Today’s Date: 11/16/11 Mixture Word Problems Notes on Handout

Not given to you, but expected to know: Water = 0% “Pure” Acid = 100% Remember: a) Define variable b) Write equation c) Solve Mixture Equation: Amount % = Total

How many kilograms of water should be mixed with 60 kg of a 30% acid solution to make a 10% solution? Example 1 Amount%Total Water Acid Mix 10(x + 60) 0 x x (60)(30) Tells you it’s the mix Amt of Water + Amt of Acid = Amt of Mix Asking for amt of water Multiply Amt ∙ % in each row

Amount%Total Water Acid Mix 10(x + 60) 0 x x (60)(30) Example 1 cont… Get your equation from the last column Total Water + Total Acid = Total Mix 0 + (60)(30) = 10(x + 60) 1800 = 10x = 10x 120 = x a)x is amt of water b)1800 = 10x c)120 kg of water

Example 2 How many kilograms of pure salt must be added to 8 kg of a 10% salt solution to obtain a 20% salt solution? Amount%Total Pure Salt Salt Mix 20(x + 8) 100 x x x (8)(10) Tells you it’s the mix Asking for amt of “pure” salt 100x + (8)(10) = 20(x + 8)

Example 2 cont… 100x + (8)(10) = 20(x + 8) 100x + 80 = 20x x + 80 = x = 80 x = 1 a) x is amt of pure salt b)100x + 80 = 20x c)1 kg of pure salt

Example 3 How many kilograms of a 30% acid solution must be added to 80 kg of a 20% solution to produce a 25% solution? Amount%Total Solution 1 Solution 2 Mix 25(x + 80) 30 x x x (80)(20) Tells you it’s the mix 30x + (80)(20) = 25(x + 80)

Example 3 cont… 30x + (80)(20) = 25(x + 80) 30 x = 25x x = x = 400 x = 80 a) x is amt of 30% solution b)30x = 25x c)80 kg of 30% solution

Homework #9 Two Sided Worksheet: Mixture Problems Review Sum/Difference/Complex Fractions Mix Prob #2: Mix Amt = 36 Need to find Milk Amt & Cream Amt How?