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Unformatted text preview: problems with which you should be familiar. The first is sometimes referred to as dry mixture, in which we mix dry ingredients of different values, such as nuts or coffee. Also solved by the same method are problems dealing with tickets at different prices, and similar problems. In solving this type of problem it is best to organize the data in a chart with three rows and columns, labeled as illustrated in the following example. Example: Mr. Sweet wishes to mix candy worth 36 cents a pound with candy worth 52 cents a pound to make 300 pounds of a mixture worth 40 cents a pound. How many pounds of the more expensive candy should he use? Solution: More expensive Less expensive Mixture No. of pounds x 300 – x 300 · Price per pound = 52 36 40 Total value 52x 36(300 – x) 12000 The value of the more expensive candy plus the value of the less expensive candy must be equal to the value of the mixture. Almost all mixture problems derive their equation from adding the final column in the chart. 52x + 36(300 – x) = 12000 Notice that all values were computed in cents to avoid decimals.
52 x + 10, 800 − 36 x = 12, 000 16 x = 1200 x = 75 He should use 75 pounds of the more expensive candy. In solving the second type of mixture problem, we are dealing with percents instead of prices and amounts of a certain ingredient instead of values. As we did with prices, we may omit the decimal point from the percents, as long as we do it in every line of the chart. Example: How many quarts of pure alcohol must be added to 15 quarts of a solution that is 40% alcohol to strengthen it to a solution that is 50% alcohol? Solution: Diluted Pure Mixture No. of quarts 15 x 15 + x · Percent Alcohol = 40 100 50 Amount of Alcohol 600 100x 50(15 + x) Notice that the percent of alcohol in pure alcohol is 100. If we had added pure water to weaken the solution, the percent of alcohol in pure water would have been 0. Again, the equation comes from adding the final column since the amount of alcohol in t...
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- Spring '10