SOLUTION: bobs cars radiator has a capacity of 25L. it is currently full of water mixed with antifreeze one fifth of the volume of the solution is antifreeze. the weather is getting colder a

Algebra ->  Linear-equations -> SOLUTION: bobs cars radiator has a capacity of 25L. it is currently full of water mixed with antifreeze one fifth of the volume of the solution is antifreeze. the weather is getting colder a      Log On


   

Question 846874: bobs cars radiator has a capacity of 25L. it is currently full of water mixed with antifreeze one fifth of the volume of the solution is antifreeze. the weather is getting colder and bob wants to change the mixture so that the solution will be three fifths of antifreeze. Determine how much of the solution in his radiator bob needs to drain and replace with antifreeze.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Let C = capacity and amount of volume of solution in the radiator.
Let H = fraction of antifreeze solution as pure antifreeze the radiator HAS initially;
Let W = fraction of antifreeze solution WANTED.
Let v = volume of antifreeze solution to drain and replace with pure antifreeze.
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The values of the described variables are:
Let C = 25 L
Let H = 1/5
Let W = 3/5
Let v = unknown to solve, Liters
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highlight%28%28CH-Hv%2B1%2Av%29%2FC=W%29.
If you can understand how to obtain this equation, then that is most of the solution taken care of. You then just solve for v, and substitute the values for the other variables and compute the value.
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If you do not understand the equation, then some help is needed for it. The equation simply is the ratio of the amount of pure antifreeze to the quantity of antifreeze solution to be held in the radiator.