A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour
less. If he walks 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man
and his original rate of walking.
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please see solution here.
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Answer:
Step-by-step explanation:
Let the rate the man walked = r
Le the time that he actually walked = t
thus, (r+.5) = 1 hr less rate and (r-1) = 3 hr more rate
Time taken at the faster speed = (t-1)
Time taken at the slower speed = (t+3)
Distance will be same for all three situations = D = R × T
So, rt = (r+.5)(t-1)
= rt = rt - r + .5t - .5
= r - .5t = -.5 ----(1)
For the slower speed case,
= rt = (r-1)(t+3)
= rt = rt + 3r -t - 3
= -3r + t = -3 --- (2)
Multiply equation 1 by 2
= 2r - t = -1
= -3r + t = -3 (eliminates t)
= -r = -4
Thus, actual walking rate = 4 km/h
Finding t using- r -.5t = -.5
= 4 - .5t = -.5
= -.5t = - 4.5
= t = 9
Thus, actual time = 9hrs
Hence, Distance = 4 × 9 = 36 km
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