Question

# 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 hrsCase I : When the defect in the engine occurs after covering a distance of 30 km.We have,Speed for first 30 km =x km/hrand, Speed for the remaining (y−30) km =45x km/hr∴ Time taken to cover 30 km =30xhrsTime taken to cover (y−30) km =y−304x5hrs =54x(y−30) hrsAccording to the given condition, we have,30x+54x(y−30)=yx+456030x+5y−1504x=yx+34⇒120+5y−1504x=4y+3x4x⇒5y−30=4y+3x⇒3x−y+30=0 ...... (i)Case II : When the defect in the engine occurs after covering a distance of 48 kmSpeed for first 48 km =x km/hrSpeed for the remaining (y−48) km =4x5km/hr∴ Time taken to cover 48 km =48x hrsTime taken to cover (y−48) km =⎛⎜ ⎜ ⎜⎝y−484x5⎞⎟ ⎟ ⎟⎠={5(y−48)4x} hrAccording to the given condition, the train now reaches 9 minutes earlier i.e., 36 minutes later.48x+5(y−48)4x=yx+3660⇒48x+5y−2404x=yx+35⇒192+5y−2404x=5y+3x5x⇒5y−484=5y+3x5⇒25y−240=20y+12x⇒12x−5y+240=0 (ii)Thus, we have the following system of simultaneous equations:3x−y+30=012x−5y+240=0By using cross-multiplication, we havex−240+150=−y720−360=1−15+12⇒x−90=−y360=1−3⇒x=−90−3=30 and y=−360−3=120Hence, the original speed of the train is 30 km/hr and the length of the journey is 120 km.

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