Q. 194.8( 6 Votes )

# 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 :

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

**Explanation:**

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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Solve the following pair of linear equation by cross - multiplication method:

x + 4y + 9 = 0

5x – 1 = 3y

KC Sinha - Mathematics