Example: Ascertain the percent B in a mix-ture of 600g that contains 15 g of B.Solution: 15/600 = X/100cross multiplyx = (15 x l(X))/600 or X = 1500/600x = 2.5. The parts of active ingredientper hundred parts of total mixtureor 2.5%.RATIO AND PROPORTIONRATIO is the relationship of one quantity toanother quantity of like units. Ratios are indicatedas 5:2, 4:1; these would be read as 5 to 2, 4 to 1.A ratio can exist only between units of thesame 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 consideredsimultaneously.Example: 1:3::3:9Since the ratios are equal, the proportion mayalso be written: 1:3 = 3:9.Terms of ProportionThe first and fourth terms (the terms on theends) are called the “extremes.” The second andthird terms (middle terms) are called the “means.”In a proportion the product of the meansequals the product of the extremes; therefore,when three terms are known, the fourth orunknown term may be determined.Application of ProportionThe important factor when working propor-tions is to put the right values in the right placeswithin the proportion. By following a few basicrules, you can accomplish this without difficultyand 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 thefourth position.2. Let the unit of like value to X be the thirdposition.3. If X will be smaller than the third position,place the smaller of the two leftover valuesin the second position; if X will be larger,place the larger of the two values in the sec-ond position.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 500ml of 70% alcohol to which 150 ml of water havebeen added? When adding 150 ml to 500 ml, thetotal quantity will be 650 ml; consequently, ourfour values will be 500 ml, 650 ml, 70% and X,the unknown percent. When you use the aboverules, the problem will appear as follow: X willbe in the fourth position. Since X will solve aspercent, the unit for like value for the third posi-tion will be the 70 of the original solution. Whenwe 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 150ml of water. Therefore, the smaller of the twofigures (650 and 500) will be placed in the secondposition: 500.650 remains for the first position.The proportion appears as follows:650:500::70:XMultiplying the extremes and the means, we ar-rive at:Consequently, by dividing 650 into 35,000, wewould arrive at: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 isevaporated to 400 ml.What is the percent strength?Letting X be the fourth position, and the unitof like value (15%) the third, we realize that byevaporating the solution it becomes stronger;therefore, the LARGER of the other two values(1000) will occupy the second place and 400 willbe the first position, thus:8-9