Question

A 16 liter radiator is filled with a solution of 40 % antifreeze. How much should you drain from the radiation and replace with pure antifreeze to obtain a 60 % antifreeze solution?

Answer 1

`5(1)/(3)` liters is the amount to be drained out and replaced

Solution:

40 % antifreeze solution in 16 liter radiator

Let "x" be the amount drained from radiation and replaced with pure antifreeze

To obtain a 60 % antifreeze solution

The original solution is 16 liter, 40% of which is antifreeze

You want the solution to be 60% antifreeze:

60 % x 16 =
`(60)/(100) * 16 = 9.6`

You will remove x liters of the 40% solution and replace it with x liters pure (100%) antifreeze.

`40 \% (16 - x) + 100 \% * x = 60 \% * 16`

Let us solve expression for "x"

`(40)/(100) * (16 - x) + (100)/(100) * x = (60)/(100) * 16\\\\0.4(16-x) + x = 0.6 * 16\\\\6.4 - 0.4x + x = 9.6\\\\6.4 + 0.6x = 9.6\\\\0.6x = 3.2\\\\x = 5.33\\\\x = 5(1)/(3)`

Thus
`5(1)/(3)` liters is the amount to be drained out and replaced