Example: Ascertain the percent B in a mix-
ture of 600g that contains 15 g of B.
Solution: 15/600 = X/100
x = (15 x l(X))/600 or X = 1500/600
x = 2.5. The parts of active ingredient
per hundred parts of total mixture
RATIO AND PROPORTION
RATIO is the relationship of one quantity to
another quantity of like units. Ratios are indicated
as 5:2, 4:1; these would be read as 5 to 2, 4 to 1.
A ratio can exist only between units of the
same kind, as the ratio of percent to percent,
grams to grams, dollars to dollars; in other words,
the denominates must be constant.
PROPORTION is two equal ratios considered
Since the ratios are equal, the proportion may
also be written: 1:3 = 3:9.
Terms of Proportion
The first and fourth terms (the terms on the
ends) are called the extremes. The second and
third terms (middle terms) are called the means.
In a proportion the product of the means
equals the product of the extremes; therefore,
when three terms are known, the fourth or
unknown term may be determined.
Application of Proportion
The important factor when working propor-
tions is to put the right values in the right places
within the proportion. By following a few basic
rules, you can accomplish this without difficulty
and solve the problem correctly.
In numbering the four positions of a propor-
tion from left to right, i.e., first, second, third,
and fourth, observe the following rules:
1. Let X (the unknown value) always be in the
2. Let the unit of like value to X be the third
3. If X will be smaller than the third position,
place the smaller of the two leftover values
in the second position; if X will be larger,
place the larger of the two values in the sec-
4. Place the last value in the first position.
When the proportion is correctly placed,
multiply the extremes and the means and deter-
mine the value of X, the unknown quantity.
Example: What is the percent strength of 500
ml of 70% alcohol to which 150 ml of water have
been added? When adding 150 ml to 500 ml, the
total quantity will be 650 ml; consequently, our
four values will be 500 ml, 650 ml, 70% and X,
the unknown percent. When you use the above
rules, the problem will appear as follow: X will
be in the fourth position. Since X will solve as
percent, the unit for like value for the third posi-
tion will be the 70 of the original solution. When
we add water to a solution, the strength is diluted;
consequently, the 70 percent strength of this solu-
tion will be lessened when we add the extra 150
ml of water. Therefore, the smaller of the two
figures (650 and 500) will be placed in the second
position: 500.650 remains for the first position.
The proportion appears as follows:
650: 500:: 70:X
Multiplying the extremes and the means, we ar-
Consequently, by dividing 650 into 35,000, we
would arrive at:
When 150 ml of water are added to 500 ml of 70%
alcohol, we would then have 650 ml of 53.8%
1000 ml of 25% solution is
evaporated to 400 ml.
What is the percent strength?
Letting X be the fourth position, and the unit
of like value (15%) the third, we realize that by
evaporating the solution it becomes stronger;
therefore, the LARGER of the other two values
(1000) will occupy the second place and 400 will
be the first position, thus: