Example: Ascertain the percent B in a mixture of 600g that contains 15 g of B.
Solution: 15/600 = X/100
cross multiply
x = (15 x l(X))/600 or X = 1500/600
x = 2.5. The parts of active ingredient per hundred parts of total mixture or 2.5%.
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 simultaneously.
Example: 1:3::3:9
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 proportions 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 proportion 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 fourth position.
2. Let the unit of like value to X be the third position.
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 second position.
4. Place the last value in the first position.
When the proportion is correctly placed, multiply the extremes and the means and determine 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 position 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 solution 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 arrive at:
650 X = 35,000
Consequently, by dividing 650 into 35,000, we would arrive at:
X = 53.8
When 150 ml of water are added to 500 ml of 70% alcohol, we would then have 650 ml of 53.8% solution.
Example: 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:
400 : 1000 :: 25 : X
