How do you solve a question like this?

Question

The cooling system of a certain foreign-made car has a capacity of 15 liters. If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze?

Answer 1

current ratio = 6:9
... 6 is 40% of 15l (the 9 is the rest of whatever is in the mixture)

you need a 9:6 ratio
... where 9 represents the qty of antifreeze)

This is a straight 2:3 & 3:2 switch so we can take 2/3 of 15l = 10l
We keep 10liters and trash the rest.
we add 5 liters of pure antifreeze to bring us to a 60% solution and back to 15l

Answer 2

let x be the amount of the mixture that is drained.
x is also the amount of pure antifreeze added.

15 - x is the amount left in the cooling system, and still, 40% of the mixture is pure antifreeze

after adding x amount of pure antifreeze, the total amount of the mixture is back to 15 L, in which 60% are pure antifreeze

.6(15) = 9L pure antifreeze

.40(15 - x) + x = 9
6 - .4x + x = 9
.6x = 3
x = 5 L <== answer

Answer 3

Antifreeze in 40% antifreeze solution = 0.4 * 15 liters
Now drain x liters of of the old solution, and add x liters of pure antifreeze.

40% of x liter drained is antifreeze
So antifreeze in new solution is
= 0.4*15 - 0.4*x + x
But 0.4*15 - 0.4*x + x = 0.6 * 15 ==> 60 % of new solution
or 6 + 0.6x = 9
or 0.6x = 3
or x = 3/0.6 = 30/6 = 5 litres

5 litres drained out.