Proportion is two equal ratios consideredsimultaneously. An example proportion is1:3::3:9This proportion is expressed as “1 is to 3 as 3 is to 9.” Sincethe ratios are equal, the proportion may also be written1:3 = 3:9Terms of ProportionThe 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 theproduct of the extremes; therefore, when three terms areknown, the fourth (or unknown) term may be determined.Application of ProportionThe important factor when working proportions isto put the right values in the right places within theproportion. By following a few basic rules, you canaccomplish this without difficulty and solve theproblem correctly.In numbering the four positions of a proportionfrom left to right (i.e., first, second, third, and fourth,observe the following rules):Let X (the unknown value) always be in thefourth position.Let the unit of like value to X be the thirdposition.If X is smaller than the third position, place thesmaller of the two leftover values in the secondposition; if X is larger, place the larger of the twovalues in the second position.Place the last value in the first position. Whenthe proportion is correctly placed, multiply theextremes and the means and determine the valueof X, the unknown quantity.6-15When we add water to a solution, the strength isdiluted; consequently, the 70% strength of thissolution will be lessened when we add the extra150 ml of water. Therefore, of the two remaininggiven 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 firstposition:2^{nd}position: 500 ml1^{st}position: 650 mlThe proportion appears as follows:650 : 500 :: 70 : XMultiplying the extremes and the means, we arriveat:650X = 35,000, orX = 53.8When 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 of70% alcohol to which 150 ml of water has been added?Solution: When adding 150 ml to 500 ml, the totalquantity will be 650 ml; consequently, our four valueswill be 500 ml, 650 ml, 70%, and X (the unknownpercent). Following the rules stated above, the problemwill appear as follows:4^{th}position: X (%)3^{rd}position: 70% (like value to X)Example #2: When 1000 ml of 25% solution isevaporated to 400 ml, what is the percent strength?Solution:4^{th}position: X(%)3^{rd}position: 15% (like value to X)When we evaporate a solution, it becomes stronger.Therefore, the larger of the two remaining givenvalues (1000 ml and 400 ml), will be placed in thesecond position, leaving the quantity 400 ml to beplaced in the first position:2^{nd}position: 1000 ml1^{st}position: 400 mlThe proportion appears as follows:400 : 1000 :: 15 : XMultiplying the extremes and the means, we arriveat:400X = 25,000, orX = 62.5When 1000 ml of water is evaporated to 400 ml, theresult is a 62.5% solution.

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