‫)‪Pearsonvue (KSA‬‬

‫اأسئ ه خ صه أسئ بيرس ن في لتخصص الصيدله ل حص ل ع‬

‫ترخيص مزا ل الم ن من قبل هيئ السع ديه ل تخصص الصحيه‬
‫يتك ن ااختب ر من ‪ ϭϬϬ‬سؤال اختي ر يتك ن من ‪ ϯ‬اجزاء ‪ -ϭ‬أسئ ه‬
‫ف رم ك لجي ‪ -Ϯ‬مس ئل ري ضي ‪Cases -ϯ‬‬

‫س ه ف تجميع هذه‬ ‫المذكره مج نيه ليس ل بيع المق بل ال حيد ه الدع ء لمن تع‬
‫المذكره رج ء مراجع الجر ص ح فكرةانش ء هذه المذكره كل فتره لمت بع كل م‬
‫ه جديد نظراً ل تحديث الذى يت كل فتره ع اأسئ ه من قبل شرك بيرس ن في رابط‬
‫الجر ‪:‬‬
‫‪https://www.facebook.com/groups/pearsonvue.questions‬‬

‫‪/‬‬
 1.Laws:
 Vd=dose/co
Vd = Volume of distribution
Co = Conc. of drug in plasma at zero time
 Loading dose = Vd x Css
Loading dose = Vd [C2 – C1]
- Loading dose .. is the dose needed to reach steady state
- Css = Concentration of the drug in blood at steady state
- C1 = Concentration of the drug in plasma
- C2 = Concentration of the drug needed to add to C1 to reach equired conc.

 At steady state rate of drug input=rate of elimination

 Time required to reach steady state (Tss) = 4.5 or 5 t1/2
 Half life (T1/2) = the time required for the concentration of a substance in the
body to decrease by half.

 Therapeutic index (TI) =LD50/ED50

- LD50 = Median Lethal Dose is the amount of an agent that is sufficient to kill
50 percent of a population of animals
- ED50 = Median Effective Dose is the dose that produces a quantal effect in
50% of the population
- Drugs with narrow TI = highly dangerous

 Bioavailability =AUC/conc
Bioavailability = AUC ((oral)) /AUC ((iv)) x 100
Bioavailability =plasma conc of drug by any route/plasma conc of drug by iv
AUC = Area Under Curve

 Specific gravity = Wt. of substance ((Kg)) / Wt. of equal amount of water ((L))
Specific gravity = mass unit volume of sub. / mass unit volume of water
Specific gravity = Denisty of sub. / Denisty of equal amount of water
Denisty = Mass ((gm)) / Volume ((ml)) .. or Kg/L

 mEq = Wt. ((mg)) x valency / M.wt

mEq = milliequivalent

Clearance Laws:

 Clearance (Cls) = 0.693 x Vd / T 1/2

 Vd=dose/co
 Cls=rate of elimination/drug conc
 Cls =renal cls +non renal cls
 Cls = Ke x Vd ………. Ke = elimination rate conistant
 Creatinine clearance for male =(140 – age)x weight /72 x ser.
Creatinine

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 2

  Cr.cls for female = Cr.cls for male x 0.85

Molality Laws :

 Molality ((m)) = No. of moles of solute / Kg mass of solvent

 No. of moles = Wt. of solute / M.wt
 Mass ((M)) = Denisty ((D)) x Volume ((V))

Molarity Laws :

 Molarity ((M)) = No. of moles of solute / L volume of solution

Osmolarity laws:

 Millimoles = [ wt. of sub. ((gm)) / M.wt ] x 1000

 mosm = millimoles x no. of species
Examples Of no. of species:
Ex1: NaCl = 1 Na + 1 Cl = 2 -------- Ex2: CaCl2 = 1 Ca + 2 Cl = 3
--- check problems No. 11 & 12

Some Conversions :

Weight:

 Kg = 2.2 Pound (( lb ))
 Grain = 0.065gm
Volume:

 Tea spoonful (( tsp )) = 5 ml

 Table spoonfull (( tbsp )) = 15 ml
 16 drop (( dp )) = 1 ml
 1 fluid ounce = 30 ml
 1 L = 0.22 Gallon
 1 L = 10 Decilitre
Tempreture:

 5F = 9C + 160
F = Fehrenhiet --------- C = clsius
Length:

 1fool (( ft )) = 12 inch
 1inch =2.54 cm
Others :

 PPM = Part Per Milion = mg / L

 10% w/w = 10 gm in 90 gm (( total wt. = 100 gm )) --------- w/w = gm in
gm

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 3

  10% w/v = 10 gm in 100 ml (( total voume = 100 ml )) --------- w/v = gm
in ml
 10% v/v = 10 ml in 90 ml (( total voume = 100 ml )) ----------- v/v = ml in
ml

Some Molculer weights you may use:

 HCl = 36.4
 NaCl = 58.5
 CaCl2 = 111
 Kcl = 74.5
 NH4Cl = 53.5
 MgCl2 = 95.2

Some other laws haven't be used till now but may be useful for you (( just
read )) :

 Child dose = wt (( lb )) / 150 x adult dose

Child dose = age / (age+12) x adult dose
 E = Extraction ration = drug elimination of an organ ((
eg. Liver ))
E = [arterial drug conc. - venous drug conc.] / arterial
drug conc.
Cls of liver = E x hepatic blood flow

2.Problems :
1- amount of drug is 5 mg in 1 ml what the amount of drug in 1
tsp in microgram
a)5
b ) 25
c ) 500
d ) 2500
e ) 25000

