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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 4/5 of its original speed. Consequently, the train reached its destination late by 45 minutes, had it happened after covering 18 kilometres more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey.

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Original speed = 10 Km/hr, Length of the journey = 90 Km

Original speed = 15 Km/hr, Length of the journey = 100 Km

Original speed = 25 Km/hr, Length of the journey = 110 Km

Original speed = 30 Km/hr, Length of the journey = 120 Km

answer is D.

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Detailed Solution

Concept- Firstly, Assume the speed of the train to be x and the distance of journey to be y
Then, with the help of a speed formula and form equations to calculate the speed and distance.
Given: First, after 30km with a uniform speed, it’s speed is reduced to

\frac{4}{5}

of its original speed and reaches its destination 45 min late. Second, After covering a distance of 18 km in addition to the one in first case with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to 4/5 of its original speed and it reaches its destination 9 min earlier
Assume the speed of the train to be x km/hr.
The total distance to be covered in the journey be y km.
We know that,

Speed = \frac{Distance}{Time} \Rightarrow x = \frac{y}{time} \Rightarrow Time = \frac{y}{x}

Given: the train covers the first 30 km with its original speed.

\Rightarrow

Speed for first

30 Kms = x km / hr

\Rightarrow

Time taken by train to cover

30 km = \frac{30}{x} hours

After that for the remaining journey, its speed is reduced to

\frac{4}{5}

of its original speed.

\Rightarrow

Speed of the train for the remaining

(y - 30) km = \frac{4 x}{5} km / hr

\Rightarrow

Time taken by train to cover

(y - 30) km = \frac{y - 30}{\frac{4 x}{5}} hours

\Rightarrow \frac{5 (y - 30)}{4 x} hours

But, the train reaches its destination 45 min late
⇒Total time taken for the journey = Total time taken in constant average speed + 45 minutes
We know,

1 \min = \frac{1}{60} hour

45 \min = \frac{45}{60} hour

\Rightarrow \frac{30}{x} + \frac{5 (y - 30)}{4 x} = \frac{y}{x} + \frac{45}{60}

\Rightarrow \frac{120 + 5 y - 150}{4 x} = \frac{60 y + 45 x}{60 x}

On further solving,

\Rightarrow 5 y - 30 = 4 y + 3 x

\Rightarrow y - 3 x = 30

…….-(i)
Now, for the second condition.
The speed remains constant for 18 km more than the previous case(i.e. 30+18=48 km)

\Rightarrow Speed for first 48 km = x km / hr

\Rightarrow Time taken to cover 48 km = \frac{48}{x} hours

Similarly, the speed of the train reduces to

\frac{4}{5}

of its original speed in the remaining journey.

\Rightarrow

Speed of the train for remaining

(y - 48 y) km = \frac{4 x}{5} km / hr .

\Rightarrow

Time taken by train to cover

48 Kms = \frac{48}{x} hours

.
Similarly, time taken to cover

(y - 48) km = \frac{(y - 48)}{\frac{4 x}{5}} hours

\Rightarrow \frac{5 (y - 48)}{4 x} hours

Hence, the train gets delayed by 45−9=36 min in second condition
⇒Total time taken in 2nd condition = Total time taken in constant average speed +36 min
We know,