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A train 75 m long overtook a person who was walking at the rate of 6 km/hr and passes him in \(7\frac12\) seconds. Subsequently, it overtook a second person and passed him in \(6\frac34\) seconds. At what rate was the second person travelling? 

(a) 4 km/hr 

(b) 1 km/hr 

(c) 2 km/hr 

(d) 5 km/hr

1 Answer

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Best answer

(c) 2 km/hr

Let the speed of the train be x m/s. 

Speed of the 1st man = 6 km/hr = \(\big(6\times\frac{5}{18}\big)\) m/s = \(\frac53\) m/s

\(\frac{75}{(x-\frac53)} = \frac{15}{2}\)

⇒ 150 = 15 x \(\frac{(3x-5)}{3}\) = 15\(x\) - 25

⇒ 15\(x\) = 175  ⇒ \(x\) = \(\frac{175}{15}=\frac{35}{3}\) m/s

Let the speed of the second man be y m/s. Then

\(\frac{75}{\big(\frac{35}{3}-y\big)}=\frac{27}{4}\)

⇒ 300 = 27 x \(\big(\frac{35-3y}{3}\big) = 315 - 27y\)

⇒ 27y = 15 ⇒ y = \(\frac{15}{27}\) m/s = \(\big(\frac{15}{27}\times\frac{18}{5}\big)\) km/hr = 2 km/hr.

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