SOLUTION: A certain compound is a mixture of two concentrations of iron which are 44% and 29%. How many pounds of each concentration must be used to form 50 lbs.of compound having 38% iron?

Question 1040359: A certain compound is a mixture of two concentrations of iron which are 44% and 29%. How many pounds of each concentration must be used to form 50 lbs.of compound having 38% iron?
Found 3 solutions by stanbon, ikleyn, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A certain compound is a mixture of two concentrations of iron which are 44% and 29%. How many pounds of each concentration must be used to form 50 lbs.of compound having 38% iron?
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Equation:
iron + iron = iron
0.44x + 0.29(50-x) = 0.38*50
-------------------------
44x + 29*50 - 29x = 38*50
15x = 9*50
x = 30 lbs (amt. of 44% needed)
50-x = 20 lbs (amt of 29% needed)
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Cheers,
Stan H.
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Answer by ikleyn(52796)   (Show Source): You can put this solution on YOUR website!
.
A certain compound is a mixture of two concentrations of iron which are 44% and 29%. How many pounds of each concentration must be used to form 50 lbs.of compound having 38% iron?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Equation:
iron + iron = iron
0.44x + 0.29(50-x) = 0.38*50
-------------------------
44x + 29*50 - 29x = 38*50
15x = 9*50
x = 30 lbs (amt. of 44% needed)
50-x = 20 lbs (amt of 29% needed)

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A certain compound is a mixture of two concentrations of iron which are 44% and 29%. How many pounds of each concentration must be used to form 50 lbs.of compound having 38% iron?

With any math problem, you need to have an idea of what your answer should or should not be. In this case, a 44% and a 29% are
used to form a 38% compound. The 44% compound is closer to the end result (38%) than the 29% compound is. Therefore, we know,
for certain that more of the 44% compound should be mixed to form the 38% compound.
Let the amount of 44% compound to be mixed be F
Then amount of 29% compound to be mixed = 50 - F
We then get: .44F + .29(50 - F) = .38(50)
.44F + 14.5 - .29F = 19
.44F - .29F = 19 - 14.5
.15F = 4.5
F, or amount of 44% compound to be mixed = , or
Amount of 29% compound to be mixed: 50 - 30, or
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