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

$5\frac{1}{3}$ 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 = $\frac{60}{100} \times 16 = 9.6$

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

$40 \% (16 - x) + 100 \% \times x = 60 \% \times 16$

Let us solve expression for "x"

$\frac{40}{100} \times (16 - x) + \frac{100}{100} \times x = \frac{60}{100} \times 16\\\\0.4(16-x) + x = 0.6 \times 16\\\\6.4 - 0.4x + x = 9.6\\\\6.4 + 0.6x = 9.6\\\\0.6x = 3.2\\\\x = 5.33\\\\x = 5\frac{1}{3}$

Thus $5\frac{1}{3}$ liters is the amount to be drained out and replaced