Question

A
Passenger train takes 3 hours lessfor a journey of 140 km . If the speed is
increased by 42 kmph from its normal speed then The normal speed is ?

Answer 1

original time taken by the passenger train = x
distance = 140 km
time taken by the passenger train = x-3
normal speed = a
travelled speed = a +42
speed = d/t
d = speed *t
(x-3)(a + 42) = xa
xa +42x -3a -126 =xa
42x = 3a +126 =140
3a = 140 - 126
3a = 14
a = 14/3 Kmph

Answer 2

Let the normal speed of train = x km/h
time taken to cover 140 km = $\frac{140}{x}$ h

 If the speed is increased by 42 km/h from its normal speed, journey becomes 3 hours less.
   speed = (x+42)km/h
  time  = $\frac{140}{x+42}$

$\frac{140}{x}- \frac{140}{x+42}= 3$

⇒$140( \frac{1}{x}- \frac{1}{x+42})=3$

⇒$140( \frac{(x+42)-x}{x(x+42)} )=3$

⇒$140*42 = 3x(x+42)$

⇒$3 x^{2} +126x-5880=0$

Solving the equation, we get x= 28 or -72. ignore -72 as speed can't be negative.

So the normal speed of train is 28 km/h