Jane, Chen and Boris start together to travel the same way

I think it should be (Distance)/(GCD of the speeds).
Time = (20)/(1/2) = 40 kms.

Hi

This is my solution:

time for every person to make full circle:

20 km / 4 kmh = 5 hours
20 km / 5 kmh = 4 hours
20 km / 0.5 kmh = 10 hours
20 km / 8 kmh = 2.5 hours

We should find LCM for (5, 4, 10, 2.5) but before we have to make them integer (multiple by 2)
LCM (10, 8, 20, 5) = 40

Answer is A.

Keep in mind that there are only three people, who take 4, 40/11 and 5/2 hours to go around once. In h hours, they make h/4, 11h/40 and 2h/5 laps respectively. The number of laps must be an integer in each case, so h has to be a multiple of 4 , 40 and 5. The smallest possible value of h is 40. (A)

yes, this is an excellent fast approach for this particular problem - the caveat being if there is more than one answer choice that is a multiple of the distance (the least multiple value may not necessarily be correct given other speed combinations).

kevincan, this is a nice way of figuring it out.
i was familiar with the LCM method (taking the LCM of the time it takes to complete one lap), but was stuck bcoz the LCM of 4, 40/11 and 5/2 wouldn't work in this case. your approach is super!

BTW, what is a quick way to solve the problem if it is not required that they meet at the starting point i.e. find the time it takes for them to all meet the very first time, wherever it be.

Sumande,
Why should it be D/(GCD of Speeds)? Thanks.

20/.5=10 hours???

Oops I got 200/11. Now I see why it is 40. Good question

glad you got 200/11 - am happy that somebody thought the same as I did.
but the OA is 40.
200/11 was not in the answer choices. i added it since that is what i got using the LCM approach (LCM of 4, 40/11 and 5/2 - the time it takes for each of them to complete a lap), which seemed to me as the right approach (and works for most other similar problems).

however, i later realized 200/11 is not the LCM of 4, 40/11 and 5/2 and I made an error in including it. i think now, kevincan's method is correct (although backsolving (finding multiple of 20) is another much easier method, it cannot be used in the general case).

finally, the real challenging part is a variation of this question - i.e. a good method to find when they meet (for the first time) at whatever point on the track (i.e. if NOT specified that they meet at the starting point). the above solutions will not work for this case. i tried to do it using pairwise LCM method, but took way too much time. if anyone has some insight into that, plz explain.