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43 votes
Mr. May owns "The Coffee Pot," a specialty coffee store. He wants to create a special mix using two coffees, one priced at $8.40 per pound and the other at $7.28 per pound. How many pounds of $7.28 coffee should he mix with 9 pounds of the $8.40 coffee to sell the mixture for $7.95 per pound.

User Urasquirrel
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1 Answer

14 votes
14 votes

In order to determine how many pounds of $7.28 coffe should Mr. May mix with 9 pounds of the $8.40 coffee, you first write the given situation in an algebraic way.

If x is the needed number of pounds $7.28 coffee, 7.28x is the cost of x pounds of such a coffe. This cost, added to the cost of the 9 pounds of $8.40 coffee is equal to y (total cost of both type of coffe).

Then you have:

7.28x + 9(8.40) = y

this is the total price for x pounds of $7.28 coffe and 9 pounds of $8.40.

y/(x + 9) = 7.95

this is the price per pound of the mixed coffe.

Then, you have the following system of equations:

y/(x + 9) = 7.95

7.28x + 9(8.40) = y

To solve for y and x, you proceed as follow:

solve the first equation for y:

y/(x + 9) = 7.95

y = 7.95(x + 9)

y = 7.95x + 71.55

next, you equal the previous equation with the equation 7.28x + 9(8.40) = y, and you solve for x:

7.95x + 71.55 = 7.28x + 9(8.40)

7.95x + 71.55 = 7.28x + 75.6 subtract 7.28x both sides

7.95x - 7.28x + 71.55 = 75.6 subtract 71.55 both sides

7.95x - 7.28x = 75.6 - 71.55 simplify like terms

0.67x = 4.05 divide by 0.67 both sides

x = 4.05/0.67

x = 6.04 ≈ 6

next, you replace the previous value of x in one of the equations:

y = 7.95x + 71.55

y = 7.95(6) + 71.55 = 119.25

Hence, the number of pounds of $7.28 coffee should Mr. May mix with the 9 pounds of $8.40 is 6 pounds.

User Hoford
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