Question

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

Answer 1

Let “x” be the usual speed of the train

Let T1 be the time taken to cover the distance 90 km in the speed x km/hr

Let T2 be the time taken to cover the distance 90 km in the speed (x + 15) km/hr

Time = Distance/Speed

T1 = 90/x

T2 = 90/(x + 15)

By using the given condition

T1 - T2 = 30/60

(90/x)- (90/(x + 15)) = ½

Taking 90 commonly from two fractions

90 [ (1/x) – 1/(x+15) ] = ½

(X + 15 – x)/x(x + 15) = 1/(90 x 2)

15/x² + 15 x = 1/180

15 (180) = 1(x² + 15 x)

x² + 15 x = 2700

x² + 15 x – 2700 = 0

x² + 60 x - 45 x – 2700 = 0

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

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

x – 45 = 0 x + 60 = 0

x = 45 x = -60