I tried by Listing Method below because of the looping effect going round and round unlike straight lines.
Fastest car travelled at 180km/h. For each complete round is 36km, time taken is 36/180 x 60min = 12 min. Every 12 min, fastest car completes 1 round, and its distance from start line resets to 0km-mark.
Every 12min, slowest car travelled 100 x 12/60 =20km
Fast car at Slow car travelled Fast car pass
12min 20km, 0) 20km-mark 0 time
24min 40km, 1) 4km-mark 0 time
36min 60km, 1) 24km-mark 1 time
48min 80km, 2) 8km-mark Still 1 time
60min 100km 2) 28km-mark 2 times
72min 120km 3) 12km-mark Still 2 times
84min 140km 3) 32km-mark 3 times
60min x 32 to catch up/(180-100) = 24 more min to overtake to pass 4th time.
108min 100×108/60 = 180km, 0km-mark 4th time exact
3pm … add 108min —> 4:48pm# (pls counter check solution & given method)
[or add on 12 mins in table]
96min 160km 4) 16km-mark Still 3 times
108min 180km 5) 0km-mark Exact 4th time
After first 12 mins, slowest car is 20 km “ahead” of fastest car. The fastest car will take 60x 20/(180-100) = 15 mins to catch up 1st time. So 12+15=27 mins to catch up once for both cars to be at the same exact spot on the loop.
So to catch up 4th time –> 27×4 = 108 mins , or 4:48pm#
Method 2 and 3 only came to me because I got the solution and visualising by Listing Method, so can comfortably agree with the logic and accuracy.
Method 3 :
Fast Car : Slow Car : Difference
Speed 180 : 100 :
Ratio of speed 9 : 5 : 4
Ratio of distance 9 : 5 : 4
Actual distance 1st pass 81 : 45 : 36km (Fastest car has completed 1 more round after analysis)
Actual distance 4th pass 324 : 180 : 36 x4
Time taken = D/S = 324/180 or 180/100 = 1.8hrs = 1h 48 min
Time at 4th pass is 4:48pm#