Antifreeze

A car owner needed to replace the antifreeze liquid in the radiator. On the label of the low-freezing liquid it was stated: volume parts of low-freezing liquid 2, volume parts of distilled water 1, freezing point -20°C. How many litres of low-freezing liquid and how many litres of water did he have to mix if he wanted to be sure that his engine would not be damaged down to an outside temperature of -20°C, when 4 litres of liquid fit into the cooling system?

Final Answer:

a =  2.6667 l
b =  1.3333 l

Step-by-step explanation:

$$\begin{gathered} V=4\text{l} \\ a:b=2:1 \\ a=\frac{2}{2+1}\cdot V=\frac{2}{2+1}\cdot 4=2.6667\text{l} \end{gathered}$$

$b=\frac{1}{2+1}\cdot V=\frac{1}{2+1}\cdot 4=1.3333\text{l}$

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