A boy and a dog

Let's have the dog runs for 1 round (to the top and then back to the boy). The boy also moves to the top, hence the dog will always travel more when it runs to the top than when it runs back. We call the difference between the distances: $A$ .

The difference for first, second, third, ..., n round are $A_1, A_2, A_3, ..., A_n$ . If when round n ends, the boy reaches the top, we have the following equation: $A_1 + A_2 + A_3 + ... + A_n = 100(m)$

With $x$ is the total time the dog spends running uphill, $y$ is the total time the dog spends running downhill: $\Rightarrow 3x - 5y = 100 (1)$ On the other hand, the amount of time that the boy spends to walk uphill is: $\frac{100}{1} = 100(s)$ $\Rightarrow x + y = 100$ $\Leftrightarrow x = 100 - y (2)$

From $(1)$ and $(2)$ , we have: $3(100-y) - 5y=100$ $\Leftrightarrow y=25$ $\Rightarrow x = 100 - y =75$

The total distance the dog will have run is: $3x + 5y = 3\times75 + 5\times25 = 350(m)$