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%
7 gallons
50 gallons @ 3% must be added.
5 gallons and 1 quart plus 7 gallons and 3 quarts is 12 gallons and 1 quart.
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.
19
7 over 12 as a percent = 58.33%% rate:= 7/12 * 100%= 0.5833 * 100%= 58.33%
The increase from 7 to 12 is about 71.428%
7% of 12 = 7% * 12 = 0.07 * 12 = 0.84
-2
Let x represent the amount of 12% solution and (10-x) represent the amount of 20% solution. The equation to solve is: 0.12x + 0.20(10-x) = 0.14(10). Solving for x gives x = 4, so you need 4 gallons of the 12% solution and 6 gallons of the 20% solution to make 10 gallons of the 14% solution.
NO. -0.7% is greater than -12%.