<PRE><B>Question 467328</B>
It is necessary to have a 40% antifreeze solution in the radiator of a certain car.
 The radiator now has 40 liters of 20% solution.
 How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
:
Let x = amt of original anti-freeze to be drained, also the amt of pure antifreeze to be added
:
Write a decimal amt of antifreeze equation
:
.20(40-x) + 1x = .40(40)
8 - .2x + 1x = 16
-.2x + 1x = 16 - 8
.8x = 8
x = {{{8/.8}}}
x = 10 liters removed and 10 liters of pure antifreeze to be added
:
:
Check this in the original equation
.20(40-10) + 10 = .4(40)
.2(30) + 10 = 16
6 + 10 = 16; confirms our solution of x = 10</PRE>