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A coffee distributor is blending a new coffee that will cost $4.90 per pound. It will consist of a blend of $3.00 per pound coffee, $5.00 per pound coffee, and $6.00 per pound coffee. If the distributor wants 100 pounds of blended coffee, and they have the same amount of $5.00 per pound coffee as the combined amounts of $3.00 and $6.00 per pound coffees, how many pounds of each type is needed?

1 Answer

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Final answer:

The coffee distributor requires 20 pounds of $3.00 coffee, 50 pounds of $5.00 coffee, and 30 pounds of $6.00 coffee to create a 100-pound blend that costs $4.90 per pound.

Step-by-step explanation:

To solve the problem of blending coffee to achieve a specific price per pound, we can set up a system of equations based on the given information. We know the distributor wants 100 pounds of blended coffee, which should cost $4.90 per pound. The mixture will include coffees priced at $3.00, $5.00, and $6.00 per pound respectively. Let's define the variables: x for the amount of $3.00 coffee, y for the amount of $5.00 coffee, and z for the amount of $6.00 coffee.

According to the problem, we have the following constraints:

  • The total weight is 100 pounds: x + y + z = 100
  • The total cost of the blend per pound is $4.90: 3x + 5y + 6z = 490 (since the total cost is 100 pounds times $4.90)
  • The distributor has the same amount of $5.00 coffee as the combined amounts of $3.00 and $6.00 coffees: y = x + z

Substituting the third equation into the first equation, we get 2x + 2z = 100. Dividing by 2, we simplify to x + z = 50. Using these equations, we can solve for the amounts of each type of coffee needed:

  • x = 50 - z
  • y = 50
  • z (unknown)

Substitute x and y into the second equation: 3(50 - z) + 5(50) + 6z = 490. This simplifies to 150 - 3z + 250 + 6z = 490. Combining like terms, we get 3z = 90, which means z = 30. Then, from x = 50 - z, we get x = 20. Therefore, the distributor needs 20 pounds of $3.00 coffee, 50 pounds of $5.00 coffee, and 30 pounds of $6.00 coffee.

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