Answer:
1 tsp = 5 ml
5 mg …. 1 ml
X mg …. 5 ml

X = 5 x 5/1 = 25 mg = (25 x 1000) 25000 mcg

***

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 4

2- A solution is made by dissolving 17.52 g of NaCl exactly 2000 ml.
What is the molarity of this solution?
a- 3.33
b- 0.15
c- 1.60
d-3.00 x 10 -4
e-1.6x10 -4

Answer :
Molarity=mole/volume (L)
1 Mole=molecular weight of subs. In 1 grams
No of Moles = wt / Mwt
So, molecular weight of NACL=23+34=57
So, Mole=17.52/57=0.307
So, Morality=0.307/2=0.153

***

3-5ml of injection that conc. 0.4% calculate the amount of drug?

a-0.2mg
b-2mg
c-200mg
d-2000mg
e-20mg

Answer:
0.4 gm … 100 ml
X gm … 5 ml
X = 5 x 0.4/100 = 0.02 gm = (0.02 x 1000) = 20 mg

***

4-An elixir contains 0.1 mg of drug X per ml. HOW many micrograms are
there in one tsp of the elixir

A. 0.0005 micrograms
B. 0.5 micrograms
C. 500 micrograms
D. 5 micrograms
E. 1500 micrograms

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 5

Answer :
0.1 mg in 1 ml
X mg in 5 ml

X = 0.1×5 /1 = 0.5 mg = 500 micro

***

5- sol contain D5W another one contain D50W we want to prepare sol cotain
D15W its volune is 450ml ... how much ml we need of each sol
a) D50w/D5w=10/35

Answer:
try the choices ratio in the equation :
(C1 × V1) + (C2 × V2) = (C × V)
( 50 × 10 ) + ( 5 × 35 ) = ( 15 × 45 )

Another answer :
(X) 50 ---------- 10 15 – 5 = 10
15
(Y) 5 ----------- 35 50 – 15 = 35

X / Y = 10 / 35 ---------- Y = 3.5 X
X + Y = 450 ---------- X + 3.5 X = 450
4.5 X = 450 --------- X = 450 /4.5 = 100
Y = 3.5 X = 3.5 x 100 = 350
X = amount of D50w …. Y = amount of D5w

***

6- prescription
hydrocortisone 2%
Cold cream 60gm
You have concentrations of hydrocortisone 2.5% & 1% how many grams will
you use from two concentration?

a- 20gm from 1% and 40gm from 2.5%

b- 40gm from 1% and 20gm from 2.5%
c- 30gm from both

Answer:
try the choices ratio in the equation
( C1 × V1 ) + ( C2 × V2 ) = ( C × V )
( 1 × 20 ) + ( 2.5 × 40 ) = ( 2 × 60 )

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 6

Another answer :
(X) 2.5% ---------- 1 2–1=1
2%
(Y) 1% ----------- 0.5 2.5 – 2 = 0.5

X / Y = 1 / 0.5 -------------- X = 0.5 Y

X + Y = 60 -------- 0.5 Y + Y = 60
1.5 Y = 60 ----------- Y = 60 /1.5 = 40
X = 0.5 Y = 0.5 x 40 = 20
X = amount of 2.5 % …. Y = amount of 1%

***

7-Prescription
hydrocortisone 2% w/w
Cold cream 60gm
you have hydrocortisone solu. 100 mg/ml .. how many milliliters will you use
from the solution ?

a.10 ml
b.20 ml
c.40 ml

Answer :
2% w/w = 2% x 100gm = 2 gm means the prep. needs 2 gm of
hydrocortisone
0.1 gm in 1 ml
2 gm in X ml

X = 1 x 2/0.1 = 20 ml

***

8- if we have 0.8687g cacl2 in 500 ml solvent , denisty of the solvent is 0.95

g\cm3 ....Find the molality

a- 0.0165 Molal
b- 0.0156 Molal
c- 0.0165 m
d- 0.0156 m

Answer :
Moles = mass/m.wt = 0.8687 / 111 = 0.00782
Weight = density × volume = 0.95 × 500 = 475 gm = 0.475 kg
Molality = moles / kg of solvent = 0.00782/0.475 = 0.0165 molal

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 7

 ***

9. How gm of substance X must added to 2000 gm of 10% substance X

solution in order to prepare 25% of substance x solution

a) 10000 gm
b) 400 gm
c) 40 gm
d) 10 gm
e) 0.4 gm

Answer:
(C1 × V1) + (C2 × V2) = ( C × V )
( 100% × Xgm ) + ( 10% × 2000 gm ) = ( 25% × 2000+X gm )
100X + 20,000 = 50,000 + 25X
100X - 25X = 50,000 - 20,000
75X = 30,000 ..... X = 30,000/75 = 400 gm

Another answer :
100% -------------- 15 25 - 10 = 15
25%
10% ---------------- 75 100 - 25 = 75
so the ratio between 100% : 10 % to reach 25% = 15 : 75
2000 gm ------ 75
X gm ------------ 15
X = 2000 x 15 / 75 = 400 gm

***

10- How much water (in milliliters) should be added to 250 mL of 1:500 w/v
solution of benzalkonium chloride to make a 1:2000 w/v solution

A/0.4L
B/2L
C/0.2L
D/ 0.05L

Answer:
250/500 = 0.5
250/2000 = 0.125
0.5 – 0.125 = 0.375

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 8

 ***

11-How many mOsm are present in 1 liter of sodium chloride injection

(Mwt: sodium chloride= 58.5) ?
308 mosm

Answer :

