# A passenger train takes 1 hour less for a journey of 120 km :

A passenger train takes 1 hour less for a journey of 120 km, if its speed is increase by 10 km/hour from its usual speed. What is its usual speed?

(A) 50 km/hour

(b) 40 km/hour

(c) 35 km/hour

(d) 30 km/hour

**Answer: (d) 30 km/hour**

**Solution:-**

Let the usual speed of a passenger train be X km/hr and time taken by train be Y hr.

Time taken by train with original speed (Y) = 120/X

X Y = 120

X = 120/Y

Train 1 hour less for a journey, if its speed is increase by 10 km/hr.

Then, (Y + 10) = 120 /(X – 1)

(X – 1) (Y + 10) = 120

X Y + 10 X – Y – 10 = 120

Put the value of X = 120/Y,

Then, 120 Y/Y + 10 x 120/Y – Y – 10 = 120

120 + 1200/Y – Y – 10 = 120

1200 – Y^2 – 10 Y = 0

Y^2 + 10 Y – 1200 = 0

Y^2 + 40 Y – 30 Y – 1200 = 0

Y (Y + 40) – 30 (Y + 40) = 0

(Y – 30) (Y + 40) = 0

Y – 30 = 0

Y = 30

Then, the usual speed of the train is 30 km/hr.

Hence,the correct answer is option (d) 30 km/hr.