Question

A hiker walks a certain distance. If he had gone 1 km/h faster, he would have walked the distance in 4/5 of the time. If he had walked 1 km/h slower, he would have taken 2.5 hours longer to travel the distance. Find the distance.

Answer 1

Let the time taken by the hiker be t hrs.

and Distance be x km and speed of hiker be y km/h

then Distance =speed $\times$ time

$\Rightarrow x=y\times t~~~~~-(1)$

Case1:- when hiker travel 1 km/h faster, Time taken=$\dfrac{4}{5}$ t hrs.

then Distance=speed$\times$ time

$\Rightarrow x=(y+1)(\dfrac{4}{5}t)~~~~~~-(2)$

Case 2:- When hiker travels 1 km/h slower, time taken= t+2.5

then $x=(y-1)(t+2.5)~~~~~-(3)$

Substitute eqs.(1) in eqs.(2) and eqs.(3)

$\Rightarrow yt=(y+1)(0.8t) \Rightarrow y=0.8y+0.8\\\Rightarrow y-0.8y=0.8\Rightarrow 0.2y=0.8\Rightarrow y=4$ km/h

Substituting the value of y in eqs.(1) and (3) and solving -

$x=4t,x=(4-1)(t+2.5)$

$\Rightarrow 4t=3(t+2.5)\Rightarrow 4t=3t+7.5\Rightarrow t=7 .5hrs$

Hence The distance $x=4t=4\times 7.5=30km$