SOLUTION: 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
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Question 1051498: 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. Found 2 solutions by Boreal, addingup:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! bean A is 0.08/pound
bean B is 0.32/pound
A+B=110
.08A+0.32(B)=29.7 (that comes from the mixture of 110 pounds at 0.27/pound)
A=110-B
0.08(110-B)+0.32B=29.7
8.8-0.08B+0.32B=29.7
0.24B=20.9
B=87.1 pounds@0.32=27.87
A=22.9 pounds@0.08=1.83
That is 110 pounds with a dollar value of $29.70, which is $0.27 per pound.
This is reasonable if one can weigh beans to the nearest tenth of a pound.
You can put this solution on YOUR website! The total is 110
Let the beans that sell at 0.08 be x. Therefore, the beans that sell at 0.32 will be 110-x
:
0.08x+0.32(110-x) = 0.27(110)
0.08x+35.20-0.32x = 29.70
-0.24x = -5.5 Divide both sides by -0.24 and remember -/- = +
x = 22.92 pounds of the 0.08 beans and
110-22.92 = 87.08 pounds of the 0.32 beans.
:
Check your numbers, I have the feeling that there is one bad number in the problem as you wrote it because it's unusual to get a fraction like 22.92 pounds, it's usually a round number like 23.
If you find a number that is different, plug the number into my work and calculate the new answer. Or send me a note.
