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Introduction to Mixture Applications The following is designed to help you understand the basics of one of the popular application problems in introductory algebra: mixture problems. Recall that percent means per hundred, so... Example 1:

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The jar on the right is marked in 10 equal divisions. 10 9 8 7 6 5 4 3 2 1 Fill the jar with acid up to the “2” mark … Example 2:

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The jar on the right is marked in 10 equal divisions. 10 9 8 7 6 5 4 3 2 1 Fill the jar with acid up to the “2” mark … … and add water to the top of the glass… Example 2:

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The jar on the right is marked in 10 equal divisions. 10 9 8 7 6 5 4 3 2 1 Fill the jar with acid up to the “2” mark … … and add water to the top of the glass… Example 2:

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The jar has 2 units of acid … 10 9 8 7 6 5 4 3 2 1 … and 8 units of water.

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10 9 8 7 6 5 4 3 2 1 … for a total of 10 units of mixed solution.

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10 9 8 7 6 5 4 3 2 1 The acid represents 2 units … … out of a total of 10 units, or …

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10 9 8 7 6 5 4 3 2 1

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10 9 8 7 6 5 4 3 2 1 This means that the acid represents 20% of the total liquid in the jar. Figured another way, the amount of acid is

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10 9 8 7 6 5 4 3 2 1 Mix 2 units of acid with only 6 units of added water. Example 3:

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10 9 8 7 6 5 4 3 2 1 2 units of acid 6 units of water Mix 2 units of acid with only 6 units of added water. Example 3:

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10 9 8 7 6 5 4 3 2 1

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10 9 8 7 6 5 4 3 2 1 This means that the acid represents 25% of the total liquid in the jar. Figured another way, the amount of acid is

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10 9 8 7 6 5 4 3 2 1 Mix acid and water again. Example 4:

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10 9 8 7 6 5 4 3 2 1 Mix acid and water again. Example 4:

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10 9 8 7 6 5 4 3 2 1

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10 9 8 7 6 5 4 3 2 1 Mix acid and water again. Example 5:

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10 9 8 7 6 5 4 3 2 1 Mix acid and water again. Example 5: If the solution is 60% acid, determine the number of units of acid

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10 9 8 7 6 5 4 3 2 1

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10 9 8 7 6 5 4 3 2 1 Assume the jar has x units of solution that is 40% acid. Example 6: Write an expression for the amount of acid in the jar.

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