Question

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. Find the distance covered by the man and his original rate of walking.

Answer 1

Given: Let distance = d km and original speed = v km/h.

Step-wise calculation:

1. Original time = `d/v`.

2. If speed is `(v + 1/2)`, time becomes `d/(v + 1/2) = d/v - 1`.

Multiply through: `dv = d(v + 1/2) - v(v + 1/2)` 

⇒ `v(v + 1/2) = (1/2) d`

Hence `d = 2v(v + 1/2) = 2v^2 + v`.

3. If speed is (v – 1), time becomes `d/(v - 1) = d/v + 3`.

Multiply through: dv = d(v – 1) + 3v(v – 1) 

⇒ d = 3v(v – 1)

= 3v² – 3v

4. Equate the two expressions for d:

2v² + v = 3v² – 3v 

⇒ 0 = v² – 4v 

⇒ v(v – 4) = 0

v > 0 so v = 4 km/h

5. d = 2v² + v

= 2(4)² + 4

= 32 + 4

= 36 km

6. Check: original time = `36/4` = 9 h; at 4.5 km/h time = `36/4.5` = 8 h (1 hour less); at 3 km/h time = `36/3` = 12 h (3 hours more). All conditions satisfied.

Distance = 36 km; original walking rate = 4 km/h.