Question 0b5cb

Oct 16, 2016

$\frac{10}{9} = 1.111 \ldots$ litres

Explanation:

The amount of antifreeze in the radiator at the start is 10% of $5$ litres, that is

$0.1 \cdot 5 = 0.5$ litres.

If we remove $x$ litres of liquid, then assuming the antifreeze is uniformly mixed in, 10% of the liquid we remove will be antifreeze, so we will also be removing $0.1 x$ litres of antifreeze. Then we will add in $x$ litres of antifreeze. Thus, after the removal and addition of antifreeze, the total amount of antifreeze will be

$0.5 - 0.1 x + x = 0.5 + 0.9 x$ litres.

Our final goal is to have the antifreeze constitute 30%# of the $5$ litres of liquid, that is

$0.3 \cdot 5 = 1.5$ litres.

Thus, we need to drain off and replace $x$ such that the total is $1.5$ litres, i.e.,

$0.5 + 0.9 x = 1.5$

$\implies 0.9 x = 1.5 - 0.5 = 1$

$\implies x = \frac{1}{0.9} = \frac{1}{\frac{9}{10}} = \frac{10}{9} = 1.111 \ldots$

So $\frac{10}{9} = 1.111 \ldots$ litres should be drained off.