Applied Problem 1

Applied Problem 1 - 100 % x L 100/1 L 100 x 30 % 6 L 30/1 L...

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Rebekah Unruh Honors Algebra 9-13-07 I. Problem Description How much pure acid should be added to 6 liters of 30 % acid to increase the concentration to 50 % acid? II. Math Concepts For this problem, the distance formula, “distance equals rate multiplied by time,” or d = rt can be used. This formula will then be used to create a linear equation to solve the problem. III. Construction of the Mathematical Model The unknown in the problem is the amount x (in liters) of pure acid. From the problem description we know that x must be added to 6 liters of 30 % acid, so x + 6 is the amount (in liters) of 50 % acid that will be created. By creating a table with the amounts in liters and the amounts of acid, using the distance formula, values can be calculated to create the linear equation for the problem. Strength t (liters of acid) r (parts per liter) d (parts of acid)
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Unformatted text preview: 100 % x L 100/1 L 100 x 30 % 6 L 30/1 L 180 50 % x + 6 L 50/1 L 50 ( x + 6) Since the pure acid added to the 30 % acid must equal the 50 % acid, the governing equation for the problem is: 100x + 180 = 50(x + 6) 1 IV. Solution of the Mathematical Problem Using the distributive property on the above equation gives 100x + 180 = 300 + 50x Subtract 50x from both sides of the equation 50x + 180 = 300 Subtract 180 from both sides of the equation 50x = 120 Divide both sides of the equation by 50 to solve for x x = 2.4 For the amount of 50 % acid, adding 6 to 2.4 gives 8.4. V. Interpretation of the Mathematical Results In the problem, x was the amount of pure acid needed to add to 6 liters of 30 % acid to create 50 % acid. Therefore, adding 2.4 liters of 100 % acid to 6 liters of 30 % acid will create 8.4 liters of 50 % acid. 2...
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Applied Problem 1 - 100 % x L 100/1 L 100 x 30 % 6 L 30/1 L...

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