Question : 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.
Doubt by Jayant
Solution :
Let the original speed of the train be x km/h
Distance covered = 360 km
Time taken (t1) = Distance/speed
t1=360/x
New Speed = (x+5) km/h
Time Taken (t2) = 360/(x+5)
ATQ
360/x - 360/(x+5) = 48/60
360[1/x - 1/(x+5)] = 4/5
360[(x+5-x)/x(x+5)]=4/5
360[5/(x²+5x)]=4/5
360×5×5=4(x²+5x)
90×25=x²+5x
2250=x²+5x
x²+5x-2250=0
x²+(50-45)x-2250=0
x²+50x-45x-2250=0
x(x+50)-45(x+50)=0
(x-45)(x+50)=0
x+50 = 0
x = -50
x is speed and we know speed can't be negative. so rejected.
x-45=0
x=45
Hence, the original speed of the train is 45 km/h.