Question

asap
.A passenger train takes one hour less for a journey of 150km.If its speed is increased by 5km/hr

from its usual speed. Find the usual speed of the train.

Answer 1

Given :-

• Total distance of the journey = 150 km.
• If the train increases its speed by 5 km/hr, it would reach its destination 1 hour earlier.

To Find :-

• The usual speed of the train.

Solution :-

Let the usual speed of the train be x km/hr. Then, the increased speed of the train would be (x + 5) km/hr.

We know,

Time = Distance/Speed

Hence,

• Original time taken (with usual speed) = 150/x hrs.
• New time taken (with increased speed) = 150/(x + 5) hrs.

According to the question, the difference between the original time and new time is 1 hour.

$\implies\rm{\dfrac{150}{x}-\dfrac{150}{x+5}=1}$

$\implies\rm{150\left(\dfrac{1}{x}-\dfrac{1}{x+5}\right)=1}$

$\implies\rm{150\left(\dfrac{x+5-x}{x(x+5)}\right)=1}$

$\implies\rm{150\times{}5=x^2+5x}$

$\implies\rm{x^2+5x-750=0}$

By splitting the middle term, we get,

$\implies\rm{x^2-25x+30x-750=0}$

$\implies\rm{x(x-25)+30(x-25)=0}$

$\implies\rm{(x-25)(x+30)=0}$

$\implies\rm{x=25\:or\:-30}$

Neglecting negative value, we get,

$\implies\rm{x=}\:\bold{25}$

Hence,

• Usual speed of the train = 25 km/hr.

Answer 2

Hey User!

GIVEN THAT :

• A passenger train takes 1 hour less for completing it's journey of 150 km if it's speed is increased by 5 km/hr from its usual speed.

To find:

• Usual speed.

Formula used:

• Time = Distance/speed

Let the usual speed of the train be x km/hr

So, ATQ, we have

150/x - 150/(x+5) = 1

(150x + 750 - 150x )/(x^2 +5x) = 1

x^2 +5x = 750

x^2 +5x -750 = 0

x^2 +30x -25x -750 = 0

x(x+30)-25(x+30) = 0

(x+30)(x-25) = 0

so, x = -30 or x=25

Since, speed can't be negative.

so, x can't be -30 km/hr

therefore, x = 25 km/hr.

So, usual speed of train is 25 km/hr.

Hope this helps