10 gallons.
t = number of liters of 30% acid solution s = number of liters of 60% acid solution t+s=57 .30t+.60s=.50*57 t=57-s .30(57-s)+.60s=.50*57 .30*57 -.30s +.60s = .50*57 .30s = .50*57 - .30*57 = .20*57 s = .20*57/.30 = 38 liters of 30% solution t = 57 - s = 57-38 = 19 liters of 60% 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.
To get a 2% acid solution, you need to dilute the 50% acid solution with water. Since the final volume is 2 gallons, you will need to mix 2 gallons of water with the 2 gallons of 50% acid solution to get a 2% acid solution.
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
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
20 is what percent of 50= 20 / 50= 0.4Converting decimal to a percentage:0.4 * 100 = 40%
50 Pints = 6.25 Gallons
25 gallons
50 percent of 20 cents = 10 cents
2 gallons, 10/5 = 2
70gallons