A man takes 6 hours to go to a place and come back by walking both the ways. He could have saved 2 hours by riding both the ways. The distance covered in the whole journey is 20 miles. What is the average speed for the whole journey if he goes by walking and comes by riding?

4 miles/hr

Speed of the man while walking = \(\frac{ distance }{ time }\) = \(\frac{20}{6}\) = 3.33 miles/hr

Time taken while riding = 6 - 2 = 4hrs

Speed by riding = \(\frac{20}{4}\) = 5 miles/hr

Required average speed = \(\frac{total\;distance}{total\;time}\)

Time taken by walking half way = \(\frac{10}{speed\;of\;walking}\) = \(\frac{10}{3.33}\)

                                               = 3 hours

Time taken by riding half way = \(\frac{10}{speed\;of\;riding}\) = \(\frac{10}{5}\)

                                           = 2 hours 

Total time taken for the whole journey = 5 hours

Thus, average speed = \(\frac{20}{5}\)  = 4 miles/hr