Question 40643
Let onward speed = x
Let distance = d
on ward time = d/x = 5 (given) ..equation 1
..
return

speed = x+5
distance = d
return journey time = d/(x+5)= 4 (given)..equation 2
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from eq 1 .. we get d/x=5, 5x=d

substitute d=5x in eqn 2

d/(x+5) = 4
5x/(x+5)=4
5x = 4(x+5)
5x = 4x+20
5x-4x = 20
x = 20.
Onward speed = 20
Return speed = 25 ..
good luck..