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IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 113 | P a g e
Improved and Novel Methods Of Pharmaceutical Calculations
R. Subramanya Cheluvamba Hospital, Karnataka, Mysore-570 001, India.
Abstract It i s a review and r esearch article wherein general equations are derived for some problems and new rules are framed on
percentage calculations, ratio and proportions and on variation of surface area with respect to size of the particle. Percentage
calculations are projected from novel angles. Dimensions or units of quantities are retained up to the result. An easy method
for solution of r atio and proportions are suggested. Problems solved by alligation method in the books are solved by simple
algebraic method with little new rules.
Keywords: axis, decimal, percentage, serially, syntax,
1. Introduction This article is review of books [1] & [2]. section-2, contains the review on basics of percentage calculations. A new general
equation is furnished at the end of this section to ease out solving of such problems. In most of the books same mistake is
committed as in [3] of [1]. Hence, proof of these general equations are given to confirm the mathematical statements.
In section 3, problems solved by dimensional analysis are solved by other methods that are easier than dimensional
analysis.[4]. In the second ratio of set of equivalent ratios of first method, different dimensions are changed into same kind of dimensions i.e., ml, which has made the solution of problem easier and shorter. Similarly second method is also easier
and shorter.
A general equation for the total surface area with respect to the decrease in the particle size is derived. Accordingly, total
surface area is inversely proportional to the size of the particle. Because the equation belongs to magic series, increase in
surface area w.r.t the decrease in size of the particle is a wonder which is shown in example problems.
Section 5 gives a synoptic view of basic measurements.
Section 6 contains new ideas and laws on percentage calculations. Some of the problems are solved by using simple
mathematical operators l ike addition, subtraction, appropriate factor, and by simple division. In most of the books
proportional method is used for solving these problems. In this paper while solving alligation problems new methods are
used. And, algebraic solutions with multiple results are shown.
Section-7, gives a simple equation for conversion of percentage to milligram per ml and vice versa.
In section-8, new method i s furnished to solve the problems on dilution
In section-9, some problems are solved in different and easy methods than in books.
2.Review on Fundamentals of Percentage Calculations:
π. π. πΆπππ£πππ‘ 3
8 π‘π πππππππ‘[3]
Given calculation in the book is 3
8Γ 100 = 37.5%
The above equation does not satisfy the meaning of the symbol β=β (equals) used in between the expressions
3
8Γ 100 and 37.5%.
βΆ π΄ππ‘π’ππ π£πππ’π ππ 3
8Γ 100 = 37.5
Aππ37.5 = 37.5 Γ 100
100
βΆ = 3750%
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 114 | P a g e
The correct method of solution is as following
βΆ3
8= 0.375
βΆ = 0.375 Γ100
100
βΆ =37.5
100
βΆ = 37.5%
2.2. . A 1: 1000 solution has been ordered..You have a 1% solution, a 0.5% solution, a 0.1% solution in stock.
Will one of these work to fill the order. [4]
ππππ’π‘πππ: 1: 1000 =1
1000=
1
10 Γ 100= 0.1%
In comparison with solution of [4]in [1],a few steps are reduced.
2.3. Express 0.02% as ratio strength.
ππππ’π‘πππ: 0.02% =0.02
100=
2100100
=2
10000=
1
5000= 1: 5000
2.4. . Give the decimal fraction and percent equivalent of 1
45
βΆ1
45=
1
45Γ
100
100
βΆ =2.22
100
βΆ = 2.22%
2.5. General Equation:
βΆ π₯% =π₯
100β β β β β β β β β β(1)
ππππππ π₯ ππ π ππππ ππ’ππππ : π»ππππ, π₯ = 100π₯% ---------------- (2)
2.6. Proof of (1) and (2):
πΏππ‘ π ππ π ππππ ππ’ππππ, βΆ ππππ, π = π Γ 1
βΆ π΄ππ π, π = π Γ100
100
βΆ =100π
100
βΆ π = 100π%
βΆ π΄ππ, π% =π
100
:πΉπ’ππ‘πππ, ππ¦ ππππ π ππ’ππ‘πππππππ‘πππ π€π πππ£π 100π% = π
: 100 Γπ
100= π
: πΆπππ πππ’πππ‘ππ¦, π = π
βΆ β
3. Another way for dimensional analysis: 3.1. How many fluidounces (fi.oz) are there in 2.5 liters (L)? [5]
.....Method-1:
: πΊππ£ππ, 1 πππ’ππππ’πππ = 29.57 ππ
βΆ 2.5πΏ
1ππ. ππ§=
2500ππ
29.57 ππ=
2500
29.57= 84.55
Method -2:
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 115 | P a g e
1 ππ. ππ§. = 29.57 ππ
.....Multiplying both sides by the factor2500
29.57 that modifies 29.57 ml to required quantity, we get
βΆ 1ππ. ππ§.Γ2500
29.57= 29.57ππ Γ
2500
29.57
βΆ 84.55ππ. ππ§. = 2.5πΏ
4. General equation to find the total surface area of particles. (Imaginary and ideal case).[6]. 4.1. A general equation is derived to find the total surface area of cubes when a cube is divided serially along the xy, yz and
zx βplanes. When a cube is divided along the planes of axes x, y and z, it gets divided into eight equal cubes at every time i.e.