 Note ; normally conc. of NaCl injection = 0.9%

that means 0.9 gm in 100 ml ..... that means 9 gm in 1 L
 Step 1.
millimoles = wt (gm) / Mwt (gm) × 1000 = 9 /58.5 ×1000 = 154
Note ; millimole = wt (mg) / Mwt (gm)
 Step 2.
mOsm = millimoles x no. of dissosation particles =154 × 2 =308 mosm

***

12-A solution contains 448 mg of KCl (MW=74.5) and 468 mg of NaCl (MW =
58.5) in 500mL. What is the osmolar conc. of this solution ?
0.056 Osm/l

Answer :

 For ( KCl )
0.448 gm in 500ml
X gm in 1000 ml ...... X= 0.896 gm
moles= 0.896/74.5 = 0.012
Osm= moles × no. of dissosation particles =0.012 × 2= 0.024
 For NaCl
0.468 gm in 500 ml
X gm in 1000 ml ..... X= 0.936 gm
moles= 0.936 /58.5 = 0.016
Osm= 0.016 × 2= 0.032
 Total osmalar conc. of sol. = 0.032 + 0.024 = 0.056 Osm/l

***

13. A Patient weighting 80 Kg is supposed to receive a drug at a dose of

2mg/kg/day. What is the dose that the patient should take for each day:

A. 80 mg ..... B. 160 mg ..... C. 240 mg ..... D. 320 mg ..... E. 400 mg

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 9

 ***

14. Drug X is a given to a 70 Kg patient at an infusion rate of 0.95 mg/kg/hr.

How much drug we need for a 12-hr infusion bottle

A. 798 mg ..... B.66.5 mg ..... C. 665 mg ..... D. 84 mg

***

15. how many gm of water add to 5% KCL soln to make 180 gm of

solution(w\w)?
171 gm

Answer:
5gm--------------100
Xgm--------------180

X= 5x180/100=9 gm
So, the amount of water is:- 180 - 9 =171 gm

***

16. hypoparathyroid patient with tingling and numbness has the following lab
result so what is value of calcium correlative to albumin when below 45

Result normal value

Calcium 1.6 2.25-2.6
Albumin 34 18-56

a.2.3 b-1.5 c-2.5

N.B: 2.3 is a Conistant value you have to know

***

17. in clinic patient prescriped with a 500mg dose of aspirin , initial plasma
conc is 100mg .. With half life 6 hours calculate total body clearance ?

a.0.5 L/hr
b.5 L/hr
c.50 L/hr

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 10

Answer:
Vd = dose / initial conc = 500/ 100 = 5L ..... T1-2 = 6 hr
Cl = 0.693 Vd / T1-2 = 0.693 × 5 / 6 = 0.5775 L/hr

***

18. - aminophylline (80%theophylline) was prescriped for asthmatic patient in

a dose of 500mg , half life =6.93 hours how many hours will it take to reach
below 2 % ?
42 hr

Answer:
(80%) ...T1... (40%) ...T2... (20%) ...T3... (10%) ...T4... (5%) ...T5... (2.5%)
...T6... (1.25%)

Time = 6 × T1/2 = 6 × 6.93 = 41.5 hr

***

19. Drug aminophylline (80% theophylline) in 500ml sln . Half life 6 h .what is
the concn of theophylline after 1 day ?
5%

Answer:
1 day = 24 hr = 4 T1-2
(80%) ....T1... (40%) ...T2... (20%) ...T3... (10%) ...T4... (5%)

***

20.For 1 litre of NaCl 3% calculate the osmolarity m.wt=58.5

1026

Answer:
3% means 3gm in 100 ml ... that means 30gm in 1L
No. of moles = wt / Mwt = 30 / 58.5 = 0.513 mole
Osm = no. of mole × no. of dissosation particles = 0.513 × 2 = 1.026
1.026 x 1000 = 1026 mosm

***

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 11

 21. If we give 250 ml of a drug and the area under curve was 112mg/hr/L and
after that we give 500 ml and the area under curve was 56 mg/hr/ml
The bioavilability decreased by
A-25%
b-50%
c-75%

Answer:
250ml ........ 112
500 ml .........X

X= 122×500 / 250 = 224

But real auc was = 56
So the bioavilability decreasing = 56/224 ×100 = 25%

***

22. drug A taken IV and drug B taken orally

the AUC of A =300 and Auc of b =225
what is biovalbility of drug

A. 85%
B. 90%
C. 75%
D. 80%

Answer:
Bioavailability= auc oral /auc iv ×100 = 225/300 × 100 = 75%

***

23.T 1/2 .. in frist line is ....

A .1/k
B . 0.693/ k

***

24. a drug is given as iv infusion in a rate of 2mg/hr ,its T1-2 = 2hr , how much
mg of the drug we need to reach steady state

A. 4mg
B. 16mg
C .20mg
D. 40mg

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 12

Answer :
We reach steady state after 5 T1-2 = 5 × 2 = 10hr
2mg ...ever... 1 hr
Xmg ...after... 10 hr
X = 2×10/1 = 20mg

***

25. a drug with T1/2 = 72hr , the body will recive complete dose after ;

A. 1 day
B. 2days
C. 1week
D. 2weeks

Ans: We will reach Steady state after 5 half-life = 5×72= 360hr = 2weeks

***

26. A patient takes levofloxacin 250mg/ml , the pharmacist has levoflaxacin

injection 500mg / 20 ml , the concentration needs to be dilated for patient ..
which of the following concentration is more accurate:

A/ 10 ml
B/ 15 ml
C/ 7.5 ml

Answer :
500 mg in 20 ml
250 mg in X
X = 20 x 250 / 500 = 10 ml

***

27. priscription for a child contain Omeprazol syr. 10 mg/ml twice daily for a
week .. you have Omeprazol capsul 20 mg in your pharmacy,
how many capsules are needed to prepare solution with concantration 2
mg/ml ??
7 cap.

