A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Explanation

Let the original speed of the train `=x km//h`
Then, the increased seed of the train `=(x+5)km//h` [by given condition]
and distance `=360 km`
`=360/x-360/(x+5)=4/5` [ `:' "time" =("Distance")/("Speed")` and 48 min `=48/60h=4/5h`]
`implies (360(x+5)-360x)/(x(x+5))=4/5` [ `:' 48 min =48/60 h =4/5h`]
`implies (360x+1800-360x)/(x^(2)+5x)=4/5`
`implies 1800/(x^(2)+5x)=4/5`
`implies x^(2)+5x=(1800xx5)/4=2250`
`implies x^(2)+5x-2250=0`
`x^(2)+(50x-45x)-2250=0`
`x^(2)+50x-45x-2250=0` [by factorisation method]
`implies x(x+50)-45(x+50)=0`
`implies (x=50)(x-45)=0`
NOw `x+50=0impliesx=-50`
which is not possible because speed cannot be negatie and `x-45=0impliesx=45`.
Hence the original speed of the train `=45km//h`