2) A dog that is at point $A$ chases a hare that is, at the initial moment, $30 \mathrm{~m}$ ahead of $A$. The dog gallops with strides of $2 \mathrm{~m}$, while the hare escapes making strides of $1 \mathrm{~m}$. For every 2 strides of the dog, the hare makes 3. Where will the dog catch the hare?

(A) At $30 \mathrm{~m}$ from point $A$

(B) at $60 \mathrm{~m}$ from point $A$

(C) at $120 \mathrm{~m}$ from point $A$

(D) at $600 \mathrm{~m}$ from point $A$

(E) the dog will never catch the hare.
2) The answer is $\mathbf{( C )}$.

Two strides of the dog cover $2 \cdot 2=4 \mathrm{~m}$, in the same time 3 strides of the fleeing hare cover $3 \cdot 1=3 \mathrm{~m}$. Thus, every time the dog travels $4 \mathrm{~m}$, it gains 1 on the hare, which is then caught at a point that is $\frac{4 \cdot 30}{1}=120 \mathrm{~m}$ from point $A$.