Okay, this is the answer that I get. Let me know if I am right. let the distance from my home to the office be "d" kilometres. On the first day, distance = d, speed = 10 and let time taken = t1. Therefore, 10 = d/t1. So, t1 = d/10.
On the second day, distance (it will be the same) = d, speed = 15 and let time taken be t2. Therefore, 15 = d/t2. So, t2 = d/15.
On the first day I reached at 1pm, and on the second day I reached at 11am. So the difference is 2 hours. Therefore, t1 - t2 = 2 (t1 is obviously greater than t2 because speed is less).
Therefore, d/10 - d/15 = 2.
Therefore, d = 60 kilometres.
Consider any of the two equations: t1 = d/10 or t2 = d/15 to determine the starting time. Let's take t1 = d/10. So, t1 = 60/10. Therefore, t1 = 6 hours. So on the first day it took me six hours to reach office. I had reached at 1 pm, which means I had started at 7am. (The second equation will give me the same result, obviously. Let's try it. t2 = 60/15. So, t2 = 4 hours. On the second day it took me four hours. I had reached office at 11am which means I had started at 7am).
So, I start at 7am and wish to reach office at 12 noon. This means I have to cycle for five hours. Now I can easily calculate the speed with which I should cycle. Distance = 60 and time = 5, so speed = 60/5 or 12 kmph.
By the way, if I have to cycle 60 kilometres to office and then 60 kilometres back home every day, I need to be a Superman.