Question

Difficulty: HardSystems of Linear Equations

A distributor plans to mix two coffee blends, Blend A and Blend B, to create a custom mixture. The table below shows the distribution of Colombian and Ethiopian coffee beans in each blend by weight:

Coffee BlendColombian BeansEthiopian Beans
Blend A60%40%
Blend B20%80%

The distributor wants the final custom mixture to contain exactly 14 kilograms of Colombian coffee beans and 16 kilograms of Ethiopian coffee beans. How many kilograms of Blend A should the distributor use to create this mixture?

Answer: 20 kg

Answer

The distributor should use 20 kilograms of Blend A.
By setting up a system of linear equations based on the percentage of each bean type in Blend A (AA) and Blend B (BB), we get 0.60A+0.20B=140.60A + 0.20B = 14 and 0.40A+0.80B=160.40A + 0.80B = 16. Solving this system yields A=20A = 20 and B=10B = 10. Therefore, 20 kilograms of Blend A are required.

Step-by-Step Solution

1
Define variables for the unknowns and write the system of equations.
Let AA be the number of kilograms of Blend A, and let BB be the number of kilograms of Blend B. The system of equations is:
For Colombian beans: 0.60A+0.20B=140.60A + 0.20B = 14
For Ethiopian beans: 0.40A+0.80B=160.40A + 0.80B = 16
This translates the verbal and tabular constraints of the problem into mathematical equations.
2
Simplify the system by multiplying both sides of each equation to eliminate decimals.
Multiply the first equation by 5:
3A+B=70    B=703A3A + B = 70 \implies B = 70 - 3A
Multiply the second equation by 5:
2A+4B=80    A+2B=402A + 4B = 80 \implies A + 2B = 40
Working with integers makes the algebraic manipulation easier and reduces arithmetic errors.
3
Substitute the expression for BB into the simplified second equation to solve for AA.
A+2(703A)=40A + 2(70 - 3A) = 40
A+1406A=40A + 140 - 6A = 40
5A=100-5A = -100
A=20A = 20
This isolates the variable AA, which represents the required kilograms of Blend A.

Key Concept

Systems of Linear Equations
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