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How many quarts of pure antifreeze must be added to 4 quarts of a 20 antifreeze solution to obtain a 40 antifreeze and solution?
Wiki User
∙ 6y ago
You need 1 1/3 quarts of pure antifreeze.
Wiki User
∙ 6y ago
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How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How many liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
60% solution contains 6/10 x 30 ie 18 litres 18 + A = 3/4 (30 + A) 72 + 4A = 90 + 3A 4A - 3A = 90 - 72 A = 18 ie add another 18 litres, giving 36 litres out of 48 which is the required 75%.
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 liters of a 7 percent solution of salt should be added to a 19 percent solution in order to obtain 456 liters of a 15 percent solution?
This is an algebra problem. There will be 2 equations and 2 unknowns. Let x = amount of 7% solution, and y = amount of 19% solution. .07x + .19y = .15*456. x + y = 456; or x = 456-y which will substitute into the first equation. .07(456-y) +.19y = .15*456. 31.92-.07y+.19y = 68.4. .12y = 36.48. y = 304L (amount of 19% solution to add). x = 456-y. x = 456 - 304. x = 152L (amount of 7% solution to add).
What must be added to a to obtain 7a?
UM 6a? Can you be a little more specific maybe? But 6a plus a is 7a
How many quartz of pure antifreeze must be added to 6 quartz of a 10 percent antifreezesolution to obtain a 20 percent antifreeze solution?
4.2 quarts
A car's radiator has a capacity of 15 quarts and it currently contains 12 quarts of a 30 percent antifreeze solution. How many quarts of pure antifreeze must be added to strengthen the solution 40 per?
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.
How many quarts of pure antifreeze must be added to 6 quarts of a 30 percent antifreeze solution to abtain a 40 percent antifreeze solution?
If my math is correct it would take an additional ( .6 of a U.S. quart of antifreeze ) to increase a 30 % antifreeze volume to 40 % if the total volume of the mixture is 6 quarts
How many qt of pure antifreeze must be added to 3 qt of a 10 percent antifreeze solution obtain a 20 percent antifreeze solution?
0.6 of a pint.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60 percent antifreeze?
2 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 35 and 8203 antifreeze?
Approx 1.86 gallons.
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
How much pure 100 percentage antifreeze should be added to five quarts of 20 percentage solution to raise its concentration to 25 percentage antifreeze solution?
I'd go buy a bottle of pre-mixed antifreeze, and just fill it up with that. Less hassle, works great.
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 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 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 liters of pure antifreeze must be added to 30L of 60 percent antifreeze solution to obtain a 75 percent solution?
60% solution contains 6/10 x 30 ie 18 litres 18 + A = 3/4 (30 + A) 72 + 4A = 90 + 3A 4A - 3A = 90 - 72 A = 18 ie add another 18 litres, giving 36 litres out of 48 which is the required 75%.