Answer:
10 mg/ml twice daily for a week = 140
20 _____1
140 _____ X

X=140/20=7

***

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 13

28.Drug 500mg and 300mg eleminated outside the body and t1/2=5hr and
another drug same first one but with conc 1000mg .. how many hrs it take to
eliminate 600mg ot of the body?
5 hrs

Answer :
CLs=rate of elimination /drug conc
CLs1=300/500=0.6
Vd=t1/2×cls/0.693=5×0.6/0.693=4.3
CLs2=600/1000=0.6
t1/2=0.693×vd/cls=0.693×4.3/0.6=5 hrs

***
29. HOW can prepare 100 ml of 12% MgCl by taking?
a-12ml of MGCL dissolve in 100 ml water
b-12 gm of MGCL dissolve in 100 ml water
c-12ml of MGCL dissolve in 1000 ml water
d-90.5 ml of MGCL dissolve in 100 ml water

Note ; w/v = g/ml ..... ex ; 4% w/v means 4 gm in 100 ml

***
30. man 40 years and 80 kg sr ce 0.5 mg\dl find creatinie clearance mg\ml :
a.222
b.232
Answer :
Cr.cl for male = (140 – age)x weight /72 x ser. Creatinine
=(140 – 40 ) x 80 / 72 x 0.5 = 222

N.B : The same data for female the answer is : 189

Cr.cl for female = Cr.cl for male x 0.85 = 222 x 0.85 = 188.7

***

31.15 g of drug is added in 150mg of a solvent. Then what is the total

concentration of drug in the final mixture:
a- 6.01%
b- 9.10%
c- 10%
d- 15%

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 14

Answer:
15 + 150 = 165
15 g in 165
X g in 100
X = 100 x 15 / 165 = 9.10

***

32. A bag containing 250 ml of 25000 IU heparin

The patient weigh 70 kg should recieve 10 IU/kg/hr...calculate the amount in
ml the the patient should recieve in one hour...
7 ml

Answer:
10 iu for 1 kg
X iu for 70 kg
X = 70 x 10 /1 = 700 iu

250 ml of 25000 iu
X ml of 700 iu
X = 700 x 250 / 25000 = 7 ml

***

33.Patient with prescription of Captopril 50 mg per tab with a dose of 100 mg

daily for 4days and you only have the 25 mg tab .. How many tablets you will
dispense ?

16 Tab

Answer :
100 mg daily for 4 days = 400 mg
400/25 = 16 tab

***

34.A problem with the following data

Dose = 1000
Initial conc =10
Elimination rate constant=0.1
Calculate total clearance ??

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 15

a-250
b-200
c-150
d-100
e-10 litre

Answer:
Cl= vd × kel
Vd=dose/conc=1000/10=100
Cl=0.1×100=10

***

35.Problem with the following data :

Density = 1.75 g/cm3
Mass = 15 gm
Calculate the Volume ?

a.11
b.10
c.8.52

Answer:
Denisty = mass / volume
volume = 15 / 1.75 = 8.57

***

36.Prescription contain :
Clindamycin 1.5%
dilultion with alcohol up to 300 ml
you have a bottle 100 ml of 10% clindamycin
how many millelitres will you use ?

a.7.5
b.45

Answer:
1.5 ..... 100
X ..... 300
X=4.5

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 16

10 ...... 100
4.5 ..... X
X=45

***

37.A drug with Conc. 400 m and T1/2 = 12 hr.s

the concentration will decrease after 1 day by …

a.10%
b.25%
c.75%
d.90%

Answer:
24 hr.s = 2 half lives
(400) …T1 … (200) … T2 … (100)
so you lose 300 of the drug
( 300 / 400 ) x 100 = 75%

***
38. A drug should be given 50 ml of 2 meq/ml , but available concentration is
10 meq/ml, How many ml should dispense to patient?
a.5 ml
b.10 ml
c.15 ml
d.20 ml
e.25 ml
Answer:
2mg -----1ml
X mg-----50ml
X = 50 x 2 =100ml
10 mg-------1ml
100 mg------ X
X = 100 x 1 / 10 =10 ml

***
39. 30gm of 1% hydrocortisone mixed with 40 gm 2.5% hydrocortisonen what
is the concentration of the resulting solution?
a) 3%

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 17

b) 1.85%
c)10%
d) none of the above

Answer :
C1.V1 + C2.V2 = C3.V3
30gm × 1% = 0.3gm
40gm × 2.5% = 1gm
So, 1.3 gm is in 70 gm
So, the con. =1.3/70=1.857%

***
40. if we have 90% of substance X solution , 50% of substance X solution ,
how mixing both to give 80% of substance X solution ?
a- 3 : 1
b-1:3
c-10:30
d- 5:9
Answer :
We should try all answer with that equation
(C1×V1) + (C2×V2) = (C×V)
(90% × 3) + (50% × 1) = (80% × 4)
( 270 ) + ( 50 ) = ( 320 )
( 320 ) = ( 320 ) so the answer is 80%
Another answer :
90% 50%
80%
30 10
So .. 90/50 to reach 80 % equal 30/10 = 3/1

***
41. - prep. contain coal tar 30 part ... petroleum 15 part ... adeq. to 150 part ...
what conc. of coal tar in 500 ml:
100 part

Answer:
30 part present in 150ml of prep.
X part present in 500ml of prep.

