12 L of a 25% antifreeze solution. How many liters(L) will have to be replaced with pure antifreeze to make a new solution of 40% antifreeze

Question

Empahsize on the word replaced

Answer 1

You have 12 L of a 25% antifreeze solution.

This means you have 75% water.

You now have 3 L of antifreeze and 9 L of water.

You want the new solution to be 40% antifreeze and 60% water.

You want 60%*12=7.2 L of water in your new mixture.

You need to drain enough mixture so that you lose 9-7.2=1.8 L of water.

Don't forget that you lose water and antifreeze when you drain the mixture.

Let x=# of liters you want to drain. These liters you are draining are made up of 75% water.

How many liters of a mixture that is 75% water must you drain to lose 1.8 L of water ?

75%*x=1.8 L

0.75x=1.8

75x=180

x=180/75

x=2 30/75=2 2/5=2.4 L of solution that must be drained.

check: You drained 2.4 L of which 25% is antifreeze, or 1/4 of 2.4=0.6 L of antifreeze lost.

You now have 3-0.6=2.4 L of antifreeze and you are going to add 2.4 L of pure antifreeze for a total of

2.4+2.4=4.8 L of antifreeze. (you drained 2.4 L of solution and are replacing it with pure antifreeze)

If you have 12 L of a mixture and 4.8 L are antifreeze, what percent is antifreeze ?

4.8/12=48/120=4/10=40/100=40%

You also lost 75%*2.4=1.8 L of water leaving you with 9-1.8=7.2 L of water which is 7.2/12=72/120=6/10=60/100=60% water

 

Answer 2

The original solution is 12L, 25% of which is antifreeze.

You want the solution to be 40% antifreeze: 0.40*12 = 4.8L

You will remove x liters of the 25% solution and replace it with x liters pure (100%) antifreeze.

 

25%*(12-x) + 100%*x = 40%*12

0.25(12-x) + x = 4.8L

 

You should be able to solve for x now!

 

If you have questions, please comment.