Question 253008
how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of 25% solution to give 30% solution?
:
Let x = amt of 80% alcohol required to accomplish this
:
A simple "amt of alcohol equation"
.80x + .25(6) = .30(x+6)
:
.8x + 1.5 = .3x + 1.8
:
.8x - .3x = 1.8 - 1.5
:
.5x = .3
x = {{{.3/.5}}}
x = .6 gal of 80% solution required
:
:
Check solution in original equation
.8(.6) + .25(6) = .3(.6 + 6)
.48 + 1.5 = .3(6.6)