Question

A man walks a certain distance with certain speed.if he walks 1/2km an hour faster he takes 1hr less.But,if he walks 1km an hour slower, he takes 3hr more.Find the distance covered by the man and his original rate of walking

Answer 1

The distance traveled by him is:
:
Let r = the rate the man walked
then
(r+.5) = 1 hr less rate
and
(r-1) = 3 hr more rate
:
:
let t = time he actually walked
then
(t-1) = time at the faster rate
and
(t+3) = time at the slower rate
:
distance for all three situations is the same (dist = rate * time):
;
actual dist = faster time dist
rt = (r+.5)(t-1)
rt = rt - r + .5t - .5
rt - rt + r - .5t = -.5
r - .5t = -.5
:
actual dist = slower time dist
rt = (r-1)(t+3)
rt = rt + 3r -t - 3
rt - rt - 3r + t = -3
-3r + t = -3
:
multiply the 1st equation by 2 and add to the above equation
2r - t = -1
-3r + t = -3
---------------eliminates t
-r = -4
r = 4 km/h is the actual walking rate
:
Find t using: r -.5t = -.5
4 - .5t = -.5
-.5t = -.5 - 4
-.5t = - 4.5
Multiply eq by -2
t = 9 hrs actual time
:
Dist = r*t
4 * 9 = 36 km
:
:
see if this is true; find the time of each scenario
36/4.5 = 8 hrs (1h less)
36/3 = 12 hrs (3hr more)


hope this helped u