Example

Solving a Party Mix Mixture Problem

Apply the seven-step problem-solving strategy to a mixture word problem involving a known total amount and two ingredients with given per-unit costs.

Problem: Orlando is mixing nuts and cereal squares to make a party mix. Nuts sell for $7 a pound and cereal squares sell for $4 a pound. Orlando wants to make 3030 pounds of party mix at a cost of $6.50 a pound. How many pounds of nuts and how many pounds of cereal squares should he use?

  1. Read the problem and identify the ingredient types: nuts (value $7 per pound) and cereal squares (value $4 per pound). The total mixture weight is 3030 pounds at $6.50 per pound.
  2. Identify what to find: the number of pounds of nuts and the number of pounds of cereal squares.
  3. Name the unknowns using a single variable and the known-total technique. Let xx = the number of pounds of nuts. Then the number of pounds of cereal squares is 30โˆ’x30 - x. Organize in a table:
TypePoundsPrice per pound ($)Total Value ($)
Nutsxx777x7x
Cereal squares30โˆ’x30 - x444(30โˆ’x)4(30 - x)
Party mix30306.506.5030(6.50)=19530(6.50) = 195
  1. Translate into an equation. The value of the nuts plus the value of the cereal squares equals the value of the party mix: 7x+4(30โˆ’x)=1957x + 4(30 - x) = 195

  2. Solve the equation:

  • Distribute 44: 7x+120โˆ’4x=1957x + 120 - 4x = 195
  • Combine like terms: 3x+120=1953x + 120 = 195
  • Subtract 120120 from both sides: 3x=753x = 75
  • Divide both sides by 33: x=25x = 25 Find the number of pounds of cereal squares: 30โˆ’25=530 - 25 = 5.
  1. Check: 7(25)+4(5)=1957(25) + 4(5) = 195 175+20=195175 + 20 = 195 195=195195 = 195 โœ“

  2. Answer: Orlando should use 2525 pounds of nuts and 55 pounds of cereal squares.

0

1

Updated 2026-05-02

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.2 Solving Linear Equations - Intermediate Algebra @ OpenStax

Algebra

Related