## 188. Math Puzzle $1 / 81$

A service car (passenger car) was driven $1992 \mathrm{~km}$ in one month; of this, $379 \mathrm{~km}$ was city driving. The fuel consumption limit is eleven liters per $100 \mathrm{~km}$ for city driving and ten liters per $100 \mathrm{~km}$ for long-distance driving.

At the beginning of the month, there were still 20 l of fuel in the tank, and an additional 190.4 l were added. At the end of the month, there were still $30 \mathrm{l}$ in the tank.

How much fuel did the driver save compared to the limit through exemplary driving?
To calculate fuel savings, the actual fuel consumption must be compared with the specified limit.

a) Determination of the specified limit for $379 \mathrm{~km}$ city driving and $1992 \mathrm{~km}-379 \mathrm{~km}=1613 \mathrm{~km}$ highway driving:

Since a fuel consumption of $11 \mathrm{l} / 100 \mathrm{~km}$ is specified for city driving, the following proportion results:

$$
11: x=100: 379, \quad 100 x=4169, \quad x=41.69
$$

Thus, the driver was allowed to consume 41.69 l of gasoline for city driving.

For highway driving, the result is:

$$
10: x=100: 1613, \quad 100 x=16130, \quad x=161.30
$$

Thus, 161.30 l of fuel could be consumed.

In total, this results in a consumption limit of 41.69 l + 161.30 l = 202.99 l.

b) The actual fuel consumption is determined as follows: $20 \mathrm{l}+190.4 \mathrm{l}-30 \mathrm{l}=180.4 \mathrm{l}$.

c) Thus, the fuel savings amount to $202.99 \mathrm{l}$ - 180.4 l = 22.59 l.

Therefore, the driver was able to save almost 23 l of gasoline in one month through exemplary driving behavior.