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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 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
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 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
How many gallons of a 50 percent antifreeze solution must be mixed with 90 gallons of 30 percent antifreeze to get a mixture that is 40 percent antifreeze?
Suppose x gallons of 50% antifreeze is added to 90 gallons of 30% antifreeze. Total active ingredient in mixture = 0.5*x + 0.3*90 = 0.5*x + 27 Total volume of mixture = x + 90 gallons Concentration of mixture = (0.5*x + 27)/(x + 90) = (50x + 2700)/(x + 90) percent This should be 40% So 40 = (50x + 2700)/(x + 90) That is, 40x + 3600 = 50x + 2700 so that 900 = 10x and x = 90 So, add 90 gallons of the 50% antifreeze. In fact, for this particular example, there is a short cut. You are adding antifreeze of two concentrations to get a result whose concentration is EXACTLY halfway. This requires the same amount of the two concentrations to be added together.
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 80 percent antifreeze?
0.25 gallons of water (or 1 quart)
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 much water should be added to 30 gallons of a solution that is 70 percent antifreeze in order to get a mixture that is 60 percent antifreeze?
First off do not use tap water. Use only distilled water. You will need to add 3 gallons of distilled water to the solution to get 60% antifreeze and 40% distilled water or a 60/40 mix.
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 many gallons of 3 percent salt solution must be mixed with 50 gallons of 7 percent salt solution to obtain a 5 percent salt solution?
50 gallons @ 3% must be added.
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 gallons of a 50 percent antifreeze solution must be mixed with 90 gallons of 30 percent antifreeze to get a mixture that is 40 percent antifreeze?
Suppose x gallons of 50% antifreeze is added to 90 gallons of 30% antifreeze. Total active ingredient in mixture = 0.5*x + 0.3*90 = 0.5*x + 27 Total volume of mixture = x + 90 gallons Concentration of mixture = (0.5*x + 27)/(x + 90) = (50x + 2700)/(x + 90) percent This should be 40% So 40 = (50x + 2700)/(x + 90) That is, 40x + 3600 = 50x + 2700 so that 900 = 10x and x = 90 So, add 90 gallons of the 50% antifreeze. In fact, for this particular example, there is a short cut. You are adding antifreeze of two concentrations to get a result whose concentration is EXACTLY halfway. This requires the same amount of the two concentrations to be added together.
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
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 quarts of pure antifreeze must be added to 6 quarts of a 10 percent antifreeze solution to obtain a 20 percent antifreeze solution?
To find out how many quarts of pure antifreeze must be added to 6 quarts of a 10% antifreeze solution to obtain a 20% solution, we can set up the equation. The initial amount of antifreeze in the solution is (0.10 \times 6 = 0.6) quarts. Let (x) be the amount of pure antifreeze to add. The final solution will be (6 + x) quarts, and we need the total antifreeze to equal (0.20(6 + x)). Setting up the equation (0.6 + x = 0.20(6 + x)) and solving gives (x = 1.2) quarts. Thus, 1.2 quarts of pure antifreeze must be added.
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.
