6. Dima went to school in the morning, but after walking exactly half the distance, he realized he had forgotten his mobile phone at home. Dima estimated (he had an A in mental arithmetic) that if he continued walking at the same speed, he would arrive at school 3 minutes before the first bell, but if he ran home for the phone and then ran to school, he would arrive 3 minutes after the bell. Dima decided to run home, but he got out of breath while running (he had a C in physical education), and walked from home to school at his usual speed. As a result, he was 15 minutes late for the first class! How many times faster is the speed at which he runs compared to the speed at which he walks?
Answer: 2.

Solution: Let $x$ be the time it takes for Dima to walk from home to school, $y$ be the time it takes for Dima to run from home to school, and $T$ be the remaining time until the bell rings (at the moment Dima noticed the loss). Then the conditions of the problem can be written as $\left\{\begin{array}{l}\frac{x}{2}=T-3 \\ \frac{y}{2}+y=T+3 \\ \frac{y}{2}+x=T+15\end{array}\right.$ Solving this, we get that $x=24, y=12, T=15$, from which it follows that he runs twice as fast as he walks.