Since the answer being sought is regarding time, we want to think of everything in regards to time in this problem. Remember that rate•time = distance so, that means time = distance/rate. Also, since most of the units are given in meters and seconds, we will use those units in our problem.
Please note that we are not told that Thomas and Jimmy leave at the same time. This is a critical assumption. Obviously, if they left at different times, they can arrive at the school differently than if they didn't, no matter how long either walks or runs. We will assume they both leave at the same time.
The time that Thomas takes (we'll call it tT) is his (running distance)/(running rate) + (walking distance)/(walking rate) or tT = (running distance)/9 + (walking distance)/3. Let's call the distance he runs "x". That means the distance he walks is 1800 - x. So the equation becomes:
tT = x/9 + (1800 - x)/3
The time that Jimmy takes (we'll call it tJ) is his (running distance)/(running rate) + (walking distance)/(walking rate) or tJ = 1400/10 + (2200 - 1400)/4 or tJ = 1400/10 + 800/4 = 340 seconds
Since we want Jimmy to arrive 5 minutes before Thomas, we can make Thomas's time equal to Jimmy's time plus 5 minutes. Since our unit of measure is seconds, we want to adjust that to be Jimmy's time plus 300 seconds (60 seconds per minutes times 5 minutes). So:
tT = x/9 + (1800 - x)/3 = 340 + 300
x/9 + (1800 - x)/3 = 340 + 300 [combine like terms to get:
x/9 + (1800 - x)/3 = 640 [multiply both sides by 9 to get:
(9)x/9 + (9)(1800 - x)/3 = (9)640 [simplify to get:
x + 3(1800 - x) = 5760
x + 5400 - 3x = 5760
-2x = 360
x = -180
In other words, it is not possible for Thomas to get to school 5 minutes after Jimmy no matter how long he runs. Again, assuming they both leave at the same time and Thomas walks the entire way, he will take 1800/3 = 600 seconds. Jimmy takes 340 seconds. That's a difference of 260 seconds (less than 5 minutes). So this scenario doesn't work.