SOLUTION: A radiator hold 15 liters. How much pure antifreeze must be added to a mixture that is 10% antifreeze to make enough of a 25% mixture to fill the radiator?
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Question 286884: A radiator hold 15 liters. How much pure antifreeze must be added to a mixture that is 10% antifreeze to make enough of a 25% mixture to fill the radiator? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of pure antifreeze that's added to the 10% mixture
Then 15-x=amount of 10% antifreeze that is used
Now we know that the amount of pure antifreeze in the in the antifreeze that's added (x) plus the amount of pure antifreeze in the 10% mixture(0.10(15-x)) has to equal the amount of pure antifreeze in the final mixture (0.25*15), so:
x+0.10(15-x)=0.25*15
x+1.5-0.10x=3.75
0.90x=3.75-1.5
0.90x=2.25
x=2.5 liters-----------------amount of pure antifreeze needed
CK
2.5+0.10(15-2.5)=0.25*15
2.5+1.25=3.75
3.75=3.75
Hope this helps----ptaylor
