Question

a train Travels a distance of 300 km speed of train is increased by 5 kilometre per hour the journey would be take to our left find the Original speed of train​

Answer 1

Answer:

Speed of the train =x km /hr

Time taken =t hrs

xt=300

∴t=x300

Now, (x+5)(t−2)=300

(x+5)(x300−2)=300

(x+5)(300−2x)=300x

300x−2x2+1500−10x=300x

x2+5x−750=0

x2+30x−25x−750=0

(x+30)(x−25)=0

x=−30,x=25

Neglect the negative value, hence, x=25 km/hr

Answer 2

Appropriate Question

A train Travels a distance of 300 km. If the speed of train is increased by 5 kilometre per hour, the journey would have take 2 hour less. Find the Original speed of train.

Given

• Distance of the train = 300 km.
• If speed of the train is increased by 5 km/hr, the journey would take 2 hour less.

To find

• Original Speed of the train

Solution

Let the original speed of the train be x km/hr.

Time taken by the train to cover the distance :-

Time = $\sf\dfrac{distance}{speed}$

$: \implies \sf{Time = \dfrac{300}{x}}$ hr

∴ Time taken by the train = $\sf{\dfrac{300}{x}}$ hr

If the speed of the train is increased by 5 km. Then, new speed of the train = x + 5 km/hr.

According to the question,

$:\implies \sf \dfrac{300}{x} - \dfrac{300}{x + 5} = 2 \\ \\ \\ : \implies \sf \dfrac{300(x + 5) - 300x}{x(x + 5)} = 2 \\ \\ \\ : \implies \sf \dfrac{300x + 1500 - 300x}{ {x}^{2} + 5x } = 2 \\ \\ : \implies \sf 300x + 1500 - 300x = 2 ({x}^{2} + 5x) \\ \\ \\ : \implies \sf \cancel{300x} + 1500 - \cancel{300x} = 2{x}^{2} + 10x \\ \\ \\ : \implies \sf 1500 = 2{x}^{2} + 10x \\ \\ \\ : \implies \sf 0 = 2{x}^{2} + 10x - 1500$

⟶ 2x² + 10x - 1500 = 0

⟶ 2(x² + 5x - 750) = 0

⟶ x² + 5x - 750 = 0

⟶ x² + 30x - 25x - 750 = 0

⟶ x(x + 30) - 25(x + 30) = 0

⟶ (x - 25)(x + 30) = 0

⟶ x - 25 = 0

⟶ x = 25

⟶ x + 30 = 0

⟶ x = - 30 Reject - ve

The value of x = 25

∴ The original speed of the train = 25 km/h.