After covering a distance of 30Km with a uniform speed, there got some defect in train engine and therefore its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination 45 minutes late. If it had happened after covering 18Km of distance, the train would have reached 9 minutes earlier>Find the speed of the train and the distance of the journey.
This is my question. .
now after letting the original speed of train be $x Km/Hr$ and the time taken be y Hr, threfore distance = $xy$
speed = $x-4x/5$ Km/hr
time = $y + 45/60$
$xy=60xy/300 +45x/300 => +300xy-60xy = 45x => 240xy = 45x$ .............[i]
similarly in CASE II we get the equation:
$240xy = -9x$.........[ii]
but after solving these two equations the answer is coming to 0 which is wrong please tell me the correct solution.