Question

A train,travelling at a uniform speed for 360km,would have taken 48minutes 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

Answer:

Step-by-step explanation: let the original speed of the train be X km/h

Increased speed=X+5

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

90/X -90/X+5=1/5

90(X+5) -90(X)/X2+5X=1/5

450X+2250-450X=X2+5X

X2+5X-2250=0

X2+50X-45X=0

X(X+50)-45(X+50)=0

(X+50) (X-45)=0

X=45km/h