pranab01 wrote:

Hi everybody plz check my reasoning so that I know Iam in the right track.

Many thanks in Advance

You are definitely right. I've used a similar approach:

\(5\frac{m}{h}=\frac{3m+3m}{1h+Xh}\)

using the definition of average speed. Then, I find Xh, i.e. the time spent in doing the return trip

\(Xh=\frac{1}{5}h=12min\)

That's where 12 appears. But it is the time spent not the speed.

Finally, to get the average speed of the return trip I just have to compute

\(\frac{3m}{\frac{1}{5}h}=3m*5h=15\frac{m}{h}\)

Answer E!