Question

How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifreeze? Use the six-step method.

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Answer 1

7X+.25*100=.6(100+X)
.7X+25=60+.6X
.7X-.6X=60-25
.1X=35
X=35/.1

X=350 GALLONS OF 70% ANTIFREEZE WILL BE NEEDED.
PROOF
.7*350+25=.6*450
245+25=270
270=270

Answer 2

Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons