Question

A train travel 360 at a uniform speed has been 5 km/h more it would have takken 1h less for the same journey find the speed of the train .

Answer 1

Answer

Speed = 40 km/hr

Step-by-step-solution

It is said that train is traveling with 360km/hr. If we increase the speed by 5 km/hr it will time as one hour less.

To find

Speed of train = ?

Solution

So, let's assume speed of train to be x km/hr

THerefore,

Time taken to travel 360 km = T1 = distance/time.

Therefore T1 = 360/x --------> (1)

Train's speed had been increased by 5km/hr

Hence ,

$\sf \implies \: Speed \: of \: train \: = (x + 5) \: km \: pr \: hr$

Time taken to travel 360 km with speed of (x + 5) km/hr

T2 = distance / speed

hence ,

T2 = 360/x+5 --------> (2)

From question ,

T1 - T2 = 1hr

$\sf \implies \: \dfrac{360}{x} - \dfrac{360}{x + 5} = \: \: 1 \: \: \: \: (from \: equation \: 1 \: and \: 2)$

$\sf \implies \: \dfrac{360(x + 5) - 360}{x(x + 5)} = 1$

$\sf \implies \: 360 + 1800 - 360x = x(x + 5)$

$\sf \implies \: 1800 = {x}^{2} + 5x$

$\bold{ \boxed{\sf \implies \: { {x}^{2} + 5 - 1800 = 0}}}$

Equation to find speed.

a is constant of x² , and here constant is 1

a = 1

b is the constant of x, qand here constant is 5

b = 5

and c is a constant ad here constant is - 1800

To find the value of x , we need to find the root of :

${\sf \implies \: { {x}^{2} + 5 - 1800 = 0}}$

$\sf \implies \: {x}^{2} + 5x - 1800 = 0$

$\sf \implies \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{20}$

$\sf \implies \: x = \dfrac{ - b \pm \sqrt{ {( - 5)}^{2} - (4 )\times (1).( - 180)} }{20}$

$\sf \implies \: x = \dfrac{ - 5 \pm \sqrt{7225} }{2}$

$\sf \implies \: x = \dfrac{ - 5 \pm 85}{2}$

$\sf \implies \: x = \dfrac{ 80}{2} = 40km \: per \: hr$

OR

$\sf \implies \: x = \dfrac{ - 5 \pm 85}{2} = \dfrac{ - 90}{2} = - 45km \: pr \: hr$

For x = - 45km /hr ( NOT POSSIBLE because speed can't be negative in any condition. so , speed of train = 40km/hr)

Answer = 40km/hr