SOLUTION: Hello professor I'm having trouble with this word problem. It's necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 60 liters of 2

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Question 657436: Hello professor I'm having trouble with this word problem.
It's necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 60 liters of 20% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
The answer in the book is 15 liters but I dont know how m get the equation. Thank you!!!!!
Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
Let's start by defining our variables.
Let x be the number of liters of the orignal 20% solution in the final mix
Let y by the number of liters of 100% antifreeze in the final mix.
We want to end up with a total of 60 liters of a 40% solution, so we have
x+%2B+y+=+60 or y+=+60+-+x
The final solution will have %2840%2F100%29%2A60+=+24 liters of antifreeze.
x liters of 20% solution will have 0.20x liters of antifreeze, and
y liters of 100% solution will have y liters of antifreeze, so we now have
0.20x + y = 24.
Substituting 60-x for y, we have:
0.20x + 60-x = 24
Combining terms:
-0.80x = -36
Dividing both sides by -0.80:
%28-0.80%2Ax%29%2F-0.80+=+-36%2F-0.80+=+45
So the final solution will have 45 liters of the 20% solution and 60-45 = 15 liters of 100% antifreeze.
Consequently, we need to drain 15 liters of 20% solution and replace it with 15 liters of 100% antifreeze.