after every set of cutting. When a cube is cut through x, y and z plane it is called, βa set of cuttingβ.
The sum of the surface area of cubes at βnthβ set of cutting is : π΄ = 2π+1 Γ 3π2 π ππ’πππ π’πππ‘π --(3)
Where, βAβ denotes the sum of surface area of cubes and βnβ the ordinal value of set of cuttings from the first cube. Hence n is
a set of whole number, i.e. π = 0,1,2,3 β¦
; when cut along zx plane
= +
= + ; when cut along yz plane
= + ; when cut along xy plane
Fig.1. A typical view of a set of cuttings of the
cubes across the zx, yz and xy planes.
ππ π‘ππ 1π π‘ππ’ππ, πππππππ ππ’ππππ ππ π ππ‘ ππ ππ’π‘π‘πππ ππ 0, π. π. , n = 0, and if he length is l units
Total surface area of first cube is : π΄ = 2π+1 Γ 3 Γ π2
βΆ = 21 Γ 3 Γ π2
: = 6π2 : And if = 1 ; π΄ = 2π+1 Γ 3π2
βΆ = 22 Γ 3 π2
: = 12 π2
:And if π = 2 ; π΄ = 2π+1 Γ 3 Γ π2
βΆ = 24π2
: And so on.
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 116 | P a g e
4.2. Derivation of equation : π΄ = 2π+1 Γ 3π2 To start with, cube has six faces and if its length is l units the total surface area of the cube, when no cut is done along the
axes is
βΆ π0= one cube Γ number of faces Γ area of one surface :i.e. π0 = 1 Γ 6 Γ π2 To make the above equation look like rest of the equations, it can be written as
βΆ π0 = 80 Γ 6 Γ π
20
2
when this cube is cut across the Co-ordinate planes we get 8 cubes of dimensions equal to π
2 π’πππ‘π each.
βΆ β΄ π1 = 8 Γ 6 Γ π
2
2
: Further, π2 = 82 Γ 6 Γ π
22
2
:Finally, equation for ππ‘π term is
βΆ ππ = 8π Γ 6 Γ π
2π
2
:To simplify this equation, it can be written as
: = 23 π Γ 2 Γ 3 Γ π
2π
2
βΆ πΉπ’ππ‘πππ, ππ = 23π+1 Γ 3π2 2β2π :Replacing ππ by A, and simplifying further we get
:π΄ = 3(2π+1π2)
βΆ = 2π+1 3π2 Thus derived.
4.3. For example when π = 0 Surface area of the cube of length 1 cm is
: 2π+1 3π2ππ2 = 2 Γ 3ππ2
βΆ = 6ππ2 This is equal to the area of 2ππ Γ 3 ππ rectangle.
Further π€πππ π = 5 ( 5 set of cuts) and π = 1ππ, βΆ ππππ, 2π+1 3π2 = 26 Γ 3
βΆ = 192 ππ2 This area is equal to the area of 12 ππ Γ 16 ππ rectangle.
And when π = 20 πππ π = 1 ππ : 2π+1 3π2 = 221 Γ 3
: = 6291456ππ2 This is approximately equal to the area of 2.5π Γ 2.51π ππππππ πππ.