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 18

so, conc. of coal tar in 500ml=30x500/150= 100 part

***
42.How many grams needed from drug in one teaspoonful , if 5 tspfull doses
contain 7.5 gm of drug ?
a) 0.0005
b) 0.5
c) 500
d) 1.5

Answer:
7.5gm in 5 tsp .....
X gm in 1 tsp .....
X = 7.5×1 /5 = 1.5 gm
N.B: 1 tsp = 5 ml

***

43.KI solu. has 0.5mg/ml dissolve in 30ml water calculate the amount of KI in
the solu. ?
15mg

Answer :
0.5 mg in 1 ml
X mg in 30 ml

X= 0.5×30 /1 = 15 mg

***

44. - the dose of drug is 0.5ml per day and the total amount of the drug Is
100ml what is the total dose ?
200

Answer :
no. of doses = amount of drug / amount of one dose = 100/0.5= 200

***

45.if we have a solvent costs 150 riyal/kg and its specific gravity =1.07 ,so the
cost for 100ml of the solvent is :
16.05 riyal

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 19

Answer :
Weight (Kg) = volume (L) × sp. Gravity …….. 100 ml = 0.1 L
wt = 0.1 × 1.07 = 0.107 Kg

1 kg cost 150 riyal

0.107 kg cost X riyal

X = 0.107×150 /1 = 16.05 riyal

***

46- A patient cholesterol level is equal to 4mM/L. This cholesterol level can be
expressed in terms of mg/dL
( molecular weight of cholesterol = 386)

A.0.0154 mg/dL
B. 0.154 mg/dL
C. 1.54 mg/dL
D. 15.4 mg/dL
E. 154 mg/dL

Answer :
Conversion from (mM) to (mg) = conc. × molecular weight
Conversion from (L) to (dL) = conc. / 10
Conc (mg/dl) = conc. (mMol /L) × mwt / 10 = 4×386 /10=154.4

***
47.drug container contain 90 mg each tablet contain 0.75mg.
how many doses ?
No. of doses = total wt / wt of one dose = 90 / 0.75 = 120 dose

***
48- How need prepare benzacainamid conc. 1:1000 ,30cc of benzocainamid
solution?

a-30 mg
b-50 mg
c-80 mg
d-100 mg
e-130 mg

Note : cc = cubic centimeter = cm3 = ml

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 20

Answer :
1 gm ----- 1000 ml
X gm ----- 30 ml

X = 30 x 1 / 1000 = 0.03 gm = 30 mg

***
49. The Molal concentration of 0.559 M solution is ;
(Mwt=331.23 g/mol) (density of solution =1.157g/ml)
a-1.882
b-0.882
c-0.559
d-0.575

Answer :
Mass = moles × Mwt = 0.559 × 331.23 = 185.15 gm
wt of solution = Volume × Destiny = 1000 ml × 1.157=1157 gm
so wt of solvent = 1157 - 185.15 = 971.85gm = 0.971 kg
molality = moles / kg of solvent = 0.559 / 0.971= 0.575 molal

***

50.Problem asked to calculate Plasma Osmolarity

an you have given some data
Na 140
Cl 103
Hco3 18
Bun 8
S.cl 8

Answer is : 263

N.B:

 the data of this problem isn't complete here .. 263 is the right answer
just know it
 in general .. to calculate plasma osmolarity follow this equation :
2[Na] +[Glucose]/18 +[BUN]/2.8

***

51. drug decrease after 2hr to 50% &the user takes it every 2 hr how many
hours needed to reach steady state ?

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 21

A/2-4
B/6-8
C/10-12

Answer:
Time to reach steady state ((Tss)) = 4 to 5 T1/2
4 x 2 = 8 ….. 5 x 2 = 10
N.B: if there is (( 8-10 )) if choices … choose it

***

52. 10g of a drug was dissolved in 150g of solvent, what is the final
concentration?
6.25%

Answer:
10 … 160
X …. 100
X = 100 x 10 / 160 = 6.25 %

***

53.A physician prescribed paracetamol 120mg/5ml to take 10ml every 8 hours

but the pharmacist has only paracetamol 160mg/5ml . what is the volume to
be administered to give the effect of the first dose :

a- 6.5 ml
b- 7.5 ml
c- 10 ml
d - 11 ml

Answer:
dose = 240 mg paracetamol
160 mg in 5 ml
240 mg in X ml
X = 240 x 5 / 160 = 7.5 ml

***

54.A drug with conc. 100 mg/ml .. after 1 hr. it decreased to 50 mg/ml ..
calculate its concantraion after 3 hours :

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 22

a.25
b.12.5
c.6.25

Answer :
100 .. [1hr] .. 50 .. [2hr] .. 25 .. [3hr] .. 12.5

***

55. how many gm of water add to 5% KCL soln to make 100 gm of solution
(w\w) ?
95gm

N.B: 5% (w/w) means 5gm of KCl in 95gm of water and solution total wt=100

***

56. 1000 mg of drug follow one compartment.. calculate vd ?

Time 0 hr 2 hrs 4 hrs 6 hrs 12 hrs

Conc 80 58 34 28 10
mg/ml

A.12.5 litre
B. 4 litre
C. 45 litre

Answer :
Vd = dose / initial conc.
Vd = 1000 / 80 = 12.5 L

***
57. Drug dose 1000 mg orally
Time 0 hr 2 hr.s 4 hr.s
Conc. 40 18 8

What is the Vd of the drug ?

a.55 litre
b.45 litre
c.75 litre
d.25 litre

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 23

Answer:
Vd= 1000/40 = 25 L

***

58. HOW can prepare 100 ml of 12% MgCl by taking?

a-12ml of MgCl dissolve in 100 ml water

b-12 gm of MgCl dissolve in 100 ml water
c-12ml of MgCl dissolve in 1000 ml water
d-90.5 ml of MgCl dissolve in 100 ml water
e-0.95 ml of MgCl dissolve in 100 ml water

