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A Man walks a certain distance with certain speed. If he walks 1/2 km/h faster ,he takes 1 hour less,but if he walks 1 km a hour slower, he takes 3 more hours.Find the distance covered by the man and his original rate of walking. Answer is Distance-36 and speed 4km/h
Asked by prabhakar.mahabank | 05 Jul, 2019, 06:29: PM
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 taken at the faster speed
and (t+3) = time at the slower speed

Distance for all three situations is the same (dist = rate * time):

So, rt = (r+.5)(t-1)
rt = rt - r + .5t - .5
rt - rt + r - .5t = -.5
r - .5t = -.5 (1)

Also, for the slower speed case,
rt = (r-1)(t+3)
rt = rt + 3r -t - 3
rt - rt - 3r + t = -3
-3r + t = -3 (2)

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
t = 9 hrs actual time

Hence, Dist = r*t
4 * 9 = 36 km
Answered by Sneha shidid | 06 Jul, 2019, 09:36: PM

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