Section 2.6 Percent, Rate, and Mixture Problems 157 �� ������������������������ ������������������ Write an equation that can be used to solve each problem. Solve the equation and answer the question using a complete sentence. 3a. Teresa has a collection of dimes and quarters. She has 40 less dimes than quarters. The value of her collection is $20.50. Find the number of dimes and quarters she has in her collection. Solution 3a. What is unknown? The number of dimes and quarters is unknown. Let q = number of quarters. Then q - 40 = number of dimes. What is known? The total value of the collection is $20.50. Type of Coin Number of coins × Value of coin = Total value of the coins Quarters q $0.25 0.25q Dimes q - 40 $0.10 0.10(q - 40) The total value of the collection is the value of the quarters plus the value of the dimes. 0.25q + 0.10(q - 40) = 20.50 Express relationship. 0.25q + 0.10q - 4 = 20.50 Apply the distributive property. 0.35q - 4 = 20.50 Combine like terms. 0.35q - 4 + 4 = 20.50 + 4 Add 4 to each side. 0.35q = 24.50 Simplify. 0.35q 0.35 = 24.50 0.35 Divide each side by 0.35. q = 70 Simplify. So, Teresa has 70 quarters and 70 - 40 = 30 dimes in her collection. 3b. Louisa is a chemist and she needs to conduct an experiment with a solution that is 30% copper sulfate. She does not have a solution with this concentration but she does have 40 mL of a 25% copper sulfate solution and she also has a 60% copper sulfate solution. How many milliliters of the 60% copper sulfate solution does she need to mix with the 40 mL of the 25% solution to obtain a 30% copper sulfate solution? Solution 3b. What is unknown? The amount of the 60% copper sulfate solution is unknown. Let x = the amount of the 60% solution. What is known? She needs a solution that is 30% copper sulfate. She has 40 mL of a 25% copper sulfate solution. Type of Solution Number of mL × Solution strength = Total mL of copper sulfate 25% 40 mL 25% or 0.25 0.25(40) 60% x 60% or 0.60 0.60x 30% 40 + x 30% or 0.30 0.30(40 + x) The amount of copper sulfate in the 25% solution and the 60% solution equals the amount of copper sulfate in the 30% solution. This gives us the following equation. 60% solution x mL 25% solution 40 mL 30% solution x mL 40 mL 40 + x
hendricks_beginning_algebra_1e_ch1_3
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