***

59. How many grams of drug used to prepare 150 ml solution ,, if one tsp
contains 7.5 mg of drug
a. 4 gm
b. 0.225 gm
c. 2.25 gm
Answer:
7.5 mg in 5 ml
X mg in 150 ml
X = 150 x 7.50 / 5 = 225 mg = ((225/1000)) 0.225 gm

***

60. Patient takes dose 20 mg/kg/day

what is the dose if patient weight 60 pound ?
545 mg/day

Answer:
you have to know .. 1 kg = 2.2 pound (lb)

20 mg -------- 2.2 lb
X mg ---------- 60
X = 60 x 20 / 2.2 = 545.45 mg/day
***
61.A child was prisciped a drug with dose 65 mg/kg/hr .. his body weight =
35.2 pound
Calculate the dose ..
a.1.040 gm
b.10.40 gm
Answer:
35.2 pound = 15.97 kg = about 16 kg

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 24

65 mg … 1 kg
X mg … 16 kg
X = 16 x 65 = 1040 mg = 1.040 gm

***
62.Calculate the Specific gravity of a substance of volume = 121.92 ml & wt =
107.5
A/1.88 s.g.
B/2.88 s.g.
C/0.88 s.g.
D/8.8 s.g.
Answer:
Denisty = wt. / volume
= 107.5 / 0.12192 = 881.7
Sp. Gravity = denisty Of substance / den. Of water = 881.7 / 1000 = 0.88

***

63. The ppm concentration of a 6.35x1 0-6M solution of sucrose (Mwt of

sucrose is 342.3 g/mole) is:

A. 2.174 × 10-3ppm
B.2.174 ppm
C.2.174 × 10-6 ppm

Answer :
ppm concentration = mass in mg / volume in liters
Molar conc means no. of mole in 1 liter .... then volume= 1L
mass = moles × Mwt = 6.35x10-6 x 342.3 = 2.174x10-3gm = 2.174 mg
Then 2.174 mg is in 1L = 2.174 ppm

***

64. A 500 infusion bottle contains 11.729 mg of potassium chloride (KCI).

How many mEq of KCI are present? ( Mwt of KCI = 74.6)

A. 0.1571 mEq
B. 1571 mEq
C. 6.37 mEq
D. 0.00637 mEq

Answer :
mEq = wt (mg) × valency / Mwt = 11.729 ×1 / 74.6 mEq = 0.1572

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 25

 ***
65. Fifty micrograms equals:
a-50000 ( nanogrames )
b- 0.05 ( milligrams )
c- 0.0005 g
d- a and b
e- a and c
Note; ... mc-g = 1000 nano-g ... milli-g = 1000 mc-g ... g = 1000 mg

***
66. a 2 mg/L solution , according ppm
a-2 ppm
b-0.002 ppm
c-0.000002 ppm

Note ; ppm = mg / L
ppm : part per milion

***

67. What is The Specific gravity of substance has Weight=Y & The volume is
X?
Y/X

Answer :
The Specific gravity =Density of the substance/Density of water
Density of water = 1 .... Density of substance = weight/volume
So, the sp. gravity of sub. =weight (Y) /volume(X)/1 = Y/X

***
68. drug decrease to 50% of its plasma conc. after 2hr .. we have dose A
given each 2hr and dose B given each 4 hour … in dose B what is the plasma
conc. at steady state ?
A/0.25
B/0.5
C/2

***

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 26

69.Calculate C av .ss
1gm vancomycin for patient 78 kg Taken by infusion rate 12 hr /7 day
T 1/2 =8
Vd = 1 k/l
A. 3
B. 5
C. 17
D.19
We can't find the right answer .. try to solve it 

***

70.Paitents on treatment with acyclovir and famcyclovir .. group that treated

by acyclovir show recurrence by 27% and who treated by famcyclovir show
recurrence by 25%
the ques. is how many patients should take famcyclovir over than who take
acyclovir per year to reach equivilant results ?
The answer is : cannot be calculated because of low information

***

71.Patient's dose of some drug is 0.5 mg daily and Vd = 500 L .. his body
elimination rate is 110.16 Litre per day … in the last day about 80 % of the
drug was in his blood
Calculate half life ..
3 days
Answer:
Cl=0.693 x vd / T1.5
T1/2 = 0.639 x 500 / 110.16 = 3.14 day

***
72.Problem with data :
drug 10 mg/ml and t1/2= 3 hrs
how much hrs needed to reach steady state??
12 – 15
Answer:
Time required to reach steady state (Tss) = 4 – 5 t1/2
4x3=12 ……. 5x3=15

***

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 27

73. drug t1/2= 2h .. dose A taken every 2h and dose B taken every 4h
compare plasma concentration a to b ..
a.1/2
b.2

***

74. A half life of a drug decrease by 50% , after how hours will the time
needed to decrease to 2%
a.2 …. b.10 …. c.5 …. d.12
Answer :
100% .. [T1] .. 50% .. [T2] .. 25% .. [T3] .. 12.5% .. [T4] .. 6.25% .. [T5] .. 3.1%
.. [T6]
1.5% so we need 6 half lives to reach below 2% ......... T1/2 = 2 h.
2 x 6 = 12 h.

***

75.A problem with thin curve and ask for therapeutic range
answer : 8/2 = 4
- in other exams the same curve with LD50 = 20 & ED50 = 5
so TI = LD50/ED50 = 20/5 = 4

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 28

 ***

76.which drug has higher bioavailability ?

1.A … 2.B … 3.C … 4.D

N.B : bioavilability measured by comparing plasma level
higher plasma level = higher bioavailability

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 29

 ***

77.Which drug of the following has the safest margine ?

