The cooling system of a car has a capacity of 20 liters. If the system is currently filled with a mixture that is 60% antifreeze, how much of this mixture should be drained and replaced with pure antifreeze so that the system is filled with a solution that is 70% antifreeze?
.60(20-x) + x = .70*20
To solve this problem, we need to determine how much of the mixture needs to be drained and replaced with pure antifreeze in order to achieve a solution with 70% antifreeze.
Let's break down the problem step by step:
Step 1: Determine the amount of antifreeze in the original mixture.
Since the original mixture is 60% antifreeze and the total capacity of the cooling system is 20 liters, the amount of antifreeze in the original mixture is 0.6 * 20 = 12 liters.
Step 2: Determine the amount of antifreeze needed in the final solution.
In the final solution, we want to achieve a concentration of 70% antifreeze. Since the total capacity of the cooling system remains 20 liters, the amount of antifreeze needed is 0.7 * 20 = 14 liters.
Step 3: Determine the amount of the original mixture that needs to be replaced.
We want to find out how much of the original mixture should be drained and replaced, so let's call this amount x.
Step 4: Setting up the equation:
The amount of antifreeze that will be drained from the original mixture is 0.6x (since 60% of x is antifreeze) and the amount of pure antifreeze that will be added is x (since we're replacing an equal amount of mixture with pure antifreeze).
So, the equation becomes:
12 - 0.6x + x = 14
Step 5: Solving the equation:
Combine like terms:
-0.6x + x = 14 - 12
0.4x = 2
Divide both sides by 0.4:
x = 2 / 0.4
x = 5
Therefore, 5 liters of the original mixture should be drained and replaced with pure antifreeze to achieve a solution with 70% antifreeze.