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 117 | P a g e
5. MEASURMENT Table-1:.Measurement Of Length, Mass And Volume (SI Units)
Name of the unit Expressed in terms
of m/L/g
Values expressed in
scientific notation
Fractional notation in
m/L/g
1 Kilo-meter/liter/gram. (km/kL/kg) 1000 m/L/g 103 m/L/g
1 hecto-meter/liter/gram. (hm/hL/hg) 100 m/L/g 102 m/L/g
1 deka-meter/liter/gram. dam/daL/dag 10 m/L/g 101 m/L/g
1 meter/liter/gram. m/L/g 1 m/L/g 100 m/L/g
1 deci-meter/liter/gram. dm/dL/dg 0.1m/L/g 10β1 m/L/g 1
10
π‘π
π/πΏ/π
1 centi-meter/liter/gram. cm/cL/g 0.01 m/L/g 10β2 π/πΏ/π 1
100
π‘π
π/πΏ/π
1 milli-meter/liter/gram (mm/mL/mg) 0.001 m/L/g 10β3 π/πΏ/π 1
1000
π‘π
π/πΏ/π
1 micro-meter/liter/gram
(πππ/πππΏ/πππ) 0.000001 m/L/g 10β6 m/L/g
1
1000000
π‘π
m/L/g
1 nano-meter/liter/gram. (nm/nL/ng) 0.000000001
m/L/g 10β9 m/L/g
1
1000000000
π‘π
m/L/g
Note: all measurements are basically compared with m/L/g in this table
5.2. Add 0.5 kg, 50 mg, 2.5 dg, reduce the result to grams.[7]
: =0.5kg+50mg+2.5dg
βΆ = 0.5 Γ 1 ππ + 50 Γ 1ππ + 2.5 Γ (1ππ) -----------(4)
Conversion factors are
: 1kg=1000g, 1mg=1
1000π, 1dg=
1
10g
: Substituting the above values in (4), we get
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 118 | P a g e
: = 0.5 Γ 1000 π + 50 Γ 1
1000π + 2.5 Γ (
1
10π)
βΆ = 0.5 Γ 1000 π + 50 Γ 1
1000π + 2.5 Γ (
1
10π)
βΆ = 500π + 0.05π + 0.25 π
βΆ = 500.3 π
6. SYNTAX OF PERCENTAGE (%) AND PARTS PER n (PPn)CALCULATIONS Percent means βfor every hundredβ. Symbolically it is written as %. And mathematically it is written as the ratio
π₯ 100 ππ π₯
100 . The bar between two numbers represents the phrase βfor everyβ. It can also be read as βParts per centβ the
word cent is Latin derivative which means, hundred. And this βParts per centβ can be abbreviated as PPc. I suggest to
write this as PPh which is the abbreviation of βParts Per Hundredβ. This kind of abbreviation helps to extend the same
idea to any quantity. For example PPt for parts per ten,. i.e., PPh for βParts Per Hundred, PPth, parts per thousand, PPtth
βParts Per Ten Thousandβ, PPhth βParts Per Hundred Thousandβ, Parts Per Million βPPmβand so on. Alpha numerically it
can also be written as ππ100, PP10, ππ102 , ππ103 β―ππ10π . Usually, comparison is done in multiple of 10.
Although general expression Parts Per n can be written as πππ, π€ππππ π ππ π πππ ππ‘ππ£π πππ‘ππππ.
πΌπ π₯ ππ π‘ππ ππ’ππππ ππ ππππ‘π π‘πππ πππππππ ππ₯ππππ π πππ ππ βΆ π₯πππ.
Mathematically xPPn can be written as following in the fractional form
The staked fractional notation of xPPn is π₯
π
The skewed fractional notation is π₯ π
The linear fractional notation is π₯ π
And the decimal notation of π₯
π ππ π , πππ ππ‘ ππ π ππππ ππ’ππππ
π΄ππ π, π¦ π’πππ‘π ππ π₯πππ =π¦
π₯π
π¦
πππ‘, π§ = π¦ π₯
π ,
ππππ, π¦ π’πππ‘π ππ π₯πππ =π§
π¦
And henceforth [y](xPPn) stands for "y units of xPPn"
π΄ππππππππππ¦, y xPPn =z
yβ β β β β β(5)
Here the numerator indicates active ingredient and the denominator indicates solution or drug.
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 119 | P a g e
Table-2: x Parts Per β10π β
π₯
10π Verbal form of the
expression Abbreviation Alphanumeric
notation
Prevailing
notation
Decimal
form
π₯
10 x Parts per ten x PPt x PP10 - 0. π₯
π₯
102
x Parts per hundred
Or
Parts per cent
Or
PERCENT
x PPh
or
PPc
x PP102 π₯% 0.0π₯
π₯
103 x Parts per
Thousand
x PPth x PP103 π₯%0 0.00π₯
π₯
104 x Parts per Ten
thousand
x PPtth x PP104 - 0.000π₯
π₯
105 x Parts per hundred
thousand
x PPhth x PP105 - 0.0000π₯
π₯
106 x Parts per million x PPm x PP106 - 0.00000π₯
In addition to this, some more basic factors are intertwined in nature, with the x PPn. Those are volume, mass and weight.
