(d) 22.92 m (approx) abd 22.5 km/hr
Let the length of the train be l m and its speed be x km/hr.
Then, relative speed of train w.r.t.1st man
= (x – 6) km/hr = (x - 6) x \(\frac{5}{18}\) m/s
Relative speed of train w.r.t. 2nd man
= (x – 7.5) km/hr = (x - 7.5) x \(\frac{5}{18}\) m/s
Length of the train = Distance travelled in both the cases
⇒ (x - 6) x \(\frac{5}{18}\) x 5 = (x - 7.5) x \(\frac{5}{18}\) x 5.5
⇒ (x - 6) x 5 = (x - 7.5) x 5.5
⇒ 5x – 30 = 5.5x – 41.25 ⇒ 0.5x = 11.25
⇒ \(x\) = \(\frac{11.25}{0.5}\) = 22.5 km/hr
∴ Length of the train = \(\bigg((22.5-6)\times\frac{5}{18}\times5\bigg)\) m
= 22.92 m (approx)