Question

Question #1 of 10 05 grams of methotrexate is atment includes methotrexate, 35 mg/m of body surface area once per week for 8 weeks ib patient, 5 ft 8 in. tall would receive during this course of therapy Round answer to the nearest tenth. Do not include units the amount a 160-l Answers 1-1 5 ° 0.5 10

Answer 1

ANSWER1.

NEEDED DOSE OF ANTI CANCER METHOTREXATE AN ANTI CANCER DRUG IS 35 mg/m2

the height of the patient = 5 ft 8 inches= 1.72 meters = 172 CM

the weight of the patient = 160 lbs = 72.575 kgs

BODY SURFACE AREA = (Mosteller)2 (square) =

height(cm) x weight(kg) 3600

   =  1.8

AMOUNT OF DRUG IS TO ADMINISTERED

= needed dose X BSA

= 35 mg / m2 X 1.8 (BSA)

, 35 MG CONVERTED INTO GRAMS= 35/ 1000

= 0.035 GRAMS

AMOUNTT OF DRUG TO BE ADMINISTERD

= O.O35 GM X 1.8 (BSA)

= 0.063 GRAM WILL BE NEEDED TO ADMINISTER ONCE IN A WEEK

ADMINISTERING THE MEDICATION UP TO 8 WEEKS ONCE IN A WEEK WILL BE

0.063 (NEEDED DOSE ) X 8 (WEEKS) = 0.504

   NEAREST TENTH IS   0.5 grams

ANSWER IS 0.5grams  

ANSWER 2

the weight of the patient is 176 lb

, to convert lb into kg it will be

the weight of the patient in kg = 79.8 kgs

the dose of theophylline is 0.5 mg / kg / hour

duration of administration = 24 hours

TOTAL DOSE NEEDED IN ONE HOUR = 0.5 X 79.8 KGS( WEIGHT)

= 39.9 MG DRUG SHOULD BE ADMINISTERED IN AN HOUR

the total dose needed IN 24 HOURS = 39.9 X 24  (hours)

= 957.6 MG IN 24 HOURS

NEAREST WHOLE NO IS 960

SO, 960 IS THE ANSWER