderate mf
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integ
mf mf mf mf mf mf mf mf mf
rated Process: Teaching/Learning
mf mf
Objective: 1, 2 mf mf
2. A patient/client was prescribed 240 W
mf mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
mf mf mf mf mf mf mf mf mf mf mf
0 mL. What portion of the Ensure remained?
mf mf mf mf mf mf mf
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B m f
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qu
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
antity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fract
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ion by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 24
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
0 = 7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
and the denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can b
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
e reduced to its lowest terms of 7/12.
mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Apply mf mf
Difficulty: Moderate mf
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integ
mf mf mf mf mf mf mf mf mf
rated Process: Teaching/Learning
mf mf
Objective: 1, 2 mf mf
1|Pag e
www.PlusBay.Plus
, 3. A patient/client consumed
mf oz. of coffee, 2/3 oz. of ice cream, and mf m f m f mf mf mf mf mf mf mf mf m f
oz. of beef broth. What is the total number of ounces consumed that should be documented for the
m f mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
patient/client?
a. 3 3/4 mf
b. 4 5/12 mf
c. 4 2/3 mf
d. 4 4/9 mf
ANS: B m f
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mu
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ltiplying the whole number by the denominator and then adding that total to the numerator. For the co
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, fin
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
d the least common denominator (LCD). For 2, 3, and 4, the LCD =
mf mf mf mf mf mf mf mf mf mf mf mf mf
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and the
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
n multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coff
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are ad
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
ded together and the fraction is reduced to the lowest terms.
mf mf mf mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Analyze mf mf
Difficulty: Difficult mf
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions In
mf mf mf mf mf mf mf mf mf
tegrated Process: Communication and Documentation Objective:
mf mf mf mf mf m
1, 2
f mf
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
the nurse document as consumed
mf mf mf mf
WWW.TBSM.WS
?
a. 360
b. 420
c. 510
d. 600
ANS: B m f
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
milliliters consumed, multiply 180 7/3 ( mf mf mf mf mf mf
). When a mixed number is present, change it to an improper fraction by multiplying the whole numb
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
er by the denominator and then adding that total to
mf mf mf mf mf mf mf mf mf
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
Format: Multiple Choice Chapt mf mf mf
er: 1 mf
Client Needs: Physiological Integrity: Basic Care and Comfort C
mf mf mf mf mf mf mf mf
ognitive Level: Analyze mf mf
Difficulty: Difficult mf
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions In
mf mf mf mf mf mf mf mf mf
tegrated Process: Communication and Documentation Objective:
mf mf mf mf mf m
1, 2
f mf
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms wer
mf mf mf mf mf mf mf mf mf mf mf mf mf mf mf
e gained since admission?
mf mf mf
a. 0.78
b. 0.88
2|Pag e
www.PlusBay.Plus
