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Test Bank - Henke's Med-Math Dosage-Calculation, Preparation, and Administration, 9th Edition (Buchholz, Chapter 1-10 | All Chapters.

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Test Bank - Henke's Med-Math Dosage-Calculation, Preparation, and Administration, 9th Edition (Buchholz, Chapter 1-10 | All Chapters.

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Test Bank
Henke's Med-Math Dosage-Calculation, Preparation, and
Administration, 9th Edition (Buchholz), Chapter 1-10 | All Chapters

, .



.


Chapter 1, Arithmetic Needed for Dosage

1. A patient/client was instructed to drink 25 oz of water within 2 hours but wasonly able to drink 15 oz. What portion of the water
remained?
A) 2/5

B) 3/5 .
C) 2/25

D) 25/25
.
Ans: A
Feedback: Subtract the quantity of water the client drank (15 oz) from the total available quantity (25 oz): 10 oz remain. To determine
tabh ierb.cpoomr/tteisot n of the water that remains, create a fraction by dividing 10 oz (remaining portion) by 25 oz (total

portion). Therefore, 10 divided by 25 = 10/25. To reduce fractions, find the largest
.
number that can be divided evenly into the numerator and the denominator (5). Tendivided by 5 (10/5) = 2; 25/5 = 5. The fraction 10/25 can be
reduced to its lowest

terms of 2/5. .


Format: Multiple Choice .
Chapter: 1
Client Needs: Physiological Integrity: Basic Care and Comfort
.
Cognitive Level: Apply
Difficulty: Moderate
Page and Header: 2, Dividing Whole Numbers; 3, Fractiaobinrbs.




.
Page 1

, .


Objective: 1, 2

2. A patient/client was prescribed 240 mL of Ensure by mouth and a supplementbut consumed only 100 mL. What portion of the
Ensure remained?
A) 5/12
.
B) 7/12

C) 100/240

D) 240/240 .
Ans: B


Feedback: Subtract the quantity of Ensure the client conasbiurbm
.coe
md
/tes(t100 mL) from the total available quantity (240 mL): 140 mL remain. To


determine the portion of the Ensure that remains, create a fraction by dividing 140 mL (remaining portion) by 240
.
mL (total portion). Therefore, 140 divided by 240 = 7/12. To reduce fractions, find the
largest number that can be divided evenly into the numerator and the denominator
(20); 140 divided by 20 (140/20) = 7; 240/20 = 12. The fractionreduced to its lowest terms of 7/12. 140/240 can be



.
Format: Multiple ChoiceChapter: 1
Client Needs: Physiological Integrity: Basic Care and comfortCognitive Level: Analyze

Difficulty: Difficult .


Page and Header: 2, Multiplying Whole Numbers; 3, FractionsIntegrated Process: Communication and Documentation
.
Objective: 1, 2
1 1
3. A patient/client consumed 2 4 oz. of coffee, 2/3 oz. of ice cream, and 1 2 oz.of
abirb.com/test
beef broth. What is the total number of ounces consum e d th at should be documented for the patient/client?
A) 3 3/4 .
B) 4 5/12

C) 4 2/3
.
D) 4 4/9




.
Page 2

, .
Ans: B
Feedback: Add the amount of ounces consumed. First, change any mixed number to a fraction by multiplying the whole number by the
denomitation and then adding thattotal to the numerator. For the coffee, 4 2 = 8 + 1 = 9/4; for the beef broth, 2 1
= 2 + 1 = 3/2. Then add: 9/4 + 2/3 (ice cream) + 3/2. When fractions have different
.
denominators, find the least common denominator (LCD). For 2, 3, and 4, the LCD =
12. Rewrite each fraction using the LCD; divide the LCD by the denominator of each fraction and then multiply that result by the numerator
ab
o firbt.hcoem/ftes
r at ction. The new fractions to be added are 27/12 (coffee), 8/12 (ice cream), and 18/12 (beef broth). After conversion of the fractions,
the numerators are addaebirdb.ctoomg/teestther and the fraction is reduced to the lowest terms.


.
Format: Multiple Choice
Chapter: 1
Client Needs: Physiological Integrity: Basic Care and comfortCognitive Level: Analyze
Difficulty: Difficult
.
Page and Header: 2, Multiplying Whole Numbers; 3, FractionsIntegrated Process: Communication and
.
Documentation Objective: 1, 2

4. A coffee cup holds 180 mL. The patient/client drank 2⅓ cups of coffee. Howmany milliliters would the nurse document as
consumeda?birb.com/test
A) 360

B) 420 .
C) 510

D) 600

Ans: B .

Feedback: The coffee cup holds 180 mL. The client drank 2⅓ cups. To estimate the total number of milliliters consumed, multiply 180 7/3

(2⅓). When a mixed number
.
is present, change it to an improper fraction by multiplying the whole number by thedenominator and then adding that total to the numerator: 2
3 = 6 + 1 = 7/3.

Therefore, 180 mL × 7/3 = 420 mL (180 ÷ 3 = 60 × 7 abir
= 4b.2co0m)/test
.




.
Page 3

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