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
