Question

A train travelling a distance of 1200 km at a constant speed, when the driver learnt that he was running late, he increased the speed by 5 km/h. Now the journey took 8 hrs less than and reached on time. Find the original speed of the train.

Answer 1

Answer:V=25km/h

Step-by-step explanation: i can explain in picture....I hope you will understand..

Answer 2

Answer:

The original speed of the train is 30 km/hr.

Step-by-step explanation:

Given : A train travelling a distance of 1200 km at a constant speed, when the driver learnt that he was running late, he increased the speed by 5 km/h. Now the journey took 8 hrs less than and reached on time.

To find : The original speed of the train.

Solution :

Let the consonant speed s and time t

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

A train travelling a distance of 1200 km at a constant speed, when the driver learnt that he was running late.

$s=\frac{1200}{t}$ .....[1]

He increased the speed by 5 km/h and the journey took 8 hrs less than and reached on time.

$s+5=\frac{1200}{t-8}$ .......[2]

From [1] and [2]

$\frac{1200}{t}=\frac{1200}{t-8}-5$

$\frac{1200}{t-8}-\frac{1200}{t}=5$

$\frac{1200t-1200t+9600}{t(t-8)}=5$

$\frac{9600}{t^2-8t}=5$

$5t^2-40t=9600$

$t^2-8t-1920=0$

$t^2-48t+40t-1920=0$

$t(t-48)+40(t-48)=0$

$(t-40)(t-48)=0$

$t=40 , t=-48$

We neglect t=-48

So, The time is t=40

Substitute in [1]

$s=\frac{1200}{40}$

[tex]s=30/tex]

The original speed of the train is 30 km/hr.