Question

A man walks a certain distance and rides back in 3 3/4hours; he could ride both ways in 2 1/2 hours. How many hours would it take him to walk both ways?

Answer 1

Let the speed of his walk and ride be w and r respectively.
Then,According to your question--
d/w + d/r = 3+3/4 (Since the going time by walk + the returning time by ride = 2+1/2)
d/r + d/r = 2+1/2 (Since the sum of the 'time' of rides = 2+1/2)
we have to find=>d/w + d/w
so first we need to calculate d/w by taking the first two equations)
d/w + d/r = 3+3/4 (where 3+3/4 = 15/4)
d/w =15/4 - d/r
Now take the second equation--
d/r + d/r = 2+1/2 (where 2+1/2 = 5/2)
2d/r = 5/2
2d = 5r/2
d = 5r/4
Now we again take own of our previous calculated equations and substitute d(only from the RHS since we need d/w)--
d/w = 15/4 - d/r
d/w = 15/4 - 5r/4 × 1/r (r cancels out)
d/w = 15/4 - 5/4
d/w = 10/4 or 5/2
now we are required to calculate d/w + d/w,so--
d/w + d/w (we substitute d/w with 5/2)
5/2 + 5/2 = 10/2 = 5 hours.
So our final answer is 5 hours.