Question

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

Answer 1

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

Answer 2

Here is your answer.....