4. A kilo of sausages was placed on a straight line between a dog in a kennel and a cat. The animals simultaneously rushed to the sausages. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and ate an equal amount. It is known that the cat could eat all the sausages in one minute and also run from the starting point to the dog's kennel in one minute. To whom were the sausages placed closer, and by how many times?
Answer: 1.4 times closer to the dog than to the cat.

## Solution:

Let $v$ be the running speed of the dog, $u$ be the eating speed of the cat, and the volume of sausages eaten by each animal be 1.

Then, $2 v$ is the running speed of the cat, and $2 u$ is the eating speed of the dog.

Let the distance from the cat to the sausages be $x$, and the distance from the dog's kennel to the sausages be $-y$.

The total time from the start of movement to the end of eating sausages is the same for both the cat and the dog, i.e., $x / 2 v + 1 / u = y / v + 1 / 2 u$.

According to the condition, we have: $2 / u = (x + y) / 2 v$.

Substituting $1 / u$ and $1 / 2 u$ in the first equation with $(x + y) / 4 v$ and $(x + y) / 8 v$ respectively, derived from the second equality,

we get $x / 2 v + (x + y) / 4 v = y / v + (x + y) / 8 v$.

Multiplying by $8 v$, we have: $4 x + 2(x + y) = 8 y + (x + y)$.

Combining like terms, we finally get: $5 x = 7 y$, hence the answer.

## Instructions for Checking:

Answer without justification but with verification - 2 points.

Answer without verification - 1 point.