Question

A train travelling at a uniform speed of 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

let original speed of train = x km/h 

we know, 
time = distance/speed 

first case 
———————
time taken by train = 360/x hour 

second case 
—————————— 

time taken by train its speed increase 5 km/h = 360/( x + 5) 

question says that 

time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour 

360/x - 360/(x +5) = 48/60 = 4/5 

360{ 1/x - 1/(x +5) } = 4/5 

360 ×5/4 { 5/(x² +5x )} =1

450 x 5 = x² + 5x 

x² +5x -2250 = 0

x = { -5±√(25+9000) }/2 
=(-5 ±√(9025) )/2 
=(-5 ± 95)/2 
= -50 , 45 

but x ≠ -50 because speed doesn't negative 

so, x = 45 km/h 

hence, original speed of train = 45 km/h

Answer 2

45 km/hr

take speed as xkm/h

therfore 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

so x=45 or x=-50(dicarded)  hence speed of train=45km/hr

plzz. ask if u dont understand........