www.PlusBay.Plus
,www.PlusBay.Plus
,Chapter 1: Arithmetic Needed for Dosage
dy dy dy dy dy
MULTIPLE CHOICE dy
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15 oz
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
. What portion of the water remained?
dy dy dy dy dy dy
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A d y
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing 10
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce fracti
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ons, find the largest number that can be divided evenly into the numerator and the denominator
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
(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.
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Apply dy dy
Difficulty: Moderate dy
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
dy dy dy dy dy dy dy dy dy
ated Process: Teaching/Learning
dy dy
Objective: 1, 2 dy dy
2. A patient/client was prescribed 240 W
dy mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
dy dy dy dy dy dy dy dy dy dy dy
0 mL. What portion of the Ensure remained?
dy dy dy dy dy dy dy
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B d y
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qua
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ntity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fractio
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
n by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 240 =
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator and t
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
he denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can be redu
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ced to its lowest terms of 7/12.
dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Apply dy dy
Difficulty: Moderate dy
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
dy dy dy dy dy dy dy dy dy
ated Process: Teaching/Learning
dy dy
Objective: 1, 2 dy dy
1|Page
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, 3. A patient/client consumed
dy oz. of coffee, 2/3 oz. of ice cream, and dy d y d y dy dy dy dy dy dy dy dy d y
oz. of beef broth. What is the total number of ounces consumed that should be documented for the p
d y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
atient/client?
a. 3 3/4 dy
b. 4 5/12 dy
c. 4 2/3 dy
d. 4 4/9 dy
ANS: B d y
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mult
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
iplying the whole number by the denominator and then adding that total to the numerator. For the coffe
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
e, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, find
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
the least common denominator (LCD). For 2, 3, and 4, the LCD =
dy dy dy dy dy dy dy dy dy dy dy dy
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coffee),
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
y8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are added t
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ogether and the fraction is reduced to the lowest terms.
dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Analyze dy dy
Difficulty: Difficult dy
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
dy dy dy dy dy dy dy dy dy
egrated Process: Communication and Documentation Objective:
dy dy dy dy dy dy
1, 2 dy
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
the nurse document as consumed?
dy dy dy dy
WWW.TBSM.WS
a. 360
b. 420
c. 510
d. 600
ANS: B d y
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of mi
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
lliliters consumed, multiply 180 7/3 (dy dy dy dy dy dy
). When a mixed number is present, change it to an improper fraction by multiplying the whole numbe
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
r by the denominator and then adding that total to
dy dy dy dy dy dy dy dy dy
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Analyze dy dy
Difficulty: Difficult dy
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
dy dy dy dy dy dy dy dy dy
egrated Process: Communication and Documentation Objective:
dy dy dy dy dy dy
1, 2 dy
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms were
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
gained since admission?
dy dy dy
a. 0.78
b. 0.88
2|Page
www.PlusBay.Plus
,www.PlusBay.Plus
,Chapter 1: Arithmetic Needed for Dosage
dy dy dy dy dy
MULTIPLE CHOICE dy
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15 oz
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
. What portion of the water remained?
dy dy dy dy dy dy
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A d y
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing 10
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce fracti
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ons, find the largest number that can be divided evenly into the numerator and the denominator
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
(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.
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Apply dy dy
Difficulty: Moderate dy
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
dy dy dy dy dy dy dy dy dy
ated Process: Teaching/Learning
dy dy
Objective: 1, 2 dy dy
2. A patient/client was prescribed 240 W
dy mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
dy dy dy dy dy dy dy dy dy dy dy
0 mL. What portion of the Ensure remained?
dy dy dy dy dy dy dy
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B d y
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qua
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ntity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fractio
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
n by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 240 =
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator and t
y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
he denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can be redu
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ced to its lowest terms of 7/12.
dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Apply dy dy
Difficulty: Moderate dy
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
dy dy dy dy dy dy dy dy dy
ated Process: Teaching/Learning
dy dy
Objective: 1, 2 dy dy
1|Page
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, 3. A patient/client consumed
dy oz. of coffee, 2/3 oz. of ice cream, and dy d y d y dy dy dy dy dy dy dy dy d y
oz. of beef broth. What is the total number of ounces consumed that should be documented for the p
d y dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
atient/client?
a. 3 3/4 dy
b. 4 5/12 dy
c. 4 2/3 dy
d. 4 4/9 dy
ANS: B d y
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mult
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
iplying the whole number by the denominator and then adding that total to the numerator. For the coffe
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
e, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, find
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
the least common denominator (LCD). For 2, 3, and 4, the LCD =
dy dy dy dy dy dy dy dy dy dy dy dy
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coffee),
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy d
y8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are added t
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
ogether and the fraction is reduced to the lowest terms.
dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Analyze dy dy
Difficulty: Difficult dy
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
dy dy dy dy dy dy dy dy dy
egrated Process: Communication and Documentation Objective:
dy dy dy dy dy dy
1, 2 dy
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
the nurse document as consumed?
dy dy dy dy
WWW.TBSM.WS
a. 360
b. 420
c. 510
d. 600
ANS: B d y
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of mi
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
lliliters consumed, multiply 180 7/3 (dy dy dy dy dy dy
). When a mixed number is present, change it to an improper fraction by multiplying the whole numbe
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
r by the denominator and then adding that total to
dy dy dy dy dy dy dy dy dy
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
Format: Multiple Choice Chapt dy dy dy
er: 1 dy
Client Needs: Physiological Integrity: Basic Care and Comfort Co
dy dy dy dy dy dy dy dy
gnitive Level: Analyze dy dy
Difficulty: Difficult dy
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
dy dy dy dy dy dy dy dy dy
egrated Process: Communication and Documentation Objective:
dy dy dy dy dy dy
1, 2 dy
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms were
dy dy dy dy dy dy dy dy dy dy dy dy dy dy dy
gained since admission?
dy dy dy
a. 0.78
b. 0.88
2|Page
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