1.A … 2.B … 3.C … 4.D

N.B : safest margine = higher therapeutic index

***

Summary of the important problems :

1.Molarity of 17.52 NaCl solution : 0.15

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 30

2.Cold cream with two concentrations : 20gm from 1% and 40gm from 2.5%

3.Cold cream (( how many ml uses )) : 20 ml

4.Ca correvted to albumin : 2.3

5.Osmolarity of NaCl : 1026

6.AUC bioavailability ((112, 500)) : 25%

7.AUC bioavailability ((300, 225)) : 75%

8.Levofloxacin : 10 ml

9. Omeprazol : 7 cap.

10.Crcl of Male, 40 y, 80 kg with Scr: 0.5 mg/dL: 222ml/min

11.the same problem but for female : 189ml/min

12.Heparin bag : 7 ml

13.Captopril : 16 tablets

14.Clindamycin : 45

15. Plasma Osmolarity : 263

16.Paracetamol : 7.5 ml

17 .gm of water add to 5% KCL (( w/w )) : 95 gm

***
Don’t Forget to Study the other Files (Cases &Pharma
Questions)
Good luck 

Copyright https://www.facebook.com/groups/pearsonvue.questions Page 31

Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Scanned by CamScanner
Answer ( C )
20 g in 100 ml
8.4 g in X ml
X ml = ( 8.4 x 100 ) / 20
= 42 ml

The Answer is ( D )
The total dose for 3 days = {( 800 mg x 2) x 3}
= 4800 mg
200 mg in 1 tablet
4800 mg in x tablets
X tablets ={ (4800 x 1 ) / 200 }
= 24 tablets
The Answer is ( B )

The Answer is ( C )
12.5 mg in 5 ml
X mg in 60 ml
X mg = { ( 60 x 12.5 ) / 5 }
= 150 mg

50 mg tablets in 1 stock
150 mg in x stock
X = { (150 x 1) / 50 }
= 3
 The Answer is ( B )
40 mg for 1 m²
X mg for 1.5 m²
X mg = { ( 1.5 x 40 ) / 1 }
= 60 mg
160 mg in 16 ml
60 mg in x ml
X ml = { ( 60 x 16 ) / 160 }
= 6 ml

The Answer is ( C )
0.2 % means
0.2 g in 100 ml
Xg in 5 ml
X g = { ( 5 x 0.2 ) / 100 }
= 0.01 g = 0.01 g x 1000 = 10 mg
The Answer is ( B )
10 % means
10 g in 100 ml
Xg in 10 ml
X g = { ( 10 x 10 ) / 100 }
= 1g
1mEq = 0.0735 g
X mEq = 1 g
X mEq = { ( 1 x 1 ) / 0.0735 }
= 13.605 g
The Answer is ( D )
The dose for 1 day for 1 kg
20 mg x 4 times = 80 mg
80 mg for 1 kg
X mg for 28 kg
X mg = { (28 x 80 ) / 1 }
= 2240 mg for one day
The stock
200 mg in 5 ml
2240 mg in X ml
X ml = { ( 2240 x 5 ) / 200 } = 56 ml
The Answer is ( A )

The Answer is ( C )
0.25 g in 100 ml
Xg in 10 ml
X g = {(10 x 0.25) / 100 } = 0.025 g = 25 mg
0.1 mg in 1 ml
25 mg in X ml
X ml = 250 ml
250 – 10 = 240 ml
The Answer is ( D )

The Answer is ( C )
12.5 x 2 x 14 = 350 mg for 1 kg
350 mg x 20 = 7000 mg
The bottle is 125 mg in 5 ml
7000 mg in X ml
X ml = 280 ml
One bottle = 100 ml
X bottle = 280 ml
X bottle = 3 bottle
The Answer is ( A )
One dose = 2.5 x 25 = 62.5 mg

25 mg in 1 ml
62.5 mg in X ml
X ml = 2.5 ml

The Answer is ( A )
Teaspoonful = 5 ml
7.5 mg in 5 ml
X mg in 150 ml
X mg = 225 mg = 0.225 g
The Answer is ( D )
The dose = 10 ml x 3 x 5 = 150 ml

The Answer is ( A )
10 g in 100 ml
Xg in 10 ml x = 1g
1 g in 500 ml
X g in 100 ml
X = 0.2 g in 100 ml mean 0.2 %
The Answer is (C )
5 g in 25 ml
X g in 1 ml
X g = 0.2 g = 200 mg

The Answer is ( C )
The Answer is ( A )
X°= 1000 mg C°= 10 mg/l
Vd = X° / C°
Vd = 1000 / 10 = 100 L/kg
Cl = K Vd
Cl = 0.1 x 100 = 10 L/h

The Answer is ( D )
15% mean 15 g in 100 ml
Xg in 60 ml
so, X = (60 x15)/100 = 9 g ( we need)
the stock is 30% mean 30 g in 100 ml
9 g in X ml
So, X = (9 x 100 )/30 = 30 ml
The Answer is ( B )
The total dose need = 500 x 3 x 7 = 10500 mg
So, 250 mg in 1 tablet
10500 mg in X tablets
X = (1 x 10500 )/ 250 = 42 tablets

The Answer is (B)

The dose for 15 kg child for 1 day = 75 x 15 x 4
= 4500 units
The stock 2.5 ml contain 12500 units
X ml contain 4500

