## Task 32/63

The service life of a motorcycle tire type was determined experimentally. When mounted on the rear wheel, an average of $15000 \mathrm{~km}$ driving distance was achieved until complete wear, whereas when mounted on the front wheel, $25000 \mathrm{~km}$ was achieved.

a) After which driving distance should two simultaneously mounted tires be swapped (front and rear tires exchanged) so that both are completely worn out after the same driving distance?

b) What is the maximum driving distance that can be covered with two tires?

It is assumed that the wear is proportional to the driving distance.
We use the following notations (in $\mathrm{km}$ ):

$x$ distance traveled before tire rotation,

$y$ distance traveled after tire rotation until complete wear

$a=25000$ distance traveled by the front tire

$b=15000$ distance traveled by the rear tire (both until complete wear)

Then, for the wear of the front tire, we have

$$
\frac{x}{a} \cdot 100 \%+\frac{y}{b} \cdot 100 \%=100 \% \quad \text { or } \quad \frac{x}{a}+\frac{y}{b}=1
$$

and for the rear tire,

$$
\frac{x}{b} \cdot 100 \%+\frac{y}{a} \cdot 100 \%=100 \% \quad \text { or } \quad \frac{x}{b}+\frac{y}{a}=1
$$

From the symmetry of the two equations with respect to the unknowns $x$ and $y$ (if both are swapped, the original equations are obtained), it immediately follows that $y=x$. Thus, we get

$$
\frac{x}{a}+\frac{x}{b}=1 \quad \text { or } \quad x=y=\frac{a \cdot b}{a+b}=9375
$$

as the solution to question $a$. The solution to question $b$ is derived from $x+y=x+x=2 x=18750$. Therefore, the tires must be rotated after $9375 \mathrm{~km}$, and the total distance traveled is $18750 \mathrm{~km}$.