11.1. Anya left the house, and after some time Vanya left from the same place and soon caught up with Anya. If Vanya had walked twice as fast, he would have caught up with Anya three times faster. How many times faster would Vanya have caught up with Anya (compared to the actual time) if, in addition, Anya walked twice as slow
Answer: 7.

Solution: Let Ane's speed be $v$, and Vanya's speed be $V$. The distance Ane has walked is proportional to $v$, while Vanya catches up with her at a speed of $u=V-v$. When Vanya doubles his speed, this difference triples, i.e., $u+V=2V-v=3u$. From this, we get $V=2u=2v$. If Vanya's speed doubles and Ane's speed is halved, the new difference in speeds will be $3.5v$, which is 7 times the new speed of Masha. Therefore, Vanya will catch up to her 7 times faster.