Questions and Answers
Which of the following are possible mixture problem situations? (Select all that apply)
If x represents the pounds of walnuts used in the mixture, which expression would represent the total cost of the pecans used in the mixture?
0.60(90 - x)
If a dairyman has 300 pounds of milk testing 3% butterfat, how many pounds of skimmed milk must he remove to have milk testing 3.6% butterfat?
50 pounds
How much alcohol must a druggist add to a 20-ounce quantity of a 7% solution of iodine to reduce it to a 4% solution?
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If x represents the number of quarts of 36 cent oil, which equation could be used to solve the mixing oil problem?
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How many pounds of 80 cent seed should a seedman use to make a mixture of 30 pounds selling at 75¢ per pound?
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Which expression represents the number of pounds of salt in the new 25% salt solution if x represents the pounds of salt to be added?
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How much tin must be added to create 820 pounds of alloy that is 18% tin?
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What percentage of gold is in the new alloy after recasting if it consists of 80 ounces with only one-fifth of an ounce of gold in 5 ounces?
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How much silver must be added if the new alloy is to have only one-fifth of an ounce of gold in 5 ounces?
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Study Notes
Mixture Problem Situations
- Mixture problems can involve financial investments, chemical compositions, or agricultural seed types.
Cost of Mixtures
- Total cost for pecans in a mixture: 0.60(90 - x), where x is the pounds of walnuts.
Dairyman's Butterfat Problem
- To increase butterfat from 3% to 3.6% in 300 pounds of milk, 50 pounds of skimmed milk must be removed.
Reduction of Iodine Solution
- To convert a 20-ounce 7% iodine solution to a 4% solution, 15 ounces of pure alcohol must be added.
Oil Mixture Equation
- For blending oils worth 36¢ and 52¢ to create a 300-gallon mixture worth 40¢ per quart, the equation is: 0.36x + 0.52(1200 - x) = 0.40(1200).
Seed Mixture Calculation
- To prepare a 30-pound seed mixture, 25 pounds of 80¢ seeds are required to achieve a selling price of 75¢ per pound.
Saltwater Solution Expression
- For a 50-pound saltwater solution with 10% salt, the expression representing the salt in a new 25% solution after adding x pounds of salt is: 0.25(x + 50).
Tin in Alloy Problem
- Adding tin to an alloy that initially contains 16% tin must result in an 18% tin alloy of 820 pounds; approximately 19.5 pounds of tin must be added.
Gold Percentage in Alloy
- After recasting an alloy of silver and gold to include more silver and achieve a specific composition, the new alloy contains 4% gold.
Silver Addition Calculation
- To adjust a silver and gold alloy weighing 80 ounces (with 10 ounces of gold) to ensure 1/5 ounce of gold per 5 ounces in the new mixture, 170 ounces of silver must be added.
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Test your understanding of mixture problems with these flashcards. Explore various scenarios involving investments, chemistry, and agriculture. Solve practical examples to enhance your problem-solving skills.