Capacity of car radiator The capacity of my car radiator is $24$ quarts. The mixture of antifreeze and water in the radiator was $25$%? antifreeze.Since the winter is approaching I wanted a larger percentage of antifreeze. My mechanic suggested a $38$% mixture. I therefore needed to drain off some of the current mixture and replace it with pure antifreeze.How much of the current mixture do I need to drain off? I set up an equation and got $4.16$ quarts. I'm just not sure if that is the correct answer and way to best solve this problem. $x$ = amount drained by $x$ quarts The amount remaining would be $(24-x)$ quarts. The drained amount is replaced by $100$% anti-freeze meaning: $$100x + 25(24-x) = 38*24$$ $$75x=24(38-25)=24*13$$ $$x= 24 * \frac{13}{75}$$ $$X=4.16 Quarts$$ let amount drained be x quarts, amount remaining = $(24-x)$ quarts the amount drained is replaced by $100$% anti-freeze, so $$100x + 25(24-x) = 38*24$$ $$75x = 24(38-25) = 24*13$$ $$x = 24*\frac{13}{75} $$

I agree with your result.

Desired amount of antifreeze: $0.38(24) =9.12$ quarts. Amount of antifreeze currently in radiator: $0.25(24) =6.0$ quarts. Total volume of radiator: $24$ quarts.

Let $v$ equal the amount of $25\%$ mixture to drain and of pure antifreeze to add.

$$\frac{6}{24}(24-v) + v = 9.12$$

Solve for $v$ to get $v=4.16$ quarts. So drain $4.16$ quarts of the current mixture and add $4.16$ quarts of the pure antifreeze to get your desired $38\%$ mixture for the winter.

Hope this helps,

Paul Safier