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Here are some tips for Mixture Word Problems, which aligns with Minnesota 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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