π₯ π πππ¦ ππ π’πππππ π‘πππ ππ π‘ππππ ππ : π₯ π π£ π£ ππ π₯ π π€ π€ ππ π₯ π π£ π€ ππ π₯ π π€ π£ . πΌπ π‘ππ π£πππ’ππ ππππ πππ‘ ππππππ πππ‘ππ πππ₯πππ πππ ππ ππ πππ π πππ π‘ππ π’πππ‘π ππ π£πππ’ππ πππ πππ π ,
ππππ,π₯
π
π£
π£ =
π₯
(π¦+π₯), πππππ’π π π = (π¦ + π₯)
πππππππππ¦ , π₯ π π€ π€ =π₯ π
π¦+π₯ π , π₯ π π£ π€ =
π₯ ππ
π¦+π€π‘ .ππ π₯ ππ π
Only in case of, π₯ π π€ π£ =π₯ π
π ππ , here, volume of the active ingredient is not considered because
increase in volume of solution after dissolution of active ingredient will be negligible.
Example:. Peppermint spirit contains 10 % v/v peppermint oil. What is the volume of the solvent when volume of spirit is
75 ml ?[8]
:10%(π£ π£) =10 ππ
100 ππ
Factor that is required to make the numerator to 75 is 75
10, multiplying both the numerator and denominator by this factor
we get
βΆ 10 ππ Γ
7510
100ππ Γ7510
=75 ππ
750 ππ
βΆ =75 ππ
(675 + 75)ππ
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 120 | P a g e
This shows that volume of the solvent of 10% peppermint sprit is 675 ml.
6.1. Summing of x (PPn)
Case-1
If π₯1πππ1 , π₯2πππ1 , π₯3πππ1 β― , π₯ππππ1 πππ π‘ππ π πππ’π‘ππππ ππ πππππππππ‘ π π‘πππππ‘ππ πππ ππ π πππ πππ. ( π»πππ π πππ π πππ πππ‘π’πππ ππ’πππππ )
βΆ π‘πππ,π₯1
π1
+π₯2
π1
+π₯3
π1
+ β― +π₯π
π1
=π₯1 + π₯2 + π₯3 + β― + π₯π
ππ1
β β β β β β(6)
βΆ π‘π π πππππππ¦ 6 πππ‘, π₯1 + π₯2 + π₯3 + β― + π₯π = π¦
βΆ ππππ 6 , = π¦
π
π1β β β β β β(7)
βΆ π΄ππ ππ, π¦
π = π§
βΆ ππππ (6) =π§
π1
βΆ= π§ π1 π€
π€ππ
π£
π£ππ
π€
π£ππ
π£
π€
βΆ= π§πππ1 π€
π€ππ
π£
π£ππ
π€
π£ππ
π£
π€
πΌπ π1 = 100, π‘ππ πππ π’ππ‘ π€πππ ππ = π§%
Example: What is the percentage strength of the solution when equal quantities of 10%, 20%, 30% solutions are mixed
together.
π»πππ, π₯1 = 10, π₯2 = 20, π₯3 = 30, πππ π1πππ ππ = 100, π‘πππ π’π πππ 6 , π€π πππ£π
βΆ10
100+
20
100+
30
100=
10 + 20 + 30
300
βΆ =20
100
βΆ = 20%
Case-2: To find the strength of the solution to a specified base when different strength, different quantity, different
PPn solutions are mixed together.
