a. 10 mg

b. 0.03 mg

c. 30 mg

d. 300 mg

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1. (c) 1: 1000 generally interprets as 1 gm in 1000cc solution. The amount of lidocaine in 30cc of 1:1000 solution can be calculated as follows:

= 30 x 1/1000 = 0.03 gm = 30 milligrams.

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a. 346.16 cc

b. 234.43 cc

c. 153.84 cc

d. 121.12 cc

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2. (c) To solve this type of problem, we need to use alligation method.

75 20 (75%)

30

10 45 (10%)

Total parts 65 (30%)

To prepare 65 (30%) 20 parts (75%) need

To prepare 500 (30%) ?

= 500 x 20/65 = 153.84cc (75%) alcohol

If we mixed 153.84 cc of 75% alcohol with 346.16cc [500cc - 153.84] of 10% alcohol, then we can get 500cc of 30% alcohol solution.

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a. 2.2 % w/w

b. 1.85 % w/w

c. 0.25 % w/w

d. 1.75 % w/w

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3.(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.

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a. $ 150

b. $ 17

c. $ 500

d. $ 81

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4. (d) 1000 tablets of Risperdal 1mg cost $ 2250. The % mark up on prescription is 20%.

Therefore retail price of 1000 tablets would be:

= 120 x 2250/100 = $ 2700 ** For each $ 100 cost = $120 retail cost**

Price for 30 tablet would be:

= 30 x 2700/1000 = $ 81

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a. 2.2

b. 0.8

c. 5.2

d. 3.0

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5. (c) A pH of the solution can be found by the following formula:

pH = pKa + log ionize/unionize

= 2.2 + log 103

= 2.2 + 3

= 5.2

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a. 1.2 cc

b. 0.3 cc

c. 0.06 cc

d. 0.01 cc

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6. (c) 0.06cc. According to Friedâ€™s rule:

= age in months/150 x adult dose

= 15 x 325/150 = 32.5 mg

The dropper is calibrated to deliver 325 mg of Iron sulfate in 0.6 cc, therefore:

= 0.6 x 32.5/325 = 0.06cc

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a. 1 bag

b. 2 bags

c. 3 bags

d. 5 bags

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7. (c) Patient weight is 156 lbs, therefore weight in Kg would be 156/2.2 = 70.9 kg

A normal therapeutically recommended dose of drug is 10mg/kg/day, therefore dose in above patient

= 10 x 70.9

= 709 mg

Each ready-infusion-bag contains 250 mg of drug, so number of bags require to fill order would be:

= 709/250 = 2.83 = 3 bags.

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a. 2.25 gm

b. 46.08 gm

c. 35.15 gm

d. 25.35 gm

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8. (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.

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b. 3.25 gm

c. 4.81 gm

d. 2.016 gm

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9. (d) Gram equivalent weights of solute in 1 liter of solution is defined as normality, therefore 1N solution of sodium bicarbonate will contain 84 gms in 1000 cc. We want to find quantity of sodium bicarbonate in 240cc, 0.1 N solution:

1 N solution contains 84 gm

0.1 N solution contains ?

= 0.1 x 84 = 8.4 gms/1000cc.

240cc solution will contain:

= 240 x 8.4/1000 = 2.016 gm of NaHCO3

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a. 23.12 meq

b. 15.17 meq

c. 53.15 meq

d. 38.46 meq

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10. (d) An amount of sodium chloride presents in 250 cc of 0.9% NaCl,

= 250 x 0.9/100 = 2.25 gm NaCl

Total equivalents Na+ = weight in gm

equivalent wt

= 2.25/58.5 = 0.03846 equivalents

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a. 0.3

b. 0.9

c. 0.6

d. 0.4

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11. (d) The sum of probability of success and failure would be equal to 1 in Null hypothesis and can be expressed by following formula:

p + q = 1, where p = probability of success

q = probability of failure

q = 1 - p

= 1- 0.6

= 0.4

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a. 21.23 gm

b. 18.67 gm

c. 14.12 gm

d. 13.25 gm

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12. (b) We want to dispense 12 suppositories each weighing 2gm and containing 400mg of tannic acic:

Amount of coca butter = 2 gm x 12

= 24 gm

Amount of tannic acid = 0.4 gm x 12

= 4.8 gm.

Displacement value of tannic acid is 0.9, therefore:

= 4.8/0.9 = 5.33 gm of base will displace

4.8 gm tannic acid = 5.33 gm cocabutter

Amount of coca butter = 24 gm - 5.33 gm

= 18.67 gm

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120, 135, 140, 118, 175, 105, 115, 190

a. 135

b. 118

c. 127.5

d. 175

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13. (c) Median or Mode is generally expressed as a middle value of experiment, if number of values are even, then average of middle values should be considered. To find median or mode of experiment data, one should first arrange the data in ascending or descending order.

In our example,

105, 115, 118, 120, 135, 140, 175, 190

= (120 + 135)/2 = 127.50

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a. 1/8 times

b. 1/4 times

c. 8 times

d. 4 times

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14. (a) The concentration of reactant M is half in a reaction that is third order in kinetic:

dx/dt = k (a-x) (b-x) (c-x)

= k (a-x)3 where a=b=c

= k (M)3

now M = M/2

= k (M/2)3

= 1 /8 k (M)3

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a. 0.051 min-1

b. 0.025 min-1

c. 0.35 min-1

d. 0.86 min-1

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15. (b) 0.025 min-1

For the first order kinetic,

K = 2.303/t x log Co/C

= 2.303/90 x log 500/50

= 2.303/90 x log 10

= 0.025 min-1

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a. Excellent

b. Normal

c. Moderately impaired

d. Severely impaired

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16. (d) The status of renal function impairment can be expressed by creatinine clearance.

ClT = ClH + ClR

where,

ClT = Total body clearance

ClH = Hepatic clearance

ClR = Renal clearance

2100 = 300 + ClR

ClR = 1800 ml/hr

= 30 ml/min

The normal creatinine clearance generally lies between 80 to 120 ml/min. A creatinine clearance in patient is 30 ml/min which will be considered severely impaired.

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a. 250 mg

b. 325 mg

c. 991 mg

d. 1221 mg

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17. (c) Blood serum freezes at 0.520 C, and all solutions having this freezing point are isotonic with blood serum. 0.9% sodium chloride have the same freezing point that of blood serum.

FP provides by 1% Boric acid = -0.29o C

FP of blood = - 0.52o C

FP (needed) by NaCl = (0.52-0.29)

= 0.23o C

Now as we know that FP provides by 1% NaCl would be -0.58o C therefore one can say,

For FP 0.58o needs 1 % NaCl

For FP 0.23o needs ?

= (0.23 x 1)/0.58 = 0.396 % NaCl

0.396 gm of NaCl/ 100cc.

The amount of NaCl needed for 250cc,

= (250 x 0.396)/100 = 0.991 gm NaCl = 991 mg

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