3. Liangliang goes to school. If he walks at a speed of 40 meters per minute, he arrives at school on time at $8:00$; if he walks at a speed of 50 meters per minute, he arrives at school at $7:55$. The distance between Liangliang's home and the school is $\qquad$ meters.
Analysis: Travel-type profit and loss problem.
Traveling 40 meters per minute, just enough; if traveling 50 meters per minute, then short of $5 \times 50=250$ meters.
So $250 \div(50-40)=25$ minutes, the distance between Liangliang's home and school is $25 \times 40=1000$ meters.
Method two: Suitable for sixth grade.
Walking the same distance, the speed ratio is inversely proportional to the time, the speed ratio is $4: 5$, then the time ratio is $5: 4, 8: 00-7: 55=5$ minutes, then if traveling 40 meters per minute, Liangliang's time is $5 \div(5-4) \times 5=25$ minutes, so the distance between Liangliang's home and school is $25 \times 40=1000$ meters.