2-9 Weighted Averages Weighted Average – the sum of the product of the number of units and the value per unit divided by the sum of the number of units Mixture Problems – when two or more parts are combined into a whole and solved using weighted averages

Mixture Problems 0.2(12) + 0.4(x) = 0.25(12 + x) 12 0.2(12) + x 0.4(x) Ex. How many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution? Within a mixture problem there is always a Starting amount, an Added amount, and a final Total amount. Percent Amount Mixture 0.20 12 0.2(12) 20% Solution + 40% Solution 0.40 x 0.4(x) = 0.25 12 + x Final Solution 0.25(12 + x) 0.2(12) + 0.4(x) = 0.25(12 + x)

Mixture Problems Ex. How many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution? Percent Amount Mixture 20% Solution 0.2 12 0.2(12) 40% Solution 0.4 x 0.4(x) 25% Solution 0.25 12 + x 0.25(12 + x) 0.2(12) + 0.4(x) = 0.25(12 + x) 2.4 + .4x = 3 + .25x -.25x -.25x 2.4 + .15x = 3 -2.4 -2.4 .15x = .6 x = 4

Mixture Problems 0.5(3) + 1.0(x) = 0.75(3 + x) 3 0.5(3) + x 1.0(x) = Ex. How many liters of pure acid must be added to 3 liters of a 50% acid solution to obtain a 75% acid solution? Within a mixture problem there is always a Starting amount, an Added amount, and a final Total amount. Percent Amount Mixture 50% = 0.50 3 0.5(3) Acid Solution + Pure Acid 100% = 1.0 x 1.0(x) = 75% = 0.75 3 + x Final Solution 0.75(3 + x) 0.5(3) + 1.0(x) = 0.75(3 + x)

Mixture Problems Ex. How many liters of pure acid must be added to 3 liters of a 50% acid solution to obtain a 75% acid solution? Percent Amount Mixture Acid Solution 0.50 3 0.5(3) Pure Acid 1.0 x 1.0(x) Final Solution 0.75 3 + x 0.75(3 + x) 0.5(3) + 1.0(x) = 0.75(3 + x) 1.5 + x = 2.25 + 0.75x – 0.75x – 0.75x 1.5 + 0.25x = 2.25 - 1.5 - 1.5 0.25x = 0.75 0.25 0.25 x = 3

Mixture Problems 5.50(10) + 4.75(x) = 4.95(10 + x) $5.50 10 $5.50(10) Ex. How many pounds of almonds selling for $4.75 per pound should be mixed with 10 pounds of dried fruit selling for $5.50 per pound to obtain a trail mix that sells for $4.95 per pound? Within a mixture problem there is always a Starting amount, an Added amount, and a final Total amount. Price Per Unit Units (lb) Total Price $5.50 10 $5.50(10) Dried Fruit + Almonds $4.75 x $4.75(x) = $4.95 10 + x Trail Mix $4.95(10 + x) 5.50(10) + 4.75(x) = 4.95(10 + x)

Mixture Problems Ex. How many pounds of almonds selling for $4.75 per pound should be mixed with 10 pounds of dried fruit selling for $5.50 per pound to obtain a trail mix that sells for $4.95 per pound? Price Per Unit Units (lb) Total Price Dried Fruit $5.50 10 $5.50(10) Almonds $4.75 x $4.75(x) Trail Mix $4.95 10 + x $4.95(10 + x) 5.50(10) + 4.75(x) = 4.95(10 + x) 55 + 4.75(x) = 49.5 + 4.95(x) – 4.75(x) – 4.75(x) 55 = 49.5 + 0.20(x) - 49.5 - 49.5 5.5 = 0.20(x) 0.20 0.20 27.5 = x

Mixture Problems 0.9x + 0.14(16 – x) = 1.74 x 0.09x + 16 – x = 16 0.09 Ex. Nature Drinks wants to combine orange juice they sell for $0.09 per ounce with guava juice they sell for $0.14 per ounce to create an orange-guava drink. How many ounces of orange juice should they use to create a 16-ounce drink that would sell for $1.74? Price Amount (oz) Cost 0.09 x 0.09x Orange Juice + Guava Juice 0.14 16 – x 0.14(16 – x) = Mix 16 1.74 0.9x + 0.14(16 – x) = 1.74

Speed of One Vehicle On Alberto’s drive to his aunt’s house, the traffic was light and he drove the 45-mile trip in one hour. However, the return trip took him two hours. What was his average speed for the round trip? Going Returning R=d/t -R=d/t =45 miles/1 hour(45 miles/hour) =45 miles/2 hours(22.5 miles/hour)

To find the WEIGHTED average of Alberto’s speed, you have to take into consideration the fact that he did not drive the two speeds for equal amounts of time, so: Average speed = 45(1) + 22.5(2)/1+2 90/30 = 30 Alberto’s average speed was 30 miles per hour.

Distance Problems 40h + 30h = 245 40h 40 30h 30 h h Ex. Two trains leave Smithville at the same time, one traveling east and the other west. The eastbound train travels at 40 miles per hour, and the westbound train travels at 30 miles per hour. Let h represent the hours since departure. When will the trains be 245 miles apart? D = rt r t 40h 40 h Eastbound 30h Westbound 30 h 40h + 30h = 245

Distance Problems 30t = 36(t – 20) 30t 30 36(t – 20) 36 t t – 20 Ex. Mandy begins bicycling west at 30 miles per hour at 11 AM. If Liz leaves from the same point 20 minutes later bicycling at 36 miles per hour, when will she catch Mandy? D = rt r t 30t 30 t Mandy 36(t – 20) 36 Liz t – 20 30t = 36(t – 20)

Homework Assignment #19 Page 126 #14-23, 29-31, 40-41