3. (10 points) Liangliang goes to school. If he walks at a speed of 40 meters per minute, he will arrive at school on time at 8:00; if he walks at a speed of 50 meters per minute, he will arrive at school at 7:55. The distance between Liangliang's home and the school is $\qquad$ 1000 meters.
【Analysis】According to the problem, walking at a speed of 40 meters per minute is just enough; if walking at a speed of 50 meters per minute, there would be a shortage of $5 \times 50=250$ meters, so $250 \div(50-40)=25$ minutes, the distance between Liangliang's home and the school is $25 \times 40=1000$ meters; based on this, solve the problem.

【Solution】Solution: Walking at a speed of 40 meters per minute is just enough; if walking at a speed of 50 meters per minute, there would be a shortage of $5 \times 50=250$ meters, so $250 \div(50-40)=25$ minutes,

the distance between Liangliang's home and the school is $25 \times 40=1000$ meters;
Answer: The distance between Liangliang's home and the school is 1000 meters.
Therefore, the answer is: 1000.