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Question
A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train was slower by 6 km/h it would have taken 6 hours more than the scheduled time. Find the length of the journey.
Solution
Let the speed of the train be x km/hr.
Let the time taken to travel certain distance be y hrs.
We know that, speed × time = distance
∴ Distance = xy km
According to the first condition, if the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time.
∴ (x + 6)(y – 4) = xy
∴ xy – 4x + 6y – 24 = xy
∴ – 4x + 6y – 24 = 0
∴ 2x – 3y = –12 ......(i)
According to the second condition, if the train was slower by 6 km/hr, it would have taken 6 hours more than the scheduled time.
∴ (x – 6)(y + 6) = xy
∴ xy + 6x – 6y – 36 = xy
∴ 6x – 6y – 36 = 0
∴ x – y = 6 ......(ii)
Multiplying both sides by 2, we get
2x – 2y = 12 ......(iii)
Subtracting equation (iii) from (i), we get
2x – 3y = –12
2x – 2y = 12
– + –
– y = – 24
∴ y = 24
Substituting y = 24 in equation (ii), we get
x – 24 = 6
∴ x = 30
∴ Length of the journey = xy
= 30 × 24
= 720 km
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