Real-world math problem: A car's cooling system has an antifreeze ratio of 30% antifreeze/70% water. I want it to be a 50/50 ratio. Without draining the entire system, how much antifreeze should I drain out (if I replace that volume with a 100% antifreeze mixture) to make the system a 50/50 ratio. The capacity of the system is 11.6 quarts and is currently a 30% ratio, so it has 3.48 qt antifreeze/8.12 qt water.

Question

Real-world math problem:  A car's cooling system has an antifreeze ratio of 30% antifreeze/70% water.  I want it to be a 50/50 ratio.  Without draining the entire system, how much antifreeze should I drain out (if I replace that volume with a 100% antifreeze mixture) to make the system a 50/50 ratio.  The capacity of the system is 11.6 quarts and is currently a 30% ratio, so it has 3.48 qt antifreeze/8.12 qt water.

anonymous · Answer

Drain 3.31 quarts then fill up with 3.31 quarts 100% antifreeze and the new ratio will be 50/50

anonymous · Answer

Awesome, thank you so much.  Can I ask what equation you used to determine that so I can use it for other cars?

anonymous · Answer

First work out what 50:50 would be
$$\frac{ 11.6 }{ 2 }= 5.8$$



$$8.12water-5.8water=2.32water$$

$$\frac{ 2.32 water }{ 7 }10=3.31$$

then Check:

11.6-3.31=8.29

$$\frac{ 8.29 }{ 7 }=5.8$$

which is now 50% water (if you go back and check the beginning)

11.6 -8.29 = 3.31

which means you need to drain 3.31 quarts and then fill with 3.31 quarts 100% antifreeze to make a 50:50 ratio

anonymous · Answer

Perfect, thanks again!