The application of the equation of one variable and one degree A and B race walking around the lake, 400 meters a week, B walk 80 meters per minute, a's speed is a quarter of B, now a and B are 100 meters apart, how many minutes later a and B meet for the first time requirement: There should be process and analysis

The application of the equation of one variable and one degree A and B race walking around the lake, 400 meters a week, B walk 80 meters per minute, a's speed is a quarter of B, now a and B are 100 meters apart, how many minutes later a and B meet for the first time requirement: There should be process and analysis

solution
Because it's a circular runway, it's actually a matter of pursuit. The speed of a is one fourth of that of B. B walks 80 meters per minute, so a walks 20 meters per minute
There may be two situations when they are 100 meters apart, one is a in front of B, the other is a behind B
Setting: after X minutes, Party A and Party B met for the first time
1 equation a before B: 80x-20x = 100
X = 1 minute 40 seconds
2 equation a after B: 80x-20x = 400-100
60x=300
X = 5 minutes
A: before Party B, Party A and Party B met for the first time in 1 minute and 40 seconds. After Party B, Party A and Party B met for the first time in 5 minutes

It seems that the known quantity is not enough. Please see the problem
A, B and C drive from a to B successively. B starts 5 minutes later than C and catches up with C 45 minutes later. A starts 15 minutes later than B and catches up with C 1 hour later. Then, how long does it take a to catch up with B?
The known quantity given here is only time. I can't work out the equation. What's the solution to this problem?
Please make an equation and explain it,
Why is "time used by C divided by time used by B"? What does division mean?

I solve the problem like this: let C's speed be x, B's speed be y, and a's speed be Z. it is known that B starts 5 minutes later than C and catches up with C 45 minutes later, so (45 + 5) x = 45y. It is known that a starts 15 minutes later than B and catches up with C 1 hour later, so (60 + 5 + 15) x = 60z

100 calculation problems of linear equation with one variable
Don't be the same as the next two

3X+5X=48 14X-8X=12 6*5+2X=44 20X-50=50 28+6X=88 32-22X=10 24-3X=3 10X*（5+1）=60 99X=100-X X+3=18 X-6=12 56-2X=20 4y+2=6 x+32=76 3x+6=18 16+8x=40 2x-8=8 4x-3*9=29 8x-3x=105 x-6*5=42 x+5=7 2x+3=10 12x-...

The answer to the seventh question of the application of 6.6 one variable equation
Once in a while

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