Question

5. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr
more, it would have taken 30 minutes less for the journey. Find the original speed of
the train

Answer 1

Let the speed of train is x km/h and time taken to cover 90 km distance with it is t hours .
so, distance = speed × time
90 km = x km/h × t hours
90= xt
t = 90/x ------(1)
Again, speed is increased by 15 km/h
e.g, (x + 15) km/h
And then, time taken = (t - 1/2) hours [ 30 min = 1/2 hours ]
so, time = distance/speed
t - 1/2 = 90/(x + 15)
⇒ t = 90/(x + 15) + 1/2 -----(2)

From equation (1) and (2),
90/x = 90/(x + 15) + 1/2
⇒90/x - 90/(x + 15) = 1/2
⇒90[ 1/x - 1/(x + 15)] = 1/2
⇒90 × 15/(x² + 15x) = 1/2
⇒x² + 15x = 90 × 15 × 2 = 2700
⇒x² + 15x - 2700 = 0
⇒x² + 60x - 45x - 2700 = 0
⇒ x = -60 , 45
But x = -60 because x is speed and speed can't be negative.
so, original speed of train = 45 km/h

Answer 2

HELLO DEAR,


let the Speed of the train be x km/hr.
and time be t.
90 = xt-------( 1 )

AND,

if speed is increased by 15km/hr is (x + 15)km/hr.

90/(x + 15) = t - 1/2

90/(15 + x) + 1/2 = t---------( 2 )

90/x = 90/(x + 15) + 1/2

90/x - 90/(x + 15) = 1/2

90[1/x - 1/(x + 15)] = 1/2

(x + 15 - x)/(x² + 15x) = 1/180

15/(x² + 15x) = 1/180

x² + 15x = 2700

x² + 15x - 2700 = 0

x² + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45)(x + 60) = 0

x = 45 , x = -60[neglect]

hence, Speed of the train = 45


I HOPE ITS HELP YOU DEAR,
THANKS