CBSE Class 10 Answered

Let the original speed and time required be V and T.

The the total distance is D = VT

Time to cover distance of 30 km = 30/V

Time to cover remaining distance of D - 30 or VT - 30 = (VT-30)/(4V/5) = 5(VT-30)/4V

And it reaches 45 min late i.e. T + 45.

30/V + 5(VT-30)/4V = T + 45

This simplifies to,

VT - 180V = 30 .....(1)

Similarly for the second case, it reaches 9 min earlier, had the defect occured covering extra 18 km, hence T+45-9 = T+36

48/V +5(VT-48)/4V = T+36

Simplifying,

VT - 144V = 48 .........(2)

Solving (1) and (2),

V = 1/3 = 0.33 km/min = 0.3333x60 = 20 km/hr

and D = VT = 96 km

Hence the original speed of the train is 20 km/hr and the distance is 96 km.

Regards,

Team,

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