Question

It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

Answer 1

The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution. 
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
.8x = 14
x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

Answer 2

Answer:

17.5 litres removed and 17.5 litres of pure antifreeze added

Explanation:

Let k equal the amount of the solution to be removed

amount of antifreeze to be added

0.2(70 - x) + x = 0.4(70)

14 - 0.2x +x = 28

0.8x = 28 -14

x = 14/0.8 = 17.5

x = 17.5 litres removed and 17.5 litres of pure antifreeze added