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 5 kilometre per hour more find the original speed of the train

Answer 1

let the speed of the train be x
distance=360 km
time=distance/speed
usual time=360/x
new time=360/x+5

(360/x) - 48=360/(x+5)
(360-48x)/x=360/x+5
(360-48x)(x+5)=360x
360x+1800-48x^2-240x=360x
-48x^2-240x+1800 =0
-24x^2-120x+900=0
-8x^2-40x+300=0
-4x^2-20x+150=0
-2x^2-10x+75=0
2x^2+10x-75=0
2x^2+15x-5x-75=0
x(2x+15)-5(x-15)=0
(x-5)(2x+15)(x-15)=0
x-5=0. 2x+15=0. x-15=0
x=5. x=7.5. x=15
the speed of the train is 5km/hr or 7.5km/hr or 15km/hr

Answer 2

Answer:

Let the original speed of he train be x km/hr.

Time taken at original speed= 360/x hrs

Time taken at increased speed= 360/x+5 hrs

Now,

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

=> 360[1/x - 1/x+5] =4/5

=> x²+5x-2250=0

=> x=45 or -50 (as speed cannot be negative)

=>x=45 km/hr.