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How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
Wiki User
∙ 14y ago
How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?
Wiki User
∙ 14y ago
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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 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 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 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 much antifreeze must be added to 12 gallons of 20 percent antifreeze to make a 40 percent antifreeze solution?
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
How many gallons of a 80 percent antifreeze solution must be mixed with 90 gallons of 25 percent antifreeze to get a mixture that is 70 percent antifreeze?
Let X = gallons of 80% antifreeze Then 0.80X + .25 (90) = .70( 90 + X) .80X + 22.5 = 63 + .70X .1X = 41.5 X = 415 gallons
How many gallons of 50 percent antifreeze solution must be mixed with 70 gallons of 10 percent antifreeze to get a mixture that is 40 percent antifreeze?
Let X = gallons of 50% antifreeze .5X + .1(70) = .4 (X + 70) .5X + 7 = .4X + 28 .1X = 21 X = 210 GALLONS
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.
How many gallons of a 80 percent antifreeze solution must be mixed with 60 gallon of 15 percent antifreeze to get a mixture that's 70 percent antifreeze?
330 gallons of 80% antifreeze mixed with 60 gallons of 15% antifreeze will provide 390 gallons of 70% antifreeze. Let the multiplier of 60 gallons be x, then considering the percentages of antifreeze in the solutions: (15 + 80x) ÷ (1 + x) = 70 ⇒ 15 + 80x = 70 + 70x ⇒ 10x = 55 ⇒ x = 5.5 Therefore mix 60 x 5.5 = 330 gallons of 80% antifreeze.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
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 50 percent antifreeze solution must be mixed with 70 gallons of 25 percent antifreeze to get a mixture that is 40 percent antifreeze?
105 gallons. Let the multiple of 70 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 70 x 1.5 = 105 gallons.
