Question 211039
How much water in gallons should be added to 1 gallon of pure antifreeze to obtain a solution that is %95 antifreeze?
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Equation:
water + water = water
0*1 gal + x gal = 0.05(1+x)
x = 0.05 + 0.05x
0.95x = 0.05
x = 0.0526 (# of gallons of water that must be added) 
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You could also do it this way:
antifreeze + antifreeze = antifreeze
1 gal anti + 0*x gal anti = 0.95(1+x) gal anti
1 = 0.95 + 0.95x
0.95x = 0.05
x = 0.0526 (# of gallons of wter that must be added)
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Cheers,
Stan H.