Here are two of my math problems that I understand one moment and then don't understand the next. I hate it when my brain farts!
a. If 60 gm of 1% hydrocortisone is mixed with 80 gm of 2.5% of hydrocortisone, what is the % of hydrocortisone in final mixture?
a. 2.2 % w/w
b. 1.85 % w/w
c. 0.25 % w/w
d. 1.75 % w/w
b. How much of atropine is required to prepare 240cc in such way that when 1 teaspoonful of the solution is diluted to 1 pint gives 1 in 500 solution?
a. 2.25 gm
b. 46.08 gm
c. 35.15 gm
d. 25.35 gm
ANSWERS:
a.(b) Amount of Hydrocortisone in 60 gm, 1%
= 60/100 = 0.6 gm of hydrocortisone.
Amount of hydrocortisone in 80 gm, 2.5%
= 80 x 2.5/100 = 2 gm hydrocortisone
% amount of hydrocortisone in final mixture
= 100 x 2.6 (2gm + 0.6gm)/140 (80gm + 60gm)
= 1.85% w/w.
b. (b) To solve this kind of problem, we must first find out the amount of drug present in final solution.
Amount of atropine in 1 pint, 1 in 500 soln .
= 480 x 1 /500= 0.96 gm of atropine.
Now, 0.96 gm of drug must be present in 1 teaspoonful of drug solution, therefore we can say
5cc (1 teaspoonful) contains 0.96 gm
240cc solution requires ?
= 240 x 0.96/5 = 46.08 gm of atropine.
