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put this solution on YOUR website!How many gallons of a 12% solution must be mixed with a 20% solution to get 10 gallons of a 14% solution?
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Let x = amt of 12% solution required
:
The resulting total amt is to be 10 gal therefore:
(10-x) = amt of 20% solution
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we will use the decimal equiv of percent:
:
.12x + .20(10-x) = .14(10)
;
.12x + 2 - .20x = 1.4
:
.12x - .20x = 1.4 - 2
:
-.08x = -.6
x =
x = +7.5 gal of 12% solution required
;
;
Check solution in the original equation (2.5 gal of the 20% solution needed)
.12(7.5) + .2(2.5) = .14(10)
.9 + .5 = 1.4, confirms our solution
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