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.

Answer 1

45 km/hr
Take speed as xkm/hr
therefore eq. is 360/x-360/x+5=48/60 (time has to be converted from 48 min to 48/60 hr)
On solving we get, 1800/ xsq.+5=4/5=xsq. +5x-2250=0
Sp x=45 or x=-5(discarded) hence speed of train=45km/hr  

Answer 2

Answer:

Step-by-step explanation:

Let initial speed of train be x km/hr.

Distance travelled = 360 km.

//we know that Speed = Distance/time => Time = Distance /Speed.

Time taken by train Initially = 360/x.

If speed is increased by 5 km/hr,

Time taken by train = 360/x+5.

Difference in time taken = 48/60 hr.

=> 360/x - 360/x+5 = 48/60

=> 360(1/x - 1/x+5) = 48/60

=> 360[x + 5 - x / x(x+5)] = 48/60

=> 360 * 5/x(x+5) = 48/60

=> x(x+5) = 360 * 5 * 60 / 48

=> x² + 5x =  2250

=> x² + 5x - 2250 = 0

=> x² + 50x - 45x - 2250 = 0

=>x(x+50) - 45(x+50) = 0

=> (x - 45)(x + 50) = 0

=> x = 45 or -50.

Since x cannot be negative, x= 45 km/hr.

Thus original speed of train = 45 km/hr.