Peter is walking from the library to school at a rate of 40 meters per minute. At the same time, Paul is walking from school to the library. After they meet each other, Peter still needs 6 minutes to get to school, while Paul is 200 meters away from the library. What is the distance (in meters) between the school and the library

d = 40m/min*6min + 200m = 440 m.

To solve this problem, we can first analyze the information given. Let's denote the distance between the library and the school as "x" meters.

According to the problem, Peter is walking from the library to the school at a rate of 40 meters per minute. It takes him 6 minutes to reach the school after meeting Paul. Therefore, the time it takes for Peter to meet Paul is 6 minutes.

Let's assume that they meet after t minutes of walking. In those t minutes, Peter walks at a speed of 40 meters per minute, so he would have covered 40t meters.

At the same time, Paul is walking from the school to the library. Since they both meet, they have covered the entire distance between the library and the school (x meters) together. Therefore, Paul would have covered (x - 40t) meters in t minutes.

Since they meet after 6 minutes, we can set up the equation:
40t = x - 40t

Simplifying this equation, we get:
80t = x

Now, let's use the other piece of information given. After they meet, Peter still needs 6 minutes to reach the school, while Paul is 200 meters away from the library.

We can set up another equation for this:
x - 40(6) = 200

Simplifying this equation, we get:
x - 240 = 200
x = 200 + 240
x = 440

Therefore, the distance between the school and the library is 440 meters.