A car radiator has a 6-liter capacity. If the liquid in the radiators 40% antifreeze, how much liquid must be replaced with pure anitfreeze to bring the mixture up to a 50% solutions? If I told you the answer is 1 liter
10/3 qrts
To solve this problem, you need to calculate how much liquid needs to be replaced with pure antifreeze.
First, you need to determine the actual amount of antifreeze currently in the radiator. Since the radiator has a 6-liter capacity and the liquid in it is 40% antifreeze, you can find the amount of antifreeze by multiplying the capacity by the percentage:
6 liters x 40% = 2.4 liters of antifreeze
Next, you need to calculate how much pure antifreeze is needed to reach a 50% solution. Since the desired mixture is 50% antifreeze and you currently have 2.4 liters of antifreeze, you can find the amount needed by subtracting the current amount from the desired amount:
Desired amount - Current amount = Amount needed
50% of total capacity - 2.4 liters = Amount needed
To find the desired amount, multiply the total capacity by 50%:
6 liters x 50% = 3 liters
Then, subtract the current amount from the desired amount:
3 liters - 2.4 liters = 0.6 liters
Therefore, you need to replace 0.6 liters of the liquid in the radiator with pure antifreeze to reach a 50% solution.
However, you mentioned that the answer is 1 liter. It is possible that there is an error or additional information provided in the question that is not accounted for in this explanation. Please double-check the question or provide more information if needed.