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Answered 2008-02-29 17:06:37

I can't think you are talking drinking water pipes because anitfreeze is poisonous. Nor should you have lead pipes or lead solder in drinking water pipes. So either sleeve pipes or heating pipes then. Antifreeze has inhibitors designed to prevent corrosion.

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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.


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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.



It would take approximately 12 gallons of antifreeze to fill a semi tractor.


Find the overall balance of antifreeze: amount of antifreeze to start with + the amount of pure antifreeze to add = amount of antifreeze in the final solution Let X be the amount of pure antifreeze to add in gallons. (70 * 0.25) + X = 0.80 * (70+X) 17.5 + X = 56 + 0.8X 0.2X = 56 - 17.5 0.2X = 38.5 X = 38.5/0.2 = 192.5 gallons


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


Let X = gallons of 50% antifreeze .5X + .1(70) = .4 (X + 70) .5X + 7 = .4X + 28 .1X = 21 X = 210 GALLONS


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.


600 gallons. To solve this think about the amount of antifreeze in the solution. When 10% antifreeze is added to x times as much 80% antifreeze, the resultant percentage antifreeze will be: (10 + 80x)/(1 + x) So to obtain a 70% solution, x will need to solve: (10 + 80x)/(1 + x) = 70 ⇒ 10 + 80x = 70 + 70x ⇒ 10x = 60 ⇒ x = 6 So to 100 gallons of 10% antifreeze solution, 100 x 6 = 600 gallons of 80% antifreeze solution will be needed to make it a 70% solution. Consider adding 100 gallons of 10% antifreeze and 100 gallons of 80% antifreeze together and then taking half the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 80 gallons) ÷ 2 = 45 gallons, that is a 45% mixture. Now consider adding 100 gallons of 10% antifreeze and 200 gallons of 80% antifreeze together and then taking a third of the resultant solution, that is so that you have 100 gallons of the mixture; the amount of antifreeze in this 100 gallons is (10 gallons + 2 x 80 gallons) ÷ 3 = 562/3 gallons, that is a 562/3% mixture. If x times as much of the 80% antifreeze is added to the If 1 gallon of 10% antifreeze was added together with 80% antifreeze, the resultant mixture will still be (10+90)/2 = 45%. So if x times as much of the 80% Consider the amount of antifreeze in the solution. When two solutions of the same volume is added, the amount of antifreeze in the same volume is half the new amount. Consider the ratio of anti-freeze to water in each of the solutions in fraction form of antifreeze/water: 10% is 10/90 70% is 70/30 80% is 80/20 When two ratios are added together, the amount of antifreeze in the solution is added together and the amount of water is added together to give some odd fraction maths: 10% + 80% = (10+80)/(90+20) = 90/110 =


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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 much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?


You would use 100/3 = 33.33 gallons of the 40 percent antifreeze solution and 50/3 = 16.67 gallons of the 10 percent antifreeze solution.


Suppose x gallons of 90% antifreeze is mixed. Then total volume of mixture = x + 70 gallons and total antifreeze in mixture = 0.9*x + 0.15*70 = 0.9x + 10.5 Concentration of mixture = (0.9x + 10.5)/(x + 70) which is 80% or 0.8 So 0.9x + 10.5 = 0.8x + 56 that is 0.1x = 45.5 or x = 455 gallons


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.


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.


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.


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Suppose G gallons are required. G gallons @ 60% implies 0.60*G gallons of the active ingredient. 70 gallons @ 30% implies 70*0.3 = 21 gallons of the active ingredient. This makes a total of (G + 70) gallons containing (0.6G + 21) gallons of active ingredient. The mixture is (0.6G + 21)/(G + 70) which is to be 50% or 0.5 So 0.6G + 21 = 0.5G + 35 or 0.1G = 14 which gives G = 140 gallons


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