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Purpose: To solve word problems involving mixtures. Homework: p. 322 – 325 1- 11 all. 13 – 21 odd.

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Presentation on theme: "Purpose: To solve word problems involving mixtures. Homework: p. 322 – 325 1- 11 all. 13 – 21 odd."— Presentation transcript:

1 Purpose: To solve word problems involving mixtures. Homework: p. 322 – 325 1- 11 all. 13 – 21 odd.

2 Example # 1 A grocer makes a natural breakfast cereal by mixing oat cereal costing $2 per kg with dried fruits costing $9 per kg. How many kg of each are needed to make 60 kg of cereal costing $3.75 per kg? Use a chart to organize the information. You have 2 types plus a mixture, so you need three rows.

3 Organized Chart # of Kg x Price per kg =Total Cost cereal x2 2x fruits60 - x 9 9(60 - x ) mixture 60 3.75 225 Make your equation. (2x + 540 – 9x = 225) Solve. – 7x = - 315 x = 45 Make sure you answer the question. 45 kg of cereal and 15 kg of fruits is the answer.

4 Example #2 How many liters of water must be added to 20L of a 24% acid solution to make a solution that is 8% acid? Organize into a Chart. Water does not have acid, so it’s 0. # of L% Acid Amt. of Acid acid 20.24.24(20) waterx 00 mixture20 + x.08.08(20 + x)

5 Example #2 (cont.) Make your equation. The first mix plus water added is the total mixture. 4.8 + 0 =.08(20 + x) **Keep the decimals or multiply the equation by 100 to get rid of them. 100(4.8 + 0) = 100[.08(20 + x)] 480 = 8(20 + x) ; 480 = 160 + 8x 320 = 8x ; x = 40 40L of water must be added.

6 How many kg of water must be evaporated from 12 kg of a 5% salt solution to produce a solution that is 30% salt?.05(12) + 0 =.30(12 - x) **Get rid of decimals. 100(.05×12) = 100[.30(12 – x)] 5(12) = 30(12 – x) ; 60 = 360 – 30x -300 = -30x ; x = 10 10 kg of water must be evaporated. Kg% Salt Amt. Salt original 12.05.05(12) water x 0 0x new 12 - x 30.30(12 – x)


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