Respuesta :
Answer:
The amount of coolant solution in liters that must be drained and replace with pure water to reduce the antifreeze concentration to 50% is;
[tex]1.37\text{ liters}[/tex]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;
[tex]\begin{gathered} \text{ Antifreeze = 0.75}\times4.1\text{ liters = 3.075 liters} \\ \text{water =0.25}\times4.1\text{ liters = 1.025 liters} \end{gathered}[/tex]We want to reduce the concentration of antifreeze to 50% wihich is;
[tex]\begin{gathered} \text{ final amount of antifr}eeze=0.50\times4.1\text{ } \\ =2.05\text{ liters} \end{gathered}[/tex]The amount of antifreeze we need to remove is;
[tex]3.075\text{ liters - 2.05 liters =1.025 liters}[/tex]let x represent the amount of coolant solution we need to remove then;
[tex]\begin{gathered} 0.75x=1.025 \\ x=\frac{1.025}{0.75} \\ x=1.36666 \\ x=1.37\text{ liters (to two decimal place)} \end{gathered}[/tex]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;
[tex]1.37\text{ liters}[/tex]