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