Time and Distance [समय और दूरी]-04 [SSC (CGL)-2015, TIER-II]

Question: If a man walks at a speed of 5 km/h, he misses a train by 7 minutes. However if he walks at the rate of 6 km/h, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is [यदि एक व्यक्ति 5 किमी/घंटे की गति से चलता है, तो उसकी गाड़ी 7 मिनट पहले छूट जाती है. यदि वह व्यक्ति 6 किमी/घंटे की गति से चलता है, तो वह गाड़ी के आगमन समय से 5 मिनट पहले स्टेशन पहुँच जाता है. स्टेशन पहुँचने के लिए उसने कितनी दूरी तय की?]  [Ans. (b) 6 km]

(a) 4 km  (b) 6 km  (c) 7 km  (d) 6.25 km

Detailed Explanation:  Concept: “When Distance is constant, Time is inversely proportional to Speed”

First Method : Here distance is constant is constant, so time is inversely proportional to speed. The ratio of speed is

S₁: S₂ = 5 km/hr: 6 km/hr = 5: 6

So, T₁: T₂ = 6: 5

First time the man is 7 minutes late, while second time he reaches the station 5 minutes before, so the difference between the timing is 7 + 5 = 12 minutes

Difference = 1 ≡ 12 minutes

             T₁ =  6 ≡ ? = 72 min=  hr

             T₂ =  5 ≡ ? = 60 min = 1 hr Ans.

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 Second Method: Let “X” be the distance , then time taken at the rates of 5 km/h and 6 km/h are (X/5) hour and (X/6) hour. Since first time, the man is 7 minutes late; while second time, he reaches the station 5 minutes before, so the difference between the timing is 7 – (- 5) = 12 minutes

Therefore,  (X/5) - (X/6) = (12/60) or, (X/30) = (1/5) or, X= 6 km/h Ans

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Third Method: Let the net time taken to reach the station is “t” hours. As the man is 7 minutes late to reach the station, so time taken by the man, moving at the rate of 5 km/h is [t +(7/60) hours and distance is 5 × [t +(7/60)] km.

Similarly, second time the man reaches the station 5 minutes before the schedule time, so time taken by the man this time is [t - (5/60) and the distance described is 6 × [t - (5/60) km

Since both the time, the man covers the same distance, so

                                              Distance = 5 × [t +(7/60)] = 6 × [t - (5/60) or, t = (65/60)hr

                                              Distance = 5 × [(65/60) + (7/60)] = 6 km Ans.