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After covering a distance of $30 km$ with a uniform speed, there is some defect in a train engine and therefore, its speed is reduced to $45$ of its original speed. Consequently, the train reaches its destination late by 45 min. Had it happened after covering $18 km$ more, the trains would have reached 9 min earlier. The speed of train and the distance of journey is

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The speed of train is 50 km/hr and the distance of journey is 120km

The speed of train is 40 km/hr and the distance of journey is 110km

The speed of train is 20 km/hr and the distance of journey is 100km

The speed of train is 30 km/hr and the distance of journey is 120km

answer is D.

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Detailed Solution

Given with a uniform speed train covers 30 km and after some defect in train engine the speed of train is reduced to

45

of its original speed.
Distance travel = speed × time
Let, the speed of train be x and distance travel be y, then,
According to the given condition,

30 x + y − 30 4 / 5 x = y x + 4560 48 x + y − 48 4 / 5 x = y x + 3660 Solving above equations, we get x = 30 and y = 120

Therefore, the speed of the train is 30 km/hr and the distance of journey is 120km.
Hence, option (4) is correct.

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