πΏππ‘ π πππ1 , π πππ2 , π πππ3 , β― , π ππππ ππ π‘ππ π πππ’π‘ππππ π‘π ππ πππ₯ππ,
πππ πππ‘ ππ¦ ππ π‘ππ π ππππππππ πππ π π. π. , ππππ¦ πππ πππ‘ π§ππππ¦ ππ π‘ππ πππππ π πππ’π‘πππ,
βΆ π‘πππ, π π’π ππ π‘πππ ππ = π π₯1
π1
+ π π₯2
π2
+ π π₯3
π3
+ β― + π π₯π
ππ
βΆ =π
π₯1
π1
π+
π π₯2
π2
π+
π π₯3
π3
π+ β― +
π π₯π
ππ
π
βΆ =
ππ₯1
π1+
ππ₯2
π2+
ππ₯3
π3+ β― +
ππ₯π
ππ
π + π + π + β― + π β β β β β β 8
The required and specified base of strength is ππ¦ , then (8) will be,
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 121 | P a g e
βΆ =
ππ¦ ππ₯1π1
+ππ₯2π2
+ππ₯3π3
+ β― +ππ₯π
ππ
π + π + π + β― + π
ππ¦
β β β β β β 9
πΏππ‘,ππ¦
ππ₯1
π1+
ππ₯2
π2+
ππ₯3
π3+ β― +
ππ₯π
ππ
(π + π + π + β― + π)= π§, π‘πππ 9 π€πππ ππ
βΆ=π§
ππ¦
βΆ ππππ, 9 πππ ππ π€πππ‘π‘ππ ππ , π§ππππ¦ , ππ ππ¦ = 100, π‘ππ πππ π’ππ‘ π€πππ ππ π§%
Exmple-1:What is the percentage strength (v/v) of alcohol in a mixture of 3000 ml of 40 % v/v alcohol, 1000 ml of 60
% v/v alcohol, and 1000 ml of 70% v/v alcohol? Assume no contraction of volume after mixing.[9]
: π»πππ ππ¦ = 100 πππ π₯1 = 40, π₯2 = 60, π₯3 = 70, π΄ππ π, π = 3000, π = 1000, π = 1000
πππ π1 = π2 = π3 = 100, ππ’π‘π‘πππ π‘πππ π π£πππ’ππ ππ 9 π€π πππ‘
βΆ πππππππ‘πππ π π‘πππππ‘π =
3000 Γ 40 + 1000 Γ 60 + 1000 Γ 70 3000 + 1000 + 1000
100
Note: since π1 = π2 = π3 = 100, πππ ππ¦ = 100 , π‘πππ¦ πππ‘ πππππππππ ππ¦ πππ ππππ‘πππ.
βΆ=50
100
βΆ= 50%
Example-2: When the same problem is solved for ππ¦ = 1000, i.e., for π₯
1000 ππ πππ‘π
: π₯πππ‘π =
1000 3000 Γ 40 + 1000 Γ 60 + 1000 Γ 70
100
3000 + 1000 + 1000
1000
: =500
1000
Example-3: Same problem, when π1 = 1000, π2 = 2000, π3 = 5000 πππ ππ¦ = 100
: π₯πππ =
100 3000 Γ 40 Γ 10 + 1000 Γ 60 Γ 5 + 1000 Γ 70 Γ 2
10000
5000100
: = 3.28% This problem can also be solved by another method as following
: 3000 ππ ππ 40
1000 π π‘πππππ‘π π πππ’π‘πππ =
3000 40
1000
3000=
120
3000
: 1000 ππ ππ 60
2000 π π‘πππππ‘π π πππ’π‘πππ =
1000 60
2000
1000=
30
1000
: 1000 ππ ππ 70
5000 π π‘πππππ‘π π πππ’π‘πππ =
1000 70
5000
1000=
14
1000
Sum of these solutions are 120
3000+
30
1000+
14
1000
: =164
5000
: =164
50 Γ 100
: = 3.28%
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 122 | P a g e
Example-4: A pharmacist adds 10 ml of a 20 % (w/v) solution of a drug to 500 ml of D5W for parenteral infusion.
What is the percentage strength of the drug in the infusion solution?[10]
π·ππ‘π: π₯1 = 20 , π₯2 = 0, π ππππ π‘ππ πππ‘ππ£π ππππππππππ‘ πππ₯π‘πππ π ππ π·5π ππ πππ‘ ππππ ππππππ ππ π‘πππ πππ π,
π = 10, π = 500 πππ ππ¦ = 100, π1 = π2 = 100
π₯πππ =
100 10 Γ 20
100+
500 Γ 0100
10 + 500100
=0.392
100
= 0.392%
This is also solvable in another way as following:
10 ml of 20% solution is added to 500 ml of D5W.Here D5W is a dummy drug with respect to this problem. Hence
this statement can be written mathematically as following.
βΆ π₯πππ(π€ π£) =0
500 ππ π·5π +
20%
10ππ
To convert 20% into the quantity of active ingredient it is multiplied by 10 which is the quantity added.
