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Difficulty: EasyAlgebraic Word Problems and Modeling

A coffee shop owner creates a 3030-pound blend of coffee by mixing Bean X, which costs $8\$8 per pound, with Bean Y, which costs $14\$14 per pound. If the final mixture costs $10\$10 per pound, how many pounds of Bean X are in the mixture?

Answer: 20 pounds

Answer

The mixture contains 20 pounds of Bean X.
The quantity of Bean X required is 20 pounds. Setting up the cost equation 8x+14(30−x)=3008x + 14(30 - x) = 300 simplifies to −6x=−120-6x = -120, giving x=20x = 20.

Step-by-Step Solution

1
Define the unknown variables representing the quantities of each component.
Let xx be the weight of Bean X in pounds. The weight of Bean Y is (30−x)(30 - x) pounds.
Since the total weight of the mixture is 30 pounds, expressing Bean Y in terms of xx creates a single-variable system.
2
Set up a linear equation based on total financial value.
8x+14(30−x)=10×308x + 14(30 - x) = 10 \times 30
The total cost of Bean X plus the total cost of Bean Y equals the total cost of the combined 30-pound mixture.
3
Solve the algebraic equation for xx.
8x+420−14x=300  ⟹  −6x=−120  ⟹  x=208x + 420 - 14x = 300 \implies -6x = -120 \implies x = 20
Combining like terms isolates the variable xx to yield the required quantity of Bean X.

Key Concept

Linear mixture modeling and single-variable algebraic modeling
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Topics

Algebraic Word Problems and Modeling

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