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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?
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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 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 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 quarts of pure antifreeze must be added to 5 quarts of 20 percent antifreeze to make a 30 percent antifreeze solution?
In the original solution, you have 5 quarts total, 20% antifreeze. So 20% * 5 = 1. You have 4 qt water and 1 qt antifreeze. You will add an amount A, to this mixture. When you add, the new total volume (in quarts) is 5 + A, and the percentage is:(1 + A)/(5 + A) x 100%. So when you start out at A = zero, you have 1/5 x 100% = 20%.So set up the equation (1 + A)/(5 + A) * 100 = 30, and solve for A.Multiply both sides by (5+A) and isolate the term A. Note that since you are adding, the value of A will never be negative (so you don't have to worry about the denominator (5+A) being zero.Answer = 5/7 (quarts) = 0.714285714285155 quarts.
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 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 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 80 percent antifreeze solution must be mixed with 100 gallons of 10 percent antifreeze to get a mixture that is 70 percent antifreeze?
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 =
How many gallons of a 90 percent antifreeze solution must be mixed with 70 gallons of 15 percent antifreeze to get a mixture that is 80 percent antifreeze?
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
Is antifreeze a solution or a heterogenous mixture?
solution
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.
What type of homogeneous mixture is antifreeze?
Solution.
