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Question 150850: Solve This Problem Interactively Customize This Problem
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used?
: Solve This Problem Interactively Customize This Problem
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used?

Answer by Fombitz(1270) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call the unkown amount, x.
x*(0.06)+2(0.15)=(x+2)(0.12)
Let's multiply both sides by 100 to get rid of decimals.
6x+2(15)=12(x+2)
6x+30=12x+24
6x=6
x=1
1 ounce of the 6% solution is needed.
Question 150850: Solve This Problem Interactively Customize This Problem
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used?
: Solve This Problem Interactively Customize This Problem
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used?

Answer by ankor@dixie-net.com(3943) About Me  (Show Source):
You can put this solution on YOUR website!
A mixture containing 6% salt is to be mixed with 2 ounces of a mixture which is 15% salt, in order to obtain a solution which is 12% salt. How much of the first solution must be used?
:
How do you solve a problem "Interactively"?
:
Anyway.
:
Let x = amt of 1st solution required
:
.06x + .15(2) = .12(x+2)
:
.06x + .3 = .12x + .24
:
.3 - .24 = .12x - .06x
:
.06 = .06x
:
x = 1 oz of the 1st solution required
:
:
Check solution:
.06(1) + .15(2) = .12(1+2)
.06 + .30 = .36