Help? C3 Problem

[Hasan Naqvi]

hey guys can anyone help me with no8 in this paper? http://eiewebvip.edexcel.org.uk/Reports/Confidential Documents/0706/6665_01_que_20070614.pdf

 

[Ayesha B]

Which part?

 

[Ayesha B]

I remember this question... there's a twist to it, which i cant remember anymore.

it would make sense to sub 10 and 5 into the formula to get part (a), then sub in 10 and 1 into the formula and add this to the answer to part (a) to get part (b), but when you do that you get 14.178 (3 dp) and not 13.549...

If anyone else knows what I'm missing do let me know too.

 

[Ayesha B]

I got ittt!


So for part (a) you sub 10 into D and 5 into t and get 5.353 mg correct to 3 dp.

In part (b) you need to add the effect of the first dose and the second dose.

So you find the effect of the second dose by subbing 10 into D and 1 into t and getting 8.825 mg correct to 3 d.p.

You now need to add the effect of the first dose. But 1 hour after the second dose has been added, 6 hours have passed from the time of the first dose, not 5.

So you sub 10 into D and 6 into t this time to get 4.727 correct to 3 d.p.

Now add the two effects and you get 13.549 mg correct to 3 d.p.!

Yay!


Now onto part (c)...

 

[Ayesha B]

Okay I don't get part (c)

 

[aniekan]

Part b.

new dose = 10mg
existing dose in the blood = 5.353mg
total dose = 15.353
time = 1 hour
now x = 15.353 x e^ -(1/8)
x = 13.549

part c.

3 = 13.549 x e^-(t/8)
divide and take natural log
t= -5In(3/13.549)
t= 7.5 hours ??

 

[urwahboy]

 

[urwahboy]

forgot to take 3dps

 

[Hasan Naqvi]

owhh, i got it. thanks alot guys

 

[Ayesha B]

Part b.

new dose = 10mg
existing dose in the blood = 5.353mg
total dose = 15.353
time = 1 hour
now x = 15.353 x e^ -(1/8)
x = 13.549

part c.

3 = 13.549 x e^-(t/8)
divide and take natural log
t= -5In(3/13.549)
t= 7.5 hours ??

Hmm that's a different way of doing part b

But part c is 13.06 --> 13.1 hours.

 

[Ayesha B]

View attachment 45319

For part (c), why do you sub (10 + 5.53) into D? I don't get it :/

 

[Hasan Naqvi]

For part (c), why do you sub (10 + 5.53) into D? I don't get it :/

to add to the initial dose which is already there. 10mg is given after the 5hrs in which we had 5.53

 

[Hasan Naqvi]

http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/June 2008/6665_01_que_20080606.pdf

can anyone explain question 3c?

 

[aniekan]

Hmm that's a different way of doing part b

But part c is 13.06 --> 13.1 hours.

0.0 I made a mistake sorry

 

[Ayesha B]

to add to the initial dose which is already there. 10mg is given after the 5hrs in which we had 5.53

Mhmm, right right right. Alrighty thanks

 

[Ayesha B]

0.0 I made a mistake sorry

Its okay lol; I initially did it the same way but then when I looked at the MS I didnt get.

Now I do though

 

[umer khan666]

brother D is dose putting 10 is ok but y did u add 5.35 its the amount not the dose ?

 

[Ayesha B]

brother D is dose putting 10 is ok but y did u add 5.35 its the amount not the dose ?

5.35 is the dose that was left from the first dose after five hours!

 

[urwahboy]

To all the people who dont understand why I added 5.35... in the qs it says
"No more doses of the drug are given. At time T hours after the second dose is given, the amount of the drug in the bloodstream is 3mg."
this means that we need to take the amount of drug that was present when the second dose was given.. when second dose was give there was 5.35 mg present due to intial addition (which we calcuated in part A) and then 10 mg was added (which is the second dose) so we will add the both as when second dose was given(I am repeating again to make it clear) 5.35 was already present and 10mg was given. Adding gives 15.35 which I used.. I hope u understand now...

 

[urwahboy]

http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/June 2008/6665_01_que_20080606.pdf

can anyone explain question 3c?

cant understamd properly ... can u guys solve it? aniekan and Ayesha B