Solution

Okay, this is the answer I got. 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 start 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. Let's try it. Now, 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 at 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 the office and then 60 kilometres back home every day, I think I must look for a job nearer home!

• Union Public Service Commission - www.upsc.gov.in
• IIT-Kharagpur - www.iitkgp.ac.in
• Indian Statistical Institute - www.isical.ac.in
• Indian Institute of Technology Madras - www.iitm.ac.in
• Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
• Indian Institute of Mass Commission - www.iimc.nic.in
• IIT Bombay - www.iitb.ac.in