Question

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 1

So increased speed of train (x+5)= 50km/h
Question is
A train travels 360 km at a uniform speed. If speed would have been 5km more it would have taken 48 minutes less for the journey. Find the speed of the train

Answer 2

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​