7% of 12 gallons = 0.84 gallons.
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50 gallons @ 3% must be added.
divide 7 by 12. 7/12=58.333 percent
Suppose G gallons of 60% antifreeze are required. G gallons of 60% contain 0.6G gallons of the active ingredient. 70 gallons of 10% contain 7 gallons of the active ingredient. So, in the mixture, there are G + 70 gallons containing 0.6G + 7 gallons of the active ingredient. This is to represent 50%. So 0.6G + 7 = 0.5*(G + 70) = 0.5G + 35 0.1 G = 35 - 7 = 28 G = 280 gallons.
7%
About 58.33%