Question

If a man walks at the rate of 5 km/hr, he misses a train by only 7 minutes. However, if he walks at the
rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance
covered by him to reach the station.
​

Answer 1

Step-by-step explanation:

Let the distance to station be d kms

5 kms in 60 min

d kms in = 60/5*d = 12d min

6 kms in 60 min

d kms in = 60/10*d = 10d min

12d-10d = Difference in times = 12

(the 12 comes from 7 minutes late and 5 minutes early, if time to reach is t min, then t+5-(t-7) = 12)

2d = 12

d=6 kms

Answer 2

Answer:

6 km

Step-by-step explanation:

Given,

• If a man walks at the rate of 5 km/hr, he misses a train by only 7 minutes.
• If he walks at the  rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train.

To find,

• the distance  covered by him to reach the station.

Solution,

Let "x km" be the distance covered by him to reach the station

and "t" be the right time taken to reach the station (when the train arrives)

⇒ Speed = Distance/time taken

Distance = speed × time taken

Case (i) :

Speed of man = 5 km/hr

Time taken = (t + 7/60) hr     [ ∵ 1 min = 1/60 hr ]

Distance = x km

x = 5(t + 7/60) km

x = (5t + 7/12) km --[1]

Case (ii) :

Speed of man = 6 km/hr

Time taken = (t - 5/60) hr     [ ∵ 1 min = 1/60 hr ]

Distance = x km

x = 6(t - 5/60) km

x = (6t - 5/10) km --[2]

Thus,

 $\sf 5t+\dfrac{7}{12}=6t-\dfrac{5}{10} \\\\ 6t-5t=\dfrac{7}{12}+\dfrac{1}{2} \\\\ t=\dfrac{7}{12}+(\dfrac{1}{2} \times \dfrac{6}{6}) \\\\ t=\dfrac{7}{12}+\dfrac{6}{12} \\\\ t=\dfrac{13}{12} \ hr$

substitute t = 13/12 hr in equation [1]

$\sf x=5t+\dfrac{7}{12} \\\\ x=5(\dfrac{13}{12})+\dfrac{7}{12} \\\\ x=\dfrac{65}{12}+\dfrac{7}{12} \\\\ x=\dfrac{65+7}{12} \\\\ x=\dfrac{72}{12} \\\\ x=6 \ km$

Therefore, the distance  covered by him to reach the station is 6 km