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Difficulty: MediumFractions, Decimals, and Percentages

A coffee shop blend initially consists of a 2020-pound mixture of Arabica and Robusta beans, of which 45%45\% by weight is Arabica. The shop manager adds pure Arabica beans to the mixture to increase the proportion of Arabica beans to 56%56\% by weight. How many pounds of pure Arabica beans must be added to the mixture?

Answer: 5 pounds

Answer

5 pounds of pure Arabica beans must be added to the mixture.
To solve for the required added weight, first compute the original mass of Arabica beans: 45%45\% of 2020 pounds is 0.45×20=90.45 \times 20 = 9 pounds. Letting xx represent the pounds of pure Arabica added, the updated mass of Arabica is 9+x9 + x pounds and the updated total mass of the coffee batch is 20+x20 + x pounds. Setting the ratio 9+x20+x=0.56\frac{9 + x}{20 + x} = 0.56 gives 9+x=0.56(20+x)9 + x = 0.56(20 + x). Expanding the right side results in 9+x=11.2+0.56x9 + x = 11.2 + 0.56x. Subtracting 0.56x0.56x and 99 from both sides gives 0.44x=2.20.44x = 2.2, which simplifies to x=5x = 5 pounds.

Step-by-Step Solution

1
Determine the initial weight of Arabica beans.
Arabica weight = 0.45×20=90.45 \times 20 = 9 pounds.
The starting 2020-pound batch contains 45%45\% Arabica beans.
2
Formulate the mixture fraction with the unknown added weight xx.
9+x20+x=0.56\frac{9 + x}{20 + x} = 0.56
Adding xx pounds of pure Arabica increases both the total Arabica weight (numerator) and the total mixture weight (denominator).
3
Solve the algebraic equation for xx.
9+x=11.2+0.56x  ⟹  0.44x=2.2  ⟹  x=59 + x = 11.2 + 0.56x \implies 0.44x = 2.2 \implies x = 5
Isolating xx yields the exact amount of pure Arabica required.

Key Concept

Solving percentage mixture problems using algebraic proportions
Estimated Time:1m 30s
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Topics

Fractions, Decimals, and Percentages

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