www.PlusBay.Plus
,www.PlusBay.Plus
,Chapter 1: Arithmetic Needed for Dosage
ok ok ok ok ok
MULTIPLE CHOICE ok
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15 oz
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
. What portion of the water remained?
ok ok ok ok ok ok
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A o k
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing 10
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce fracti
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ons, find the largest number that can be divided evenly into the numerator and the denominator
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
(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.
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Apply ok ok
Difficulty: Moderate ok
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
ok ok ok ok ok ok ok ok ok
ated Process: Teaching/Learning
ok ok
Objective: 1, 2 ok ok
2. A patient/client was prescribed 240 W
ok mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
ok ok ok ok ok ok ok ok ok ok ok
0 mL. What portion of the Ensure remained?
ok ok ok ok ok ok ok
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B o k
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qua
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ntity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fractio
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
n by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 240 =
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator and t
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
he denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can be redu
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ced to its lowest terms of 7/12.
ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Apply ok ok
Difficulty: Moderate ok
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
ok ok ok ok ok ok ok ok ok
ated Process: Teaching/Learning
ok ok
Objective: 1, 2 ok ok
1|Page
ok ok ok ok ok
www.PlusBay.Plus
, 3. A patient/client consumed
ok oz. of coffee, 2/3 oz. of ice cream, and ok o k o k ok ok ok ok ok ok ok ok o k
oz. of beef broth. What is the total number of ounces consumed that should be documented for the p
o k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
atient/client?
a. 3 3/4 ok
b. 4 5/12 ok
c. 4 2/3 ok
d. 4 4/9 ok
ANS: B o k
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mult
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
iplying the whole number by the denominator and then adding that total to the numerator. For the coffe
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
e, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, find
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
the least common denominator (LCD). For 2, 3, and 4, the LCD =
ok ok ok ok ok ok ok ok ok ok ok ok
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coffee),
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
k8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are added t
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ogether and the fraction is reduced to the lowest terms.
ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Analyze ok ok
Difficulty: Difficult ok
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
ok ok ok ok ok ok ok ok ok
egrated Process: Communication and Documentation Objective:
ok ok ok ok ok ok
1, 2 ok
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
the nurse document as consumed?
ok ok ok ok
WWW.TBSM.WS
a. 360
b. 420
c. 510
d. 600
ANS: B o k
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of mi
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
lliliters consumed, multiply 180 7/3 (ok ok ok ok ok ok
). When a mixed number is present, change it to an improper fraction by multiplying the whole numbe
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
r by the denominator and then adding that total to
ok ok ok ok ok ok ok ok ok
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Analyze ok ok
Difficulty: Difficult ok
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
ok ok ok ok ok ok ok ok ok
egrated Process: Communication and Documentation Objective:
ok ok ok ok ok ok
1, 2 ok
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms were
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
gained since admission?
ok ok ok
a. 0.78
b. 0.88
2|Page
ok ok ok ok ok
www.PlusBay.Plus
,www.PlusBay.Plus
,Chapter 1: Arithmetic Needed for Dosage
ok ok ok ok ok
MULTIPLE CHOICE ok
1. A patient/client was instructed to drink 25 oz of water within 2 hours but was only able to drink 15 oz
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
. What portion of the water remained?
ok ok ok ok ok ok
a. 2/5
b. 3/5
c. 2/25
d. 25/25
ANS: A o k
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
oz): 10 oz remain. To determine the portion of the water that remains, create a fraction by dividing 10
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
oz (remaining portion) by 25 oz (total portion). Therefore, 10 divided by 25 = 10/25. To reduce fracti
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ons, find the largest number that can be divided evenly into the numerator and the denominator
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
(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.
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Apply ok ok
Difficulty: Moderate ok
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
ok ok ok ok ok ok ok ok ok
ated Process: Teaching/Learning
ok ok
Objective: 1, 2 ok ok
2. A patient/client was prescribed 240 W
ok mWLWo.
f ETnB
suSreMb.yWmSouth as a supplement but consumed only 10
ok ok ok ok ok ok ok ok ok ok ok
0 mL. What portion of the Ensure remained?
ok ok ok ok ok ok ok
a. 5/12
b. 7/12
c. 100/240
d. 240/240
ANS: B o k
Feedback: Subtract the quantity of Ensure the client consumed (100 mL) from the total available qua
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ntity (240 mL): 140 mL remain. To determine the portion of the Ensure that remains, create a fractio
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
n by dividing 140 mL (remaining portion) by 240 mL (total portion). Therefore, 140 divided by 240 =
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
7/12. To reduce fractions, find the largest number that can be divided evenly into the numerator and t
k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
he denominator (20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fraction 140/240 can be redu
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ced to its lowest terms of 7/12.
ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Apply ok ok
Difficulty: Moderate ok
Page and Header: 2, Dividing Whole Numbers; 3, Fractions Integr
ok ok ok ok ok ok ok ok ok
ated Process: Teaching/Learning
ok ok
Objective: 1, 2 ok ok
1|Page
ok ok ok ok ok
www.PlusBay.Plus
, 3. A patient/client consumed
ok oz. of coffee, 2/3 oz. of ice cream, and ok o k o k ok ok ok ok ok ok ok ok o k
oz. of beef broth. What is the total number of ounces consumed that should be documented for the p
o k ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
atient/client?
a. 3 3/4 ok
b. 4 5/12 ok
c. 4 2/3 ok
d. 4 4/9 ok
ANS: B o k
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by mult
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
iplying the whole number by the denominator and then adding that total to the numerator. For the coffe
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
e, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different denominators, find
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
the least common denominator (LCD). For 2, 3, and 4, the LCD =
ok ok ok ok ok ok ok ok ok ok ok ok
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
multiply that result by the numerator of the fraction. The new fractions to be added are 27/12 (coffee),
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok o
k8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions, the numerators are added t
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
ogether and the fraction is reduced to the lowest terms.
ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Analyze ok ok
Difficulty: Difficult ok
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
ok ok ok ok ok ok ok ok ok
egrated Process: Communication and Documentation Objective:
ok ok ok ok ok ok
1, 2 ok
4. A coffee cup holds 180 mL. The patient/client drank 2? cups of coffee. How many milliliters would
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
the nurse document as consumed?
ok ok ok ok
WWW.TBSM.WS
a. 360
b. 420
c. 510
d. 600
ANS: B o k
Feedback: The coffee cup holds 180 mL. The client drank 2? cups. To estimate the total number of mi
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
lliliters consumed, multiply 180 7/3 (ok ok ok ok ok ok
). When a mixed number is present, change it to an improper fraction by multiplying the whole numbe
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
r by the denominator and then adding that total to
ok ok ok ok ok ok ok ok ok
the numerator: 2 3 = 6 + 1 = 7/3. Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 = 420).
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
Format: Multiple Choice Chapt ok ok ok
er: 1 ok
Client Needs: Physiological Integrity: Basic Care and Comfort Co
ok ok ok ok ok ok ok ok
gnitive Level: Analyze ok ok
Difficulty: Difficult ok
Page and Header: 2, Multiplying Whole Numbers; 3, Fractions Int
ok ok ok ok ok ok ok ok ok
egrated Process: Communication and Documentation Objective:
ok ok ok ok ok ok
1, 2 ok
5. A patient/client weighed 48.52 kg on admission and now weighs 50.4 kg. How many kilograms were
ok ok ok ok ok ok ok ok ok ok ok ok ok ok ok
gained since admission?
ok ok ok
a. 0.78
b. 0.88
2|Page
ok ok ok ok ok
www.PlusBay.Plus
