rlative speed = 20-4=16miles/hr=4/15 miles/minute
in 5 minutes the cyclist cover a dis of 4/15*5 miles=4/3 miles
at a speed of 4miles / hour the hiker will need (4/3)/4 hour=1/3 hour=20 minute

my approach:

in 5 minutes, hiker's distance = 5min*1hr/60min*4mi/hr= 1/3 mile
in 5 minutes, cyclist's distance = 5 min*1hr/60min*20mi/hr=5/3 miles

difference in distance = 5/3-1/3=4/3 mile
so hiker must make up 4/3 miles

4/3 mi=4 mi/hr*t
t=1/3 hr=20 min

the above illustrates the concept of relative speed, as mentioned by phoenix, by showing that in 5 minutes both the cyclist and the hiker are advancing

In short, the cyclist (20mph) is traveling at 5 times the speed of the hiker (4mph). this means that the hiker will need to spend 5 times as much time traveling to cover the same distance.

The cyclist went on for 5 minutes. To cover that distance the hiker will have to walk 5*5=25 minutes.

The first 5 minutes, both people are traveling. The last 20 minutes, only the hiker is traveling while the cyclist waits for him. So the cyclist waits 20 minutes. The answer is C.

This is GMATPrep question 1248

Good luck,
-Patrick

All of the above approaches are sound (and Patrick's was creative) but I just want to elaborate on thephoenix's:

rlative speed = 20-4=16miles/hr=4/15 miles/minute
in 5 minutes the cyclist cover a dis of 4/15*5 miles=4/3 miles
at a speed of 4miles / hour the hiker will need (4/3)/4 hour=1/3 hour=20 minute

Because they are travelling in the same direction, we can simply subtract the rates to find either the rate at which the faster object "catches up" to the slower one or, once caught up, the rate at which the faster object "exceeds" the slower one:

20 - 4 = 16

So, after passing the hiker, for 5 minutes the cyclist exceeds the hiker at a rate of 16mph.

Because 5 is just 1/12 of an hour, the cyclist has gained (1/12)*16 = 4/3 of a mile on the hiker (this is just distance = rate*time). Because we also know the hiker's rate, we can now use distance = rate*time again to solve for time, as thephoenix did.

TAKEAWAY:

When two objects travel in the same direction, we can subtract the rates.
When two objects travel in opposite directions (meeting up or spreading apart), we can add the rates.