I don't understand the solution to this please could someone help me:

[cyan mortar]

Question: At six o’clock, a spider starts to walk at a constant speed from the hour hand anticlockwise round the rim of the clock face. When it reaches the minute hand, the spider turns around and walks round the rim in the opposite direction at the same constant speed, reaching the minute hand again after a further 20 minutes. What time does the clock read when the spider reaches the minute hand for the second time?

Solution: After the spider begins the second stage of its journey, 10 minutes pass before the end. In that time the minute hand moves through 120˚. The spider thus moves through 480˚. This implies that the spider is moving four times as fast as the minute hand. During the first stage of the journey, let the minute hand move through x˚. This spider thus moves through (180 - x)˚ in the same time. Because the spider is four times as quick as the minute hand, we have 180 - x = 4x and so x = 36. Hence the total angle that the minute hand sweeps out is 36˚ + 120˚ = 156˚, which corresponds to 26 minutes. Therefore at the end of the spider's journey the clock reads 6:26.

[fast marten]

Do you need help understanding how we find the speed of the spider, or the final derivation of how long the spider spent walking?

[tender marsh]

I have a slightly different solution

let the speed of the spider be m

so in 1 minute it passes m "minutes" in the clock face

in the 2nd journey it passes
m * 20 = 20m "minutes"

and the minute hand also goes 20 minutes further

if we assume the minute hand doesn't move then the spider will cover 60 "minutes" to go back to the minute hand

but the minute hand moves 20 minutes further instead so the spider covers 60+20=80 "minutes"

thus 20m = 80

therefore m = 4

so the spider passes 4 "minutes" / minute

now for the first journey

let the time for the spider to reach the minute hand be t minutes

which is also the number of minutes the minute hand moves

so t + 4t = 30

5t = 30
t = 6

therefore after the first journey the clock shows a time of 6:00 + 6 minutes = 6:06

hence after the 2nd journey the time shown is 6:06 + 20 minutes = 6:26

[knotty dome]

Let's do like an asian:)), (bc the solution looks so weird)
*First, let's calculate the speed of the spider.
The second journey, the spider starts at the red minute hand, then it go around and will end at the blue minute hand, all the journey takes 20 minutes.
Next, in the second journey, the spider will meet the red minute hand 2 times. The first time is when it starts, the second time is when it goes pass through it (the black dot). It means that the spider has gone 360˚. Moreover, when it meets the blue minutes hand, it has gone more 120˚ from the red one. So it goes totally 480˚ after 20 minutes, so each minute it goes 24˚
*Second, let calculate how long it takes for the spider to meet the minutes hand in the first journey.
We can see that each minute the minute hand goes 1/60 × 360=6˚. Call x(minutes) is the time that it takes for the minute hand and spider to meet. We have the equation x × 24˚+x × 6˚=180˚, hence, x=6 minutes, so in the first journey, the spider meet the minute hand at 6:06, and after 20 minutes, the spider meet the minutes hand at 6:26

[cyan mortar]

How do you know where the hands are at the start of the second leg though.