## 274. Math Puzzle $3 / 88$

Torsten needs fuel for his moped in a mixing ratio of 1:33 (oil to fuel). In a can, he still has 3 liters in a mixing ratio of 1:25, which he wants to use up. In the canister, he has enough fuel in a mixing ratio of 1:50.

How must he mix the fuels to use up the can content and achieve an oil-fuel ratio of 1:33?
Torsten must add $x$ liters of fuel in a mixing ratio of 1:50 to the existing 3 liters of 1:25 mixture. For the desired mixing ratio of 1:33, the following applies:

$$
3 \cdot \frac{1}{25}+x \cdot \frac{1}{50}=(x+3) \cdot \frac{1}{33} \quad, \quad \frac{x+6}{50}=\frac{x+3}{33}
$$

with $x \approx 2.8$

Torsten must add 2.8 liters from the can to the 3 liters in the tank and will thus obtain 5.8 liters of fuel in a mixing ratio of 1:33.