You can put this solution on YOUR website! How many gallons of a 25% acid solution must be mixed with 100 gallons of a 6% acid solution to obtain a solution that is 20% acid?
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25% solution DATA;
Amount = x gallons ; amt. of active ingredient = 0.25x gallons
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6% solution DATA;
Amount = 100 gallons ; amt. of active = 0.06*100 = 6 gallons
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Mixture DATA:
Amt. = 100+x gallons ; amt of active = 0.20(100+x)= 20 + 0.20x gallons
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EQUATION:
active + active = active
0.25x + 6 = 20+0.20x
0.05x = 14
x = 280 gallons (amt of 25% solution needed for the mixture)
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Cheers,
Stan H.
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