A and B race walking around the lake. A week around the lake is 400 meters. B walks 80 meters per minute. A's speed is 1 / 4 of B's, Now a and B are 100 meters apart. How many minutes later will they meet for the first time?

A and B race walking around the lake. A week around the lake is 400 meters. B walks 80 meters per minute. A's speed is 1 / 4 of B's, Now a and B are 100 meters apart. How many minutes later will they meet for the first time?

A's speed is 80 * 1 / 4 = 20 m / min
There should be two kinds of answers. Set X minutes later, a and B meet for the first time
1. A is 100 meters ahead of B
The answer is (80-20) x = 100
X=5/3
2. B is 100 meters in front of A
The answer is (80-20) x = 400-100
X=5

Party A and Party B race around the lake. A circle around the lake is 400 meters. Party B walks 80 meters per minute. The speed of Party A is 1.25 times that of Party B. now Party A and Party B are 100 meters apart
Party A and Party B race around the lake in the same direction. A circle around the lake is 400 meters. Party B walks 80 meters per minute. The speed of Party A is one and a quarter times that of Party B. when will party a catch up with Party B?
The first time? The tenth time?

B is 80 meters per minute, a is 100 meters per minute, the speed difference is 20 meters, so it takes 5 minutes to catch up with 100 meters, and 5 minutes for the first time
It's hard to start the second time
In the second catch-up, a walked 400 + 100 = 500 meters more than B, so it took 25 minutes to catch up. At the same time, it was 20 minutes after the first catch-up
.
In the 10th catch-up, a walked 4100 meters more than B, so it took 205 minutes and 200 minutes after the first catch-up

Party A and Party B race around the lake. A week around the lake is 400 meters. Party B walks 80 meters per minute. The speed of Party A is 1.25 times that of Party B. now Party A is 100 meters ahead of Party B. how many minutes later
How many minutes later, the two meet and solve the system of linear equations with two variables

Set the speed of meeting a after X minutes to y per minute
Y/80=1.25
(400-100)=X*(Y-80)
Simultaneous solution
X=15
Y=100
So we'll meet in 15 minutes

A and B race on the 400 meter girth ring track. It is known that B's speed is 80 meters per minute on average. A's speed is 1.25 times of B's. A is 100 meters ahead of B. now they start in the same direction at the same time. How many minutes later can a catch up with B?

Suppose that after X minutes, a can catch up with B. according to the meaning of the question, the result is: 1.25 × 80x-80x = 400-100, and the solution is: x = 15. A: after 15 minutes, a can catch up with B