βΆ=0
500 ππ π·5π +
20100 Γ 10
10ππ
=0
500 ππ π·5π +
2 π
10ππ
By direct addition, we get.
=2π
510 ππ π·5π
=2π
5.10 Γ 100ππ π·5π
=0.39 π
100ππ π·5π
= 0.39% π€ π£
Example-5:Prepare 250 ml of dextrose 7.5 % w/v using dextrose 5% (D5W) w/v and dextrose 50 % (D50W) w/v.
How many milliliters of each will be needed?[11]
Method-1: using (9)
Data: π₯ = 7.5, π = 100, π₯1 = 5, π₯2 = 50, π1 = 100, π2 = 100. ππ¦ = 100.
βΆ 7.5
100=
100 5π
100 +
50π100
π + π100
βΆ 7.5
100=
5π + 50π
100 π + π
IOSR Journal of Pharmacy
Vol. 2, Issue 1, Jan-Feb.2012, pp. 113-129
ISSN: 2250-3013 www.iosrphr.org 123 | P a g e
βΆ 500π + 5000π = 750π + 750π
βΆ 250π = 4250π
βΆ π
π=
17
1
βΆ π = 17 πππ π = 1
Check: Putting values of b and c in (9) we get.
βΆ =
100 5 Γ 17
100 +
50 Γ 1100
17 + 1100
; =7.5
100
βΆ = 7.5%
Now, to prepare 250 ml, 250
1+17 = 13.89 ππ ππ 50% π·ππ₯π‘ππ π πππ 250 β 13.89 = 236.11 ππ ππ 5% π·ππ₯π‘πππ π
are required
Example-6: In what proportion should alcohols of 95% and 50% strengths be mixed to make 70% alcohol?[12]
Method-1:This can be solve by using (9) as usual.
Method-2: Here determinant is used to solve the problem.
πΏππ‘ π₯ πππ π¦ ππ π‘ππ ππππ‘πππ π‘πππ‘ ππππππ¦ π‘ππ πππ£ππ πππ‘πππ ππ πππππππ‘πππ π‘π 70
100,
π‘πππ πππ ππ π€πππ‘π‘ππ ππ ,95π₯
100π₯+
50π¦
100π¦=
70
100
ππππ, 95π₯ + 50π¦ = 70 β β β β β β 10
π΄ππ, 100π₯ + 100π¦ = 100
π. π. , π₯ + π¦ = 1 β β β β β β(11)
ππ πππ 10 & 11 π‘ππ πππ‘πππππππ‘ β πππ ππ π€πππ‘π‘ππ ππ ππππππ€πππ,
βΆ β= 95 501 1
: =45
πΏππ‘ βπ₯ & βπ¦ ππ π‘ππ πππππππππ πππ‘ππππππππ‘π ππππππ π€. π. π‘ π π»π,
βΆ ππππ, βπ₯= 70 501 1
βΆ = 20
: π΄ππ, βπ¦ = 95 701 1
: = 25
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βΆ πππ€, π₯ =βπ₯
β πππ π¦ =
βπ¦
β
βΆ π. π. , π₯ =20
45 πππ π¦ =
25
45
βΆ π₯ =4
9 πππ π¦ =
5
9
βΆ π₯
π¦=
4
5
βΆ ππππ, π₯ = 4 πππ π¦ = 5
Example-7: A hospital pharmacist wants to use lots of zinc oxide ointment containing, respectively, 50 %,
20%, and 5% of zinc oxide. In what proportion should they be mixed to prepare a 10 % zinc oxide
ointment?[13]
Method-1:
Data: ππ¦ = 100, π₯1 = 50, π₯2 = 20, π₯3 = 5, π1 = π2 = π3 = 100, πππ ππππ’ππππ π π‘πππππ‘π ππ 10%
Using (9), we have,
βΆ 10
100=
100 50π100 +
20π100 +
5π100
π + π + π
100
βΆ 10
100=
50π + 20π + 5π
100π + 100π + 100π
βΆ 1000π + 1000π + 1000π = 5000π + 2000π + 500π
βΆ 1π = 8π + 2π
βΆ ππππ, π = 1, ππππ£π πππ’ππ‘πππ π€πππ ππ, 8π + 2π = 1
βΆ πππ€, ππ¦ π‘ππππ πππ πππππ πππ‘πππ π€π πππ ππππ π =1
16 πππ π =
1
4
ππππ, π‘ππ πππππππ‘πππ ππ π: π: π =1
16:1
4: 1
By multiplying RHS of the above equation by 16, we have,
βΆ = 1: 4: 16
Assigning different values to d we can get different proportions. For example when d=10, consequently b=1 and c=1
which is the result obtained in [14]. Also, we can get the result by giving convenient values to b and c such that it
satisfies the equation.
