Answer :

Given: A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. 

To find:  the distance covered by the man and his original rate of walking.

Solution:

Let the original speed, original time taken and distance be ‘a’, ‘t’ and ‘d’.

As we know,

distance = speed × time

⇒ d = at ------- (1)

Given, if he walks 1/2 km = 0.5 km an hour faster, he takes 1 hour less,

⇒ d = (a + 0.5)(t – 1)

⇒ d = at + 0.5t – a – 0.5

⇒ at = at + 0.5t - a - 0.5         [From 1]

⇒ 0.5t - a = 0.5

Multiply the above equation by 10 to get,

⇒ 5t - 10a = 5

⇒ 10a = 5t - 5         -------- (2)

Also if he walks 1 km an hour slower, her takes 3 more hours.,

⇒d = (a – 1)(t + 3) 

⇒ d = at + 3a - t - 3

⇒ at = at + 3a - t - 3          [From 1]

⇒ 3a - t  = 3

⇒ t = 3a  - 3     ------- (3)

Put the value of t in eq (2)

⇒ 10a = 5(3a - 3) - 5

⇒ 10a = 15a - 15 - 5

⇒ 10a - 15a = - 15 - 5

⇒ - 5a = - 20

⇒ a = 4

Putting back in (3), we get

⇒ t = 3(4) - 3 = 9 

Therefore, 

d = at
   = 4(9) = 36

Hence, speed of man  = a = 4 km/hr

distance covered by man = d = 36 km

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