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70gallons

Q: 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?

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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.

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

The original solution has 3.6 quarts of antifreeze in it. The equation then becomes (3.6 + x)/(12 + x) = 0.40, where x is the amount of antifreeze added. X is then equal to 2.

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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?

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

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.

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

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.

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.

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.

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 =

.70x+.10(60)=.60(x+60) let x= number of gallon at 70% .7x+6=.6x+36 .1x=30 x=300