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### Like us? # Coffee mixture word problem

Q: A coffee merchant has two types of coffee beans, one selling for \$3 per pound and the other for \$5 per pound. The beans are to be mixed to provide 100 pounds of a mixture selling for \$4.18 per pound. How much of each type of coffee bean should be used to form 100 pounds of the mixture?

A: Let x= the number of pounds of the \$3 beans
Let y = the number of pounds of the \$5 beans

You have two variables, so you need two equations:

Equation 1: x+y=100 (because the mixture has a total mass of 100 pounds)

Equation 2: 3x + 5y = (4.18)*100 (each quantity is cost per pound *number of pounds)

Isolate for x in Equation 1: x = 100-y, and substitute into Equation 2:

3(100-y) +5y = 418
300-3y+5y=418
2y = 118
y=59

x = 100-y
x=100-59
x=41

So you would need 59 pounds of \$5 beans, and 41 pounds of \$3 beans.