SOLUTION: 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 solut

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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?

Found 2 solutions by Fombitz, ankor@dixie-net.com:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call the unkown amount, x.
x%2A%280.06%29%2B2%280.15%29=%28x%2B2%29%280.12%29
Let's multiply both sides by 100 to get rid of decimals.
6x%2B2%2815%29=12%28x%2B2%29
6x%2B30=12x%2B24
6x=6
x=1
1 ounce of the 6% solution is needed.

Answer by ankor@dixie-net.com(22740) 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