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