Method-2:
Given, 50 %, 20 %, and 5% zinc oxide ointment and the required percentage is 10 % . Keeping 50 % and 20
% ointments as fixed. And considering 5 % ointment as the variable.
Then, let the variable base (solvent)π‘ = 100
Then, 5 = 0.05 π‘
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βΆ π. π, 50% + 20% + 0.05π‘ % = 10%
βΆ 50
100+
20
100+
0.05π‘
π‘=
10
100β β β β(12)
βΆ 70 + 0.05π‘
200 + π‘=
10
100
βΆ 100 70 + 0.05π‘ = 10(200 + π‘)
βΆ 7000 + 5π‘ = 2000 + 10π‘
βΆ π‘ = 1000
βΆ πΆπππ πππ’πππ‘ππ¦ 12 ππ ,50
100+
20
100+
50
1000=
10
100
βΆ πππππππ‘πππ ππ πππππππππ‘ππ ππ πΏπ»π ππ 100: 100: 1000
βΆ π€ππππ ππ πππ’ππ π‘π 1: 1: 10
Without using (9) answer can be found directly as following
Let all the three percentages be variables. And let π‘1 , π‘2 , π‘3 ππ π‘ππ π£πππππππ ππ’πππ‘ππ‘πππ
βΆ ππππ,0.5π‘1
π‘1
+0.2π‘2
π‘2
+0.05π‘3
π‘3
=10
100
βΆ 50π‘1 + 20π‘2 + 5π‘3 = 10π‘1 + 10π‘2 + 10π‘3
βΆ 8π‘1 + 2π‘2 = π‘3
As usual answers can be found by giving different values to π‘3 ππ π‘π π‘1& π‘2
Note: Since problem has three unknown and only two rows, square matrix cannot be formed. Hence it is not possible
to draw solution by determinant method. Although, by modifying some of the rules of determinants or matrix, answer can be obtained.
Example-8: If 50 ml of a 1:20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/v)?[14]
Method-1: Using (9).
Here two ratio strength solutions are found that are 1) 1:20 solution and 2) the diluent which can be written as 0:20.
Here base of the diluent is arbitrary hence convenient base is chosen in this case. Also, ππ¦ is also a random number which
can be chosen according to our convenience.
π·ππ‘π: π₯1 = 1, π₯2 = 0 , π = 50, π = 950, ππ¦ = 1000
βΆ π ππ‘ππ π π‘πππππ‘π =
1000 50 Γ 1
20 +950 Γ 0
20
50 + 9501000
βΆ =2.5
1000
βΆ =1
400
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In this way problems on percentage and PPn of different parameters can be solved by using (9).
7. Conversion of percentage to mg/ml and vice versa:
πΏππ‘ π₯% π ππ ππ π‘ππ πππππππ‘πππ π‘π ππ ππππ£πππ‘ππ π‘πππ
ππ, π‘πππ
: π₯% π ππ =π₯π
100ππ
: =1000π₯ ππ
100 ππ
: =10 π₯ ππ
1 ππ
βΆ π₯% π ππ = 10 π₯ ππ ππ β β β β β β( 10)
π·ππ£πππππ πππ‘π π ππππ ππ¦ 10 π€π πππ‘,
: π₯
10% = π₯ ππ ππ β β β β β β(11)
Example: A certain injectable contains 2mg of a drug per milliliter of solution. What is the ratio Strength (w/v) of
the solution?[15]
Using (11), we have.
: 2 mg /ml = 2
10%
βΆ = 0.2% =2
1000= 1: 500
8. MEANING OF 0% AND 0 PPn 0 = 0πππ Proof:-
πΊππ£ππ, 0
ππ’ππ‘ππππ¦ ππ‘ ππ¦ ππππ
βΆ = 0 π
π
βΆ =0 Γ π
π
βΆ =0
π
βΆ = 0πππ
βΆ π·ππππππ πππ‘ππ‘πππ ππ π‘πππ ππ = 0
Thus proved
0 PPn or 0 π ππ 0
π is used to denote pure base, solvent or vehicle.
