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#1




Time to reach Drug Steady State
A constant infusion of drug Y is begun. Drug Y has a Volume of Distribution of 5L, and a clearance of 0.05L/h.
Aproximately how long would it take drug Y to reach steady state? a. 70 b. 100 c. 112 d. 280 e. 400 
#3




Quote:
= 0.7 X 5 / 0.05 = 70. Answer is A. 
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neha_subh (04022013) 
#4




You forgot it takes 4 t1/2 to get to steady state, Dr. Ali.
Answer is D. 4x70=280 
#5




@q to all members
where to study pharmacokinetics n dynamics from? 
#6




kaplan+fa+uw

#7




Truth is I got this question word to word including the numbers from the qbank. I picked 70 hrs X 4 = 280 hrs, and I got wrong. The correct answer was 70 hrs alone. I reread the question and from there I came to know that all they wanted was half life alone; which in this case is 70 hours.

The above post was thanked by:  
neha_subh (04022013) 
#9




corect answer is (D):
A constant infusion of drug Y is begun. Drug Y has a Volume of Distribution of 5L, and a clearance of 0.05L/h.
Aproximately how long would it take drug Y to reach steady state? d. 280 t 1/2 = 0.7 x Vd / CL t 1/2 = 0.7 x 5 / 0.05 t 1/2=3.5 / 0.05 t 1/2 = 70 clinical steady state = 45 x t 1/2 so clinical steady state = 4 X 70 = 280
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sami trailer (03292013) 
#10




That question which I posted is a question from the Qmax question bank from FirstAid. The answer they give in their explanation is 70.
But I am also confused about this question, so I reported it, because it seems to me that they ask for steady state, and you need to multiply 70 x 4 to get 280 as most of you answered. Also, in the explanation that comes with the question, they start by writing "First you need to calculate for half time", as if a second calculation would be needed to get to the answer, so even the explanation seems incomplete. It is also known that FirstAid has some kinks, im 2 months away from the exam and im pretty sure the answer is 280, I hope I get a response from the FirstAid team. Thanks for your feedback guys. 
The above post was thanked by:  
neha_subh (04022013) 
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