Question

the radiator in a car is filled with a solution of 75% antifreeze and 25% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 4.1 liters, how much coolant in liters must be drained and replace with pure water to reduce the antifreeze concentration to 50%?

Answer 1

The amount of coolant solution in liters that must be drained and replace with pure water to reduce the antifreeze concentration to 50% is;

`1.37\text{ liters}`

Step-by-step explanation:

Given that the radiator in a car is filled with a solution of 75% antifreeze and 25% water, and the capacity of the radiator is 4.1 liters.

The initial amount of water and antifreeze in the radiator is;

`\begin{gathered} \text{ Antifreeze = 0.75}*4.1\text{ liters = 3.075 liters} \\ \text{water =0.25}*4.1\text{ liters = 1.025 liters} \end{gathered}`

We want to reduce the concentration of antifreeze to 50% wihich is;

`\begin{gathered} \text{ final amount of antifr}eeze=0.50*4.1\text{ } \\ =2.05\text{ liters} \end{gathered}`

The amount of antifreeze we need to remove is;

`3.075\text{ liters - 2.05 liters =1.025 liters}`

let x represent the amount of coolant solution we need to remove then;

`\begin{gathered} 0.75x=1.025 \\ x=(1.025)/(0.75) \\ x=1.36666 \\ x=1.37\text{ liters (to two decimal place)} \end{gathered}`

Therefore, the amount of coolant solution in liters that must be drained and replace with pure water to reduce the antifreeze concentration to 50% is;

`1.37\text{ liters}`