2 runners run on a oval track, both are at constant speed

Question

2 runners(Alice, Bob) are running on the oval track at a constant speed. The tracklength is 200 meters.

First Alice ran with such low speed that Bob passed him every 2 minutes. To run faster than Bob, Alice sped up 2 times and now she is passing him every 6 minutes. What speed did Bob ran at in km/h?

Solution:

How much was Bob faster than Alice at first speed?

1/30h is 2 minutes
x km/h * 1/30h = 0.2km
x = 6km/h

How much was Alice faster than Bob at second speed?

1/10h is 6 minutes
2x km/h * 1/10h = 0.2km
x = 1km/h

Don't know how to create equation out of this.

Answer 1

I wouldn't put $2x$ in the second equation. At that stage, you are still finding how much faster Alice is, not yet how fast Alice is. So if you omit $2$ you would find out that she was now faster by $2 km/h$.

Now it is time to set the equations. Let $A$ be Alice's original speed, $2A$ is her faster speed, let $B$ be Bob's speed, so you have:

$$B-A=6 km/h$$ $$2A-B=2 km/h$$

which gives you the solution ($A=8 km/h, B=14 km/h$).

Answer 2

You are going along the right track:

$$B_i = A_i + 6$$ Where $B_i$ is Bob's initial speed, and $A_i$ is Alice's initial speed. Likewise:

$$A_f = B_f + 2$$ Alice's final speed is 2 kph faster than Bob's, not 1 kph. In your work you divided by 2, but we are trying to get $A_f$, so there's no need to take out that factor of 2.

Now, we are given that:

$$B_f=B_i$$ $$A_f=2A_i$$

So this simplifies into just a system of 2 equations:

$$B_i = A_i + 6$$ $$2A_i = B_i + 2$$

Solving by substituting the first equation into the second, you get $A_i = 8$, so $B_i$ must be $14$kph.