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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 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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Solution

Let the original speed of the train be x km/hr and the length of the journey be y km. Then,
Time taken =yx hrs

Case I : When the defect in the engine occurs after covering a distance of 30 km.
We have,
Speed for first 30 km =x km/hr
and, Speed for the remaining (y30) km =45x km/hr

Time taken to cover 30 km =30xhrs
Time taken to cover (y30) km =y304x5hrs =54x(y30) hrs
According to the given condition, we have,
30x+54x(y30)=yx+4560
30x+5y1504x=yx+34
120+5y1504x=4y+3x4x
5y30=4y+3x
3xy+30=0 ...... (i)

Case II : When the defect in the engine occurs after covering a distance of 48 km
Speed for first 48 km =x km/hr
Speed for the remaining (y48) km =4x5km/hr
Time taken to cover 48 km =48x hrs
Time taken to cover (y48) km =⎜ ⎜ ⎜y484x5⎟ ⎟ ⎟={5(y48)4x} hr
According to the given condition, the train now reaches 9 minutes earlier i.e., 36 minutes later.
48x+5(y48)4x=yx+3660
48x+5y2404x=yx+35
192+5y2404x=5y+3x5x
5y484=5y+3x5
25y240=20y+12x
12x5y+240=0 (ii)
Thus, we have the following system of simultaneous equations:
3xy+30=0
12x5y+240=0
By using cross-multiplication, we have
x240+150=y720360=115+12
x90=y360=13
x=903=30 and y=3603=120
Hence, the original speed of the train is 30 km/hr and the length of the journey is 120 km.

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