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7x/10 (That's 7/10 times x). So if x = 10 gallons, then there are 7 gallons of bleach in the solution. (7gal/10gal = 0.70 = 70%). This answer if correct if the percentage is by volume. Otherwise, the question can not be answered without knowledge of the basis of the percentage.
What else can I help you with?
How many grams of active ingredient will you need to prepare 2.5 gallons of a 45.6 percent solution?
4314.9 grams
How many gallons of a 50 percent antifreeze solution must be mixed with 90 gallons of 25 percent antifreeze to get a mixture of 40 percent antifreeze?
135 gallons. Let the multiple of 90 gallons to add be x, then using the percentages of antifreeze in the solutions: (25 + 50x) ÷ (1 + x) = 40 ⇒ 25 + 50x = 40 + 40x ⇒ 10x = 15 ⇒ x = 1.5 Therefore add 90 x 1.5 = 135 gallons.
How many gallons of a solution that is 75 antifreeze must be mixed with 4 gallons of a 30 solution to obtain a mixture that is 50 antifreeze?
You will need 3.2 gallons.
How many gallons of a 70 percent antifreeze solution must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 60 percent antifreeze?
Suppose G gallons are required. At 70% concentration, this will contain 0.7*G gallons of antifreeze. The 70 gallons of 25% contains 17.5 gallons of the active ingredient. In total, then, there are (G + 70) gallons containing 0.7G + 17.5 gallons of antifreeze and this represents 60%. So (0.7G + 17.5) / (G + 70) = 60/100 10*(0.7G + 17.5) = 6*(G + 70) 7G + 175 = 6G + 420 G = 420 - 175 = 245. Answer: 245 gallons.
How many ml should be added to 100ml of a 25 percent solution to get produce a 20 percent solution?
add 25ml more of solution x * 20 = 100 * 25 x = 25
How many gallons of a 9 percent salt solution must be mixed with 25 gallons of a 31 percent solution to obtain a 20 percent solution?
25 gallons
How many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
How many gallons of a 90 percent antifreze solution must be mixed with 90 gallons of 10 percent antifreeze to get a mixture that is 80 percent antifreeze?
630
How many gallons of a 50 percent antifreeze solution must be mixed with 70 gallons of 30 percent antifreeze to get a mixture that is 40 percent antifreeze?
70gallons
How many gallons of a 90 percent antifreeze solution must be mixed with 80 gallons of 25 percent antifreeze to get a mixture that is 80 percent antifreeze?
614
How many gallons of a 12 percent indicator solution must be mixed with a 20 percent indicator solution to get 10 gallons of a 14 percent 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.
How many gallons of a 80 percent solution must mix with 70 gallons of 10 percent solution to get 70 percent solution?
420 gallons. Here's how to find it; If X = Gallons of 80% mix, 0.8*X + 0.1*70 = 0.7*(X+70) Which we can then solve, 0.8X + 7 = 0.7X + 49 0.1X = 42 X = 420
How many gallons of a 80 percent acid solution must be mixed with 60 gallons of a 16 percent solution to obtain a solution that is 60 percent?
Let x be the gallons of the 80% acid solution needed. The amount of acid in the 80% solution is 0.8x, and the amount in the 16% solution is 0.16*60=9.6. We want a total of (x+60) gallons of solution with 60% acid, so we have the equation 0.8x + 9.6 = 0.6(x+60). Solving for x gives x = 24 gallons.
How many gallons of a 80 percent antifreeze solution must be mixed with 70 gallons of 20 percent antifreeze to get a mixture that is 70 percent antifreeze?
70 gallons of 20% solution contains 70*0.2 = 14 gallons of antifreeze. Suppose you need G gallons of the 80% antifreeze solution. This will contain 0.8*G gallons of antifreeze. Total volume of solution = G + 70 gallons Volume of antifreeze required in this solutions to make it a 70% solution is 0.7*(G + 70) = 0.7G + 49 gallons. Volume of antifreeze = 14 + 0.8G gallons So 0.7G + 49 = 14 + 0.8G 0.7G + 35 = 0.8G 35 = 0.1G 350 = G Answer: 350 gallons.
How many gallons of a 10 percent ammonia solution should be mixed with 50 gallons of a 30 percent ammonia solution to make a 15 percent ammonia solution?
Let x represent the gallons of 10% ammonia solution. The total volume of the mixture is x + 50 gallons. The equation for the mixture is: 0.10x + 0.30(50) = 0.15(x + 50). Solving this equation gives x = 50 gallons of the 10% ammonia solution needed.
How many grams of active ingredient will you need to prepare 2.5 gallons of a 45.6 percent solution?
4314.9 grams
How many gallons of a 90 percent antifreeze solution must be mixed with 90 gallons of 15 percent antifreeze to get a mixture that is 80 percent antifreeze?
Let M equal the gallons of the 90% mixing solution. Let F equal the gallons of the final solution. So:90 + M = F.Also, the number of gallons of pure antifreeze in the final will equal the sum of the gallons of antifreeze in the two mixing parts:Original solution: ( 90 gal )*(0.15) = 13.5 gal [0.15 represents 15%]Mixing solution: M*0.90Final solution: F*0.80So 13.5 + M*0.90 = F*0.80Now you have 2 linear equations and 2 unknowns, you can solve for M & F, using your favorite method: M = 585 and F = 675. Add 585 gallons of the 90% to get 675 gallons of 80% solution.
