5. After covering a distance of 30 km with a uniform speed there is some defect in train engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, train reaches its destination late by 45 minutes. Had it happened after covering 18 km more, reached 9 minutes earlier. Find the speed of the train and the distance of covered.

Explanation

Let the initial speed of train = x km/hr
and the distance of journey = y km
`therefore` Usual time = `(y)/(x)` hr
Case I : When defect in engine, after 30 km
`therefore` Time taken in initial speed + time taken in reduced speed = `(y)/(x) + (45)/(60)`
implies `(30)/(x) + (y - 30)/((4)/(5)x) = (y)/(x) + (3)/(4)`
implies `(120 + 5(y - 30))/(4x) = (4y + 3x)/(4x)`
` implies 120 + 5y - 150 = 4y + 3x`
implies 3x - y = - 30
implies 12x - 4y = - 120 ...(1)
Case II : When defect in engine, after 48 km
`therefore (48)/(x) + (y - 48)/((4)/(5)x) = (y)/(x) + (36)/(60)` (now he is 9 minutes earlier means still he is late by 36 minutes)
`implies (192 + 5(y - 48))/(4x) = (60y + 36x)/(60x)`
`implies 15(192 + 5y - 240) = 60y + 36x`
implies 5(5y - 48) = 20y + 12x
implies 25y - 20y - 12x = 240
implies 5y - 12x = 240 ...(2)
Adding equations (1) and (2), we get
y = 120
Putting y = 120 in equation (1), we get
3x = 120 - 30 implies x = 30
Hence, initial speed of train = 30 km/hr and the distance travelled = 120 km.