Math Practice Online > free > lessons > Florida > 9th grade > Mixture Word Problems

Here are some tips for Mixture Word Problems, which aligns with Florida state standards:

Mixture Word Problems

The basic concept to understand for mixture problems is
Amount of solution × % substance = Amount of substance

So 200 mL of a 45% insecticide solution contains (200 mL)(45% insecticide) = 90 mL of insecticide.

Example 1: Salt Solutions

Solve.
Solution X is a 31% salt solution and Solution Y is a 25% salt solution. How much of each is needed to make 18 gallons of a 29% salt solution?
Solution X: gallons    Solution Y: gallons

Amount of solution
(gal)
% salt Amount of salt
(gal)
Solution Xx0.310.31x
Solution Y18 - x0.250.25(18 - x)
Together180.290.29(18)

 0.31x + 0.25(18 - x) = 0.29(18) 0.31x + 4.5 - 0.25x = 5.22 0.06x = 0.92 x = 6

x = 6 gallons of Solution X and 18 - x = 12 gallons of Solution Y are needed to make 18 gallons of 29% acid solution.
Solution X: gallons    Solution Y: gallons

Example 2: Food

Solve.
How many pounds of lima beans worth \$11.07/lb should be mixed with 49 pounds worth \$8.91/lb to make a mixture worth \$9.39/lb?
lb

Apply what you know about unit costs:   Amount of food × Unit price = Total cost of food

 Pounds Price/lb Total Cost Lima beans x \$11.07 \$11.07x Navy beans 49 \$8.91 (\$8.91)(49) Mixture x + 49 \$9.39 (\$9.39)(x + 49)

 \$11.07x + (\$8.91)(49) = (\$9.39)(x + 49) \$11.07x + \$436.59 = \$9.39x + \$460.11 \$1.68x = \$23.52 x = 14

x = 14 pounds of lima beans to add to the navy beans.   lb

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