SOLUTION: A certain automobile cooling system contains 11.0 liters of of coolant that is 15% antifreeze. How many liters of mixture must be removed so that, when it is replaced with pure an

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Question 730609: A certain automobile cooling system contains 11.0 liters of of coolant that is 15% antifreeze. How many liters of mixture must be removed so that, when it is replaced with pure antifreeze, a mixture of 25% antifreeze will result?
I know that I have to use the 100% somewhere, but I can't see how to use it. It's very odd seeing this as a subtraction problem.
I tried
15% 11-x .15
100% x 1
25% 11 .25
I need to see it in a table form so that my mind can process it.
thanks Mark
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website!
Before adjustment, 11.0 liters of mixture is in the system and after adjustment, still 11.0 liters of mixture is in the system. Give a variable, say, q, for how much material to remove and how much pure antifreeze to add.

(pure antifreeze ingredient)/(antifreeze solution)=25%, what's wanted.

You want to adjust from 15% to 25%.

Think to understand how that makes sense.

.
The steps should lead to
.

Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
A certain automobile cooling system contains 11.0 liters of of coolant that is 15% antifreeze. How many liters of mixture must be removed so that, when it is replaced with pure antifreeze, a mixture of 25% antifreeze will result?
***
let x=amt of 15% antifreeze to be removed and replaced with pure antifreeze
11-x=amt of 15% antifreeze remaining
..
15%(11-x)+100%x=25%*11
1.65-.15x+x=2.75
.85x=1.1
x≈1.294
amt of 15% antifreeze to be removed and replaced with pure antifreeze≈1.294 liters