<PRE><B>Question 88593</B>
It is 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 liter of this should be drained and replaced with 100% antifreeze to get the desired strength? Please help me with this problem.
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Let x = amt drained and amt replaced
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The remaining amt of 20% solution will = (60-x)
:
.20(60-x) + 1,00(x) = .40(60)
:
12 - .2x + 1x = 24
:
1x - .2x = 24 - 12
:
.8x = 12
:
x = 12/.8
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x = 15 liters drained and replace by 15 liters of 100% solution
:
:
Check solution, amt of 20% solution = 60-15 = 45 liters
.20(45) + 1.00(15) = .40(60)
9 + 15 = 24; proves the solution
:
Not really that hard, right??</PRE>