# The car needs 40 liters of 40% of antifreeze now it has 40 liters of 20% antifreeze how many liters need to be drained and replaced to get the desired strength?

Sep 10, 2016

A sort of cheat (not really!) way of solving material blending questions.

Remove 10 litres of original solution and replace with 10 litres of pure antifreeze.

#### Explanation:

You can either use ratio (slope) or equation of a straight line.

What you drain off from the original mix you replace with pure antifreeze. So you will always have 40 Litres.

The question boils down to: how much do you drain off and replace with pure antifreeze to achieve the required concentration.

At one end of the scale none has been drained off and replaced so the concentration is 20%. The other end of the scale is where all has been drained off and replaced with neat antifreeze.

You end up with a continuous range of concentrations between and including the two extremes. See the graph

The slope of part is the same as the slope of all so we have:

$\left(\text{antifreeze replaced")/("concentration change}\right) = \frac{40}{100 - 20} \equiv \frac{x}{40 - 20}$

$\implies \frac{40}{80} = \frac{x}{20} \implies x = 10$

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$\textcolor{b l u e}{\text{Solving by equation}}$

By inspection:

$y = \frac{100 - 20}{40} x + 20 \text{ "->" } 40 = \frac{80}{40} x + 20$

$x = \frac{40}{80} \left(40 - 20\right)$

$x = \frac{40 \times 20}{80} = \frac{\cancel{80} \times 10}{\cancel{80}} = 10$