the radiator in Johns car contains 16 L of antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so there will be a mixture of 50% antifreeze.
better read what you posted . . .
This is word for word what my teacher wrote
if the radiator already contains pure antifreeze, how would adding more antifreeze change anything?
He purposely made a problem with not enough info.
To find out how much of the mixture in John's car radiator should be drained and replaced with pure antifreeze, we can follow these steps:
Step 1: Determine the amount of antifreeze currently in the mixture:
Since the radiator contains 16 L of antifreeze, the total amount of mixture in the radiator is also 16 L.
Step 2: Set up the equation:
Let's assume x represents the amount of mixture that needs to be drained and replaced with pure antifreeze.
The amount of pure antifreeze in the drained mixture is x L.
The amount of original mixture left in the radiator is (16 - x) L.
Step 3: Calculate the new concentration:
To achieve a mixture of 50% antifreeze, we need to ensure that the amount of pure antifreeze equals 50% of the total mixture volume after the replacement.
The amount of pure antifreeze in the new mixture is:
0.5 * (16 - x) L.
Step 4: Set up the equation for the new concentration:
We need to equate the pure antifreeze in the drained mixture to the pure antifreeze in the new mixture.
x L (pure antifreeze drained) = 0.5 * (16 - x) L (pure antifreeze in the new mixture)
Step 5: Solve for x:
Now, we can solve the equation to find the value of x.
x = 0.5 * (16 - x)
x = 8 - 0.5x
1.5x = 8
x = 8 / 1.5
x = 5.33 (rounded to two decimal places)
Therefore, approximately 5.33 liters of the mixture should be drained and replaced with pure antifreeze to achieve a mixture of 50% antifreeze.