Jim and Mike stand at point P. Jim begins to cycle in a straight line

Not sure whether my way of solving is correct. Just share with you how I solve this.

Option 1:
Let x be the distance which Mike travels so that the length is as half as that Jim travels => 2x for Jim
Mike's speed is 2 and Jim's speed is 3
=> Time for Mike to travel x distance is x/2 and time for Jim to travel 2x distance is 2x/3 (in hour)
So we have the equation:
2x/3 - x/2 = 2 (Because Jim started moving first, so at point x the duration Jim traveled must be 2 hour more than that Mike traveled.
=> x = 12
=> Duration that takes Mike to travel x is : 12/2 = 6 hour

Similarly, set y as the distance which Mike travels so that the length is one-forth of the length Jim travels => 4y for Jim
The equation now is:
4/3y - 1/2y = 2
=> y = 12/5
=> Duration that takes Jim to travel y is: 4/3 * 12/5 = 3.2 hour

The gap is |3.2 - 6| = 2.8 hour = 168 minutes (Not sure why my answer does not fit the provided answer list, but my second method also provides the same result)

Option 2:
Let t be the time that Jim starts (the first person to start)

We have 2 equations:
DJ (distance Jim traveled after t hour) = 3t
DM (distance Mike traveled after t hour) = 2(t-2) (Because Mike started 2 hour later than Jim)

So the amount of time it takes Mike to cover the half distance that Jim has covered is:
DM = 1/2 DJ
=> 2(t-2) = 3/2t
=> t = 8
Because this is the number of hour that Jim traveled so we must minus 2 to get the number of hour that Mike traveled => 6

Similarly, the amount of time it takes Jim to cover four times the distance that Mike has covered:
DM = 1/4 DJ
=> 2(t-2) = 1/4*3t
=> t = 16/5 (because the question asks the duration of Jim, now we don't need to minus to get that of Mike)

So that gap is: 16/5 - 6 = 2.8 hour = 168 minutes.

Personally, I prefer the second way to solve, it takes much less time than the first one. Hope it helps.

No, that is not the case. The 't' in my solution represents the time after Mike starts moving. If you wanted to find the time including the 2 hour head start that Jim got, you'd need to add 2.

And before Mike starts moving, you could say colloquially that Jim has traveled infinitely further than Mike. But if the ratio of Jim's speed to Mike's speed is 3 to 2, then no matter what head start one of them gets, the longer they travel, the closer the ratio of their distances will get to 3 to 2 (it's Jim who is faster, by the way, not Mike). So at some point that ratio will be 4 to 1, and at some later time it will be 2 to 1, the two ratios in the question. The only ratios that would be impossible are ratios of 3 to 2 or less.

That's a very good point - in my original solution, I used Mike's time here instead of Jim's time, but your solution is perfect. Thanks for the correction! I've edited the last line of my solution to fix that.

I, too, got 168 minutes. The question itself is a pretty straightforward one but the problem is that the correct answer is not listed among the options. My solution:

(1) Let T1 hrs be the time MIKE takes to cover half the distance that JIM covers concurrently. Since Jim had a 2 hour head-start, he will have covered 3*(T1+2) miles by the time Mike covers 2*T1 miles. Thus,
2T1=(1/2)[3(T1+2)]. Solving, we get T1=6 hrs=360 minutes.

(2) Let T2 hrs be the time JIM takes to cover 4 times the distance MIKE covers concurrently.Since Mike started 2 hours later, he will have covered 2*(T2-2) miles by the time Jim covers 3*T2 miles. Thus,
3*T2=4*2*(T2-2). Solving, we get T2=(16/5)hrs=192 minutes.

T1-T2=360-192= 168 minutes.

Back-solving, this checks out:
(a) In 6 hrs, Mike travels 12 miles. Concurrently, Jim travels for 8 hrs (since he had a 2-hour head-start) and covers 8*3=24 miles which is twice the distance Mike covered. So, 360 minutes for T1 checks out.
(b) In (16/5)hrs, Jim travels (48/5)miles. Concurrently Mike travels for (16/5-2)hrs (since he started 2 hrs after Jim) and covers (6/5)*2=(12/5)miles. So the distance that Jim covers (48/5)miles is 4 time the distance that Mike covers (12/5)miles. Checks out.

On the other hand, the OA does NOT check out. Tried assuming 'x' hrs as the time for the 2nd scenario (Jim covering 4 times the distance Mike covers) which makes [x+(12/5)]hrs the time for the 1st scenario, formed equations and got different values for 'x' in the two scenarios.

Maybe, the official solution could clear things up. Expert input from Buniel, VeritasKarishma or chetan2u would be welcome.

btrg
What is the source of this question? It seems that the answer is 168, but it is not listed in the answer choices.
Thanks!