Example-1:
βΆ 0 = 0 Γ 100
100
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βΆ =0
100
βΆ = 0%
π·ππππππ πππ‘ππ‘πππ ππ 0% β’ ππ 0
Example-2: If 50 ml of a 1:20 w/v solution are diluted to 1000 ml, what is the ratio strength (w/v)?
50 ml of 1: 20 strength solution is diluted to 1000 ml. This statement can mathematically be written as,
βΆ 50
120
50+
0
950
Adding numerator and denominator directly (6), we have
βΆ =2.5
1000
βΆ =1
400
8.1. MEANING OF πΏ π πΆπΉ πΏ
π
This kind of notation may leads to confusion. Addition of ingredient is in one way simple and in another way
complex because of factors like solubility and saturation point of a solution. Hence this kind of notation is not necessary in pharmacy.
This kind of notation can be used to indicate that the drug is not in the form of solution or in any form of mixture.
And it may also imply that the solvent is completely removed.
8.2. Removal of solvent by any means. (Usually by evaporation)
Example: If a syrup containing 65% w/v of sucrose is evaporated to 85% of its volume, what percentage (w/v) of
sucrose will it contain?[16]
Method -1:
βΆ 65% π€ π£ =65 π
100ππ
This syrup is evaporated to 85% of its volume means, 15% of solvent is evaporated or 15 ml of it is evaporated if
volume of solution is 100ml.
βΆ π . π . ,65π
100ππβ
0
15ππ
By direct subtraction of numerator and denominator we have the above expression equal to
βΆ =65π
85ππ
βΆ πΉπ’ππ‘ πππ , π‘π ππππ π‘ ππ ππππ πππ‘πππ ,65π Γ
10085
85ππ Γ10085
=76.47π
100ππ= 76.47%
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Method -2:
βΆ65
100 Γ 85%
βΆ65
85
To find the percentage strength
βΆ 65 Γ
10085
85 Γ10085
βΆ 76.47
100
βΆ = 76.47%
9. Solution to Miscellaneous problems. Problem No.-1:In acute hypersensitivity reactions, 0.5mL of a 1:1000 (w/v) solution of epinephrine may be
administered subcutaneously or intramuscularly, calculate the milligrams of epinephrine given.[17]
Solution:
: 1
1000 π€ π£ =
1π
1000 ππ=
1000 ππ
1000 ππ=
0.5 ππ
0.5ππ
Hence 0.5 mg of epinephrine is given
ProblemNo.2: How many grams of 10 % w/w ammonia solution can be made from 1800g of 28% strong
ammonia solution?[17]
Solution: πΏππ‘ π ππ π‘ ππ ππππ’ππππ π£πππ’ππ ππ 10% πππππππ π πππ’π‘πππ , π‘ π ππ
βΆ π 10
100 = 1800
28
100
βΆ π = 5040
Problem.No.3: How many grams of a substance should be added to 240 ml of water to make a 4 % (w/w)
solution?[18]
Solution:
βΆ 4%(π€ π€) =4
96 + 4
To make 96 equal to 240 multiply all the elements of RHS by the factor 240
96.
βΆ =4
24096
96 24096
+ 4 24096
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βΆ =10π
240π + 10π
Conclusion: Aim of this article is to give more scientific and algebraic touch to the pharmaceutical calculations. Also, an effort is made to
make the steps of the solutions more pragmatic and easily understandable. And, an attempt is made to draw solutions from
basic datum or from data. A good r elation is maintained between every steps of the solutions.
References
Books: [1] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010)
[2] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA, 2007)
PageNo. and Section No. [3] Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.3
[4] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA,
2007),sec.2.2.3
[5] Howard C. Ansel,, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.9
[6] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.22 [7] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.30, Pr. No. 1
[8] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.85
[9] Howard C. Ansel, Pharmaceutical Calculations (Wolters Kluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.264
[10] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.95, pr.No.20
[11] Don A. Ballington, Tova Wiegand Green, Pharmacy Calculations (EMC Corporation, USA,
2007),sec.8.3.1
[12] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.265
[13] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.266
[14] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.253
[15] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p. 90
[16] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),p.253
[17] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins
530 Walnut Street,Philadelphia Pa 19106, 2010),98. pr.No.65
[18] Howard C. Ansel, Pharmaceutical Calculations (WoltersKluwer Health | Lippincott Williams & Wilkins 530 Walnut Street,Philadelphia Pa 19106, 2010),p.86