Answered

A store sells two different types of coffee beans; the more expensive one sells for $8 per pound, and the cheaper one sells for $4 per pound. The beans

are mixed to provide a mixture of 50 pounds that sells for $6.50 per pound. How much of each type of coffee bean should be used to create 50 pounds of the mixture?

Verified Answer :

SchwarzschildRadius

So,

Let the amount of the expensive coffee beans be x, and the amount of the less expensive coffee bean be y.

The number of pounds of coffee beans is 50.
x + y = 50

$8 times the amount of more expensive coffee beans plus $4 times the amount of less expensive coffee beans divided by 50 equals $6.50.
\(\frac{8x + 4y}{50} = 6.50\)

Now, all we have to do is use Elimination by Substitution.
x + y = 50

Subtract y from both sides.
x = 50 - y

Substitute 50 - y for x in the second equation.
\(\frac{8(50 - y) + 4y}{50} = 6.50\)

Distribute.
\(\frac{400 - 8y + 4y}{50} = 6.50\)

Collect Like Terms.
\(\frac{400 -4y}{50} = 6.50\)

Simplify.
\(\frac{4(100 -y)}{2*5^2} = 6.50\)
\(\frac{2(100 -y)}{5^2} = 6.50\)
\(\frac{200 - 2y}{25} = 6.50\)

Multiply both sides by 25.
\(200 - 2y = 162.5\)

Add 2y to both sides.
\(200 = 162.5 + 2y\)

Subtract 162.5 from both sides.
37.5 = 2y

Divide both sides by 2.
18.75 = y

Substitute in an earlier equation to solve for x.
x = 50 - y
x = 50 - 18.75
x = 31.25

Check.
31.25 + 18.75 = 50 lbs.

\(\frac{8x + 4y}{50} = 6.50\)
\(\frac{8(31.25) + 4(18.75)}{50} = 6.50\)
\(\frac{250 + 75}{50} = 6.50\)
\(\frac{325}{50} = 6.50\)
\(\frac{5^2*13}{2 * 5^2} = 6.50\)
\(\frac{13}{2} = 6.50\)
6.50 = 6.50 This also checks.

There were 31.75 pounds of the more expensive coffee bean and 18.75 pounds of the less expensive coffee bean.

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