4.2 quarts
2 gallons.
0.25 gallons of water (or 1 quart)
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.
10
25 gallons
0.6 of a pint.
To obtain a solution that is twenty percent antifreeze, you would add 4 parts water to 1 part antifreeze. This means that for every 1 part of antifreeze, you would add 4 parts of water. This would result in a total of 5 parts of solution, with 1 part being antifreeze and 4 parts being water, achieving a solution that is twenty percent antifreeze.
2 gallons.
0.25 gallons of water (or 1 quart)
You need 1 1/3 quarts of pure antifreeze.
You will need 3.2 gallons.
A pharmacist mixed a 20 percent solution with a 30 percent solution to obtain 100 liters of a 24 percent solution. How much of the 20 percent solution did the pharmacist use in the mixture (in liters).
Approx 1.86 gallons.
10
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%.
25 gallons
10 liters.