Let x be the liters of the 30% acid solution and y be the liters of the 60% acid solution. We can set up a system of equations: x + y = 50 (total liters) and 0.3x + 0.6y = 0.57*50 (acid content). Solving this system of equations, we find that x = 20 liters of the 30% acid solution and y = 30 liters of the 60% acid solution.
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.
50 US gallons is 189.27 litres.
No, sand covers about 20 percent pf deserts.
There are approximately 4.40 Imperial gallons in 20 liters.
You will never get it there. However much 50% you have, any amount of 20% in the mix will reduce the total below 50%. So, throw away the 20% and just use the 50% on its own.
50 percent of (100/20) = 50 percent of 5 = 2.5 (50 percent of 100) / 20 = (50 / 20) = 2.5 as well
20% of 17 gallons= 20% * 17= 0.2 * 17= 3.4 gallons
50 gallons @ 3% must be added.
20% of 190 gallons of water= 20% * 190= 0.2 * 190= 38 gallons of water
20 is what percent of 50= 20 / 50= 0.4Converting decimal to a percentage:0.4 * 100 = 40%
50 Pints = 6.25 Gallons
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
50 percent of 20 cents = 10 cents
25 gallons
2 gallons, 10/5 = 2
50% of 20 = 50% * 20 = 0.5 * 20 = 10