9.3. On a straight road between a dog in a kennel and a cat, a kilogram of sausages was placed, and the animals simultaneously rushed towards them. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and finished eating at the same time. It is known that the cat could have eaten all the sausages or run from the starting point to the dog's kennel in the same amount of time.
a) $K$ to whom the sausages were placed closer

b) How many times is the path of one animal greater than the path of the other?

Justify your answers.
# Solution:

Method 1. While the dog was eating its half of the sausages, the cat ate half as much, which is a quarter of all the sausages. The remaining sausages (in the amount of $1 / 2 - 1 / 4 = 1 / 4$) the cat ate while the dog had not yet reached the sausages. During this time, the cat could have run a quarter of the distance from its starting point to the dog's kennel, which means the dog ran $1 / 8$ of this distance. The remaining distance ($7 / 8$ of the initial distance) the dog and the cat ran together and simultaneously. During this time, the cat, running twice as fast, covered $2 / 3$ of this distance, which is $7 / 12$ of the total distance from the cat's starting point to the dog's kennel, while the dog covered the remaining part ($5 / 12$ of the entire distance). The ratio of these values, equal to $7 / 5$, gives the answer to the second part of the problem.

Method 2. Let the distance from the location of the sausages to the cat's starting point be $x$, and to the dog's kennel be $y$ (the unit of measurement for distance is not important, let's say meters). Let the speed of the dog's run (in meters per minute, for example) be $a$, then the cat's speed will be $2a$ m/min. We also introduce the speeds of eating sausages (in kg/min). Let these speeds for the cat and the dog be $b$ and $2b$ respectively. Now the conditions of the problem can be easily expressed by equations, namely: the time spent by each animal on running and eating is $\frac{x}{2a} + \frac{1/2}{b}$ for the cat and $\frac{y}{a} + \frac{1/2}{2b}$ for the dog, from which we get $\frac{x}{2a} + \frac{1}{2b} = \frac{y}{a} + \frac{1}{4b}$. The condition that the cat could have eaten all the sausages or run from the starting point to the dog's kennel in the same time is expressed by the equation $\frac{x+y}{2a} = \frac{1}{b}$. Substituting $\frac{1}{b}$ into the first equation, we get $\frac{x}{2a} + \frac{x+y}{8a} = \frac{y}{a}$. After obvious transformations, we obtain $x = 1.4y$.

Note: Part a) can also be solved as follows: suppose the sausages were placed at equal distances from the animals. Then after calculations, we get that the cat ate more than half of the sausages. This means the sausages were placed closer to the dog.

Answer: a) the sausages were placed closer to the dog; b) the cat's path is 1.4 times longer than the dog's path.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answers to both parts | 7 points |
| There are arithmetic errors in the solution, possibly leading to an incorrect answer | 6 points |
| The equation (system of equations) fully describing the condition of the problem is correctly set up but not solved | 4 points |
| Correct and justified answer in part a) with no other progress in the solution | 2 points |
| Correct answer in part b) without justification | 1 point |
| Correct answer in part a) without justification | 0 points |