X ml = (4500 x 2.5 ) / 12500 = 0.9 ml

The Answer is ( A )
The total need = 10 x 7 = 70 mg
If the capsule strength is 40 mg
so, we need 2 capsule (2 x 40) = 80 mg
2 mg in ml
80 mg in X ml
X ml = (80 x 1 ) / 2 = 40 ml
So, we need prepare 2 capsule and 40 ml water to
prepare 40 ml (2 mg / ml )

The Answer is ( B )
20 x 0.25 = 5 mg
The Answer is ( D )

The Answer is (C )

The Answer
The dose = (20 x 8 ) / 2 = 80 micrograms Not: how many
milliliter ( not correct ) this is volume not Wight
The Answer is ( A )
The dose = 2.5 x 25= 62.5 mg
The stock 25 mg in 1 ml
62.5 mg in X ml
So, X = (62.5 x 1 )/ 25 = 2.5 ml

The Answer is (D)

2.5 % W/W means 2.5 g in 100 g
Xg in 1 g
So, X = 0.025 g drug in 1 supp.
X g drug in 24000 supp. So, X = 600 g
The Answer (B)
10% means 10 g in 100 ml

The Answer (A)

40 mEq in 1000 ml
X mEq in 40 ml So, X = 1.6 mEq in 1 h
1.6 mEq for total 16 kg
X mEq for 1 kg
So, X = 0.1 mEq in 1h = 2.4 mEq in 24 hr
The Answer (B)
(% w/v) = ( g / 100ml)

The Answer (D)

10 ml = 1 g CL HCl
3 ml = X g CL HCl X = 0.3 g CL HCl = 300 mg
1 mg CL HCl = 800 mic CL
300 mg CL HCl = X mic CL X = 240000 mic CL
٠.٣=٣/١٠
٨٠٠*٣٠٠
 =٥٠٠|٢٧*٣

The Answer ( A)

The Answer ( C )
The Answer (D)

The Answer ( B)
Flow rate = (Drop factor X volume)/ Time in
minutes
= ( 30 x 20 ) / 60 = 10
The Answer (D)

The Answer (A)

The Answer ( A)

The Answer ( C)

100x10=1000/25=40/5=8

The Answer ( D)
The Answer (C)
In fed state of stomach the blood perfusion is
high
So, the hepatic clearance = (2.5 x 40)/100 = 1 L/h

The Answer ( B)
 150/15x100

The answer (D)

1.5x60=90
2.5x90=225x10=
The answer (D)

The Answer (A)

By alligation method
 5x60\50=

The answer (A)

3x8=

The answer (D)

.2x5/100=.1/1000

The answer (C )
The answer (C )

25x4=100mg/d
125x2=250ml
500x10/5=500mg
500/100=5 day

The answer (B )
 5x4=20mg/d
20x10/25=8

The answer (C )

2000mgx1/74.6=

The answer (C )
 70x250/25000=0.7x10=7ml

The answer ( B )

mass=dxv
0.92x60

The answer ( C )

The answer ( B )
The answer ( D )
K = 0.693 / t 0.5 = 0.1 h-1
Vd = X° / Cp° = 5 L
CL = K x Vd = 0.5 L/h

3x14=42/2=21

The answer ( D )
 1172/74.6

The answer ( D )

The answer ( C )

140x0.9/1=126mg
126x5/50=12.6

The answer ( B )
 10x750/5x60

The answer ( C )
Flow rate = (Drop factor X volume)/ Time in minutes

500x60x250/400000=18.75

The answer ( B )
 5t1/2=5x3=15

The answer ( B )

70x500/100=350

The answer ( C )

‫نص‬

The answer ( C )
 WT=36X20=720X4X5=14400MG
14400X5\200=360

The answer (

250x10/100=25

The answer ( B )
 ab=IV X ORAL/IV XOARAL
0.2 X340/20 X4=0.85X100=85%

The answer (B )

10KG X100ML=1000
3X50=150
1000+150=1150

The answer ( C )
The answer ( D )

The answer ( D )
Vd = X° / C° = 400 / 0.01 = 40000 ml = 40 L

2000000/1600

The answer ( B )
 0.2X10/100=0.02X1000=20

The answer ( D )

The answer ( A )

The answer ( D )
 200X100/1000=20MG

The answer ( B)

1172/74.6X100

The answer ( D )

The answer ( B ) 100X1.5/

(2.5X100=250)=60
 5X150=750

The answer ( B)

20X1000=20000X1600

The answer ( D )

The answer ( D )
 The answer ( A )

The answer ( A )

The answer ( B )
 5x1/1000x15x60x24=108
108/80=1.35=1.4ml

The answer ( B )

The answer ( B )
 12x0.95x70=

The answer ( D )

20x4=80x5=400/200=2x28
The answer ( D )

500/50=10/100=0.1

The answer ( A )
 50/2.5=20ml
20x30=600/60=10
The answer ( B )

The answer ( B )

2.5x1.5/100=0.0375gmx1000
=37.5x60
The answer ( D )
 18000x5/120

The answer ( D )

50x0.0535

The answer ( D )

The answer ( D )
The answer ( B )

The answer ( B )
The answer ( A )

The answer ( D )
The answer ( B )

The answer ( A )
Osmolarity = molarity x number of dissociated parteclis
For KCl 445 mg in 500 ml = 869 mg in 1000L /
74500=0.012x 2= 0.024 osmolar
For NaCl 468 mg in 500 ml = 936 mg in 1000 ml / 58500
= 0.016 x 2 = 0.032 osmolar
0.024 + 0.032 = 0.056
The answer ( C )

The answer ( C )

The answer ( B )
The answer ( D )

The answer ( D )

The answer ( A )
The answer ( C )
Total Mwt = 25 mg 318.04 g = 25 mg
Mwt of diclofenac = X mg 295.05 = X mg
X mg = 23.19