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Page Title: Application of Proportion
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ALTERNATE METHODS FOR SOLVING PERCENTAGE PROBLEMS
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PHARMACEUTICAL PREPARATIONS
Proportion   is   two   equal   ratios   considered simultaneously.  An example proportion is 1:3::3:9 This proportion is expressed as “1 is to 3 as 3 is to 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): ·   Let  X  (the  unknown  value)  always  be  in  the fourth position. ·   Let  the  unit  of  like  value  to  X  be  the  third position. ·   If X is smaller than the third position, place the smaller of the two leftover values in the second position; if X is larger, place the larger of the two values in the second position. ·   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. 6-15 When we add water to a solution, the strength is diluted; consequently, the 70% strength of this solution will be lessened when we add the extra 150 ml of water.   Therefore, of the two remaining given quantities (650 ml and 500 ml), the smaller (500 ml) will be placed in the second position, leaving the quantity 650 ml to be placed in the first position: 2nd position: 500 ml 1st position: 650 ml The proportion appears as follows: 650 : 500 :: 70 : X Multiplying the extremes and the means, we arrive at: 650X = 35,000,    or X = 53.8 When 150 ml of water is added to 500 ml of 70% alcohol, the result is 650 ml of 53.8% solution. Example #1: What is the percent strength of 500 ml of 70% alcohol to which 150 ml of water has been added? Solution: 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).   Following the rules stated above, the problem will appear as follows: 4th position: X (%) 3rd position: 70% (like value to X) Example #2: When 1000 ml of 25% solution is evaporated to 400 ml, what is the percent strength? Solution: 4th position: X(%) 3rd position: 15%   (like value to X) When we evaporate a solution, it becomes stronger. Therefore, the larger of the two remaining given values (1000 ml and 400 ml), will be placed in the second position, leaving the quantity 400 ml to be placed in the first position: 2nd position: 1000 ml 1st position: 400 ml The proportion appears as follows: 400 : 1000 :: 15 : X Multiplying the extremes and the means, we arrive at: 400X = 25,000,    or X = 62.5 When 1000 ml of water is evaporated to 400 ml, the result is a 62.5% solution.

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