Question

a train covers a distance of 90km at a uniform speed. it would have taken 30 minutes less if the speed had been 15km/hr more. calculate the original time of the journey ​

Answer 1

Answer:

45 kmph

Step-by-step explanation:

Let the original speed be x

time = 90/x

Increased speed = x + 15

New time = 90/x+15

Given that ,

90/x - 90/x+15 = 1/2

= 90{x+15-x/x(x+15)]= 1/2

90* 15/x^2 +15x= 1/2

x^2 + 15x = 2700

x^2+15x - 2700 = 0

x^2 + 60x - 45x - 2700 = 0

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

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

x = -60 or 45

Neglecting the negative value

x = 45kmph

Answer 2

Let the speed of train be x km/hr.

A train covers a distance of 90km at a uniform speed.

So,

$\sf{Distance\: = \:90\: km}$

Distance = Speed × Time

=> $\sf{T_1\:=\:\frac{90}{x}}$ ....(1)

The train takes 30 min less if the speed of train has been 15 km/hr. more.

=> $\sf{T_2\:=\:\frac{90}{x\:+\:15}}$ ....(2)

Also,

=> $\sf{T_1\:-\:T_2\:=\:30}$

In hours :

30 min = 1/2 hours.

So,

=> $\sf{T_1\:-\:T_2\:=\:\frac{1}{2}}$

Now, substitute value of $\sf{T_1}$ and $\sf{T_2}$ in :- $\sf{T_1\:-\:T_2\:=\:\frac{1}{2}}$.

So, that we can find the value of x i.e speed of train.

On solving we get,

x = 45, -60

As, speed of train can't be negative. So, - 60 neglected.

•°• Speed of train is 45 km/hr.