0.6 of a pint.
2 gallons.
4.2 quarts
0.25 gallons of water (or 1 quart)
You need 1 1/3 quarts of pure antifreeze.
Approx 1.86 gallons.
4 gallons Let x be the amount of antifreeze needed to be added. We know that the total amount of antifreeze in the new solution must equal the amount of antifreeze in the old solution + x: .40*(x+12)=x+.20*12 .60x=2.40 x=4 gallons
600ml.
If my math is correct it would take an additional ( .6 of a U.S. quart of antifreeze ) to increase a 30 % antifreeze volume to 40 % if the total volume of the mixture is 6 quarts
10 liters.
4 litres
The original solution has 3.6 quarts of antifreeze in it. The equation then becomes (3.6 + x)/(12 + x) = 0.40, where x is the amount of antifreeze added. X is then equal to 2.
40.8 grams