Question

A coffee mixture contains beans that sell for $0.08 per pound and $0.32 per pound. If 110 pounds of beans create a mixture that sells for $0.27 per pound, to the nearest tenth of a pound, how much of each bean is used in the mixture? Model the scenario with an equation and solve. Use complete sentences to explain whether or not your solution is reasonable.

Answer 1

Beans A= x
Beans B=(110-x)
0.08x+0.32(110-x) = 0.27(110)
Solve for x
0.08x+35.20-0.32x = 29.70
0.08x-0.32x=29.7-35.2
−0.24x=−5.5
X=5.5÷0.24
X=22.9 of Beans A

110−22.9=87.1 of Beans B

This is reasonable if one can weigh beans to the nearest tenth of a pound.

Hope it helps:-)

Answer 2

0.08x + 0.32(110 - x) = 0.27(110)

0.08x + 35.20 - 0.32x = 29.7

           -35.2                   -35.2

0.08x - 0.32x = -5.5

-0.24x = -5.5

/-0.24    /-0.24

x = 22.9 pounds of beans that sell for $0.08 per pound

110 - 22.9 = 87.1 pounds of beans that sell for $0.32 per pound

This is reasonable because the beans can be weighed to the nearest tenth.

Please rate and thank : )