Page 48 - Fpgee Question And Answer 2024 Edition

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10 = 8 + log Base
= 1500 x 0.55 Salt
= 825 calories shall be provided by dextrose 2 = log Base
Salt
The grams of dextrose required (1gm = 3.4 calories):
Base = 100: 1
825 Salt
= = 242.64 grams dextrose
3.4
1216.(c) 1 teaspoonful (5cc) of Thioridazine solution
The numbers of cc of 30% dextrose solution (30mg/cc) contains 150mg of drug. This solution is
diluted up to a mark of 480cc therefore:
required:

242.67 x 100 = 150 = 0.31 mg/ml
= = 808.82 cc dextrose solution 480
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1213. (c) A TPN solution shall provide 1100 non- 1217. (a) Weight in grams = no of cc x specific
protein calories and 45% of non-protein calories as gravity
fat, therefore:
= 500cc x 1.25 gram/cc
= 1100 x 0.45 = 625 grams
= 495 calories from fat
1218.(a) 100 mg
The remaining (1100-495 = 605) calories shall be
provided by dextrose. Since the maximum dextrose Clark’s rule = Weight in lbs x adult dose
concentration is 12.5% and each gram of dextrose 150
provides 3.4 calories, therefore:
= 20 x 750 = 100mg
605 150
= = 177.94 grams dextrose
3.4
1219.(c) Patient weight is 156lbs, therefore weight
The volumes of 12.5% solution required to provide in kg would be 156/2.2 = 70.9 kg.
177.94 gm of dextrose:
A normal therapeutically recommended dose of
177.94 x 100 drug is 10mg/kg/day, therefore the dose in the
= = 1423.52 cc dextrose solution
12.5 above patient is:

1214. (c) PKw = PKa + PKb = 10 x 70.9
= 709 mg
14 = PKa + PKb
PKb = 14 -9 = 5 Each ready-infusion-bag contains 250mg of drug, so
the number of bags required to fill the day supply
1215. (b) would be

Base
pH = pk − pk + log = 709 = 2.83  3 bags.
w
b
Salt 250

Base
pH = pk + log 1220.(a) To solve this problem, we can use the
a
Salt alligation method.

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