The radiator in Natalies car contains 6.3L of antifreeze and water.This mixture is 30% antifreeze.How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 505 antifreeze?
What is 505 antifreeze ??
That is suppose to be 50% antifreeze.
To determine how much of the mixture Natalie should drain and replace with pure antifreeze, we can follow these steps:
Step 1: Calculate the initial amount of antifreeze in the mixture.
The initial mixture consists of 6.3L, and it is 30% antifreeze. Therefore, the initial amount of antifreeze is 6.3L * (30/100) = 1.89L.
Step 2: Calculate the final amount of antifreeze needed in the new mixture.
In the final mixture, Natalie wants a mixture of 50% antifreeze, which we'll call xL. The remaining amount will be a mixture of 50% water.
Step 3: Set up an equation to represent the scenario.
The equation will show the relationship between the initial and final amounts of antifreeze. Since Natalie is not changing the total amount of liquid in the radiator, the equation is as follows:
1.89L - 0.3xL + xL = 0.505 * (6.3L)
This equation states that the initial amount of antifreeze minus the amount of antifreeze drained, plus the amount of pure antifreeze added, should equal the final amount of antifreeze required.
Step 4: Solve the equation to find the amount of mixture to drain and replace.
Simplifying the equation:
1.89L - 0.3xL + xL = 3.1815L
Combining like terms:
0.7xL = 1.2915L
Divide both sides by 0.7 to solve for x:
xL = 1.2915L / 0.7
xL = 1.845L
Hence, Natalie should drain 1.845L of the mixture and replace it with pure antifreeze to obtain a mixture of 50.5% antifreeze.