\section*{Task 3 - 320923}

When refueling an oldtimer with a two-stroke engine that requires a fuel-oil mixture of \(1: 50\), 7 liters of fuel without oil were mistakenly added first.

How many liters of the still available mixture with a ratio of \(1: 33\) need to be added now to achieve the correct mixture ratio of \(1: 50\)?

The required number of liters should be determined to one decimal place.
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Let \(V\) be the volume of the amount still to be refueled in liters.

Then, \(\frac{1}{34} \cdot V = \frac{1}{51} \cdot (V + 7)\) should hold, so \(\left(\frac{1}{34} - \frac{1}{51}\right) \cdot V = \frac{7}{51}\).

It is \(51 = 3 \cdot 17\) and \(34 = 2 \cdot 17\), so the equation, when multiplied by \(17 \cdot 6\), becomes \((3 - 2) \cdot V = V = 14\), meaning exactly (to any number of decimal places) 14 liters of the 1:33